Engineering Structures 48 (2013) 155–165
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Engineering Structures j o u r n a l h o m e p a g e : : w w w . e l s e v i e r . c o m / l o c a t e / e n g s t r u c t
Inelastic large deflection analysis of space steel frames including H-shaped cross sectional members Ahmed H. Zubydan Faculty of Engineering, Port-Said University, Port-Said, Egypt
a r t i c l e
i n f o
Article history: Received 11 April 2012 Revised 15 August 2012 Accepted 21 September 2012 Available online 24 November 2012 Keywords: Inelastic Large deflection Space frame Nonlinear analysis Residual stresses Beam–column
a b s t r a c t
This paper presents an efficient inelastic and large deflection analysis of space frames using spread of plasticity method. New accurate formulae are proposed to describe the plastic strength surface for steel wide-flange cross sections under axial force and biaxial bending moments. Moreover, empirical formulae are developed to predict the tangent modulus for cross sections under the combined forces. The tangent modulus 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 European Convention for Construction Steelwork (ECCS). A finite element program based on stiffness matrix method is prepared to predict the inelastic large deflection behavior of space frames using the derived formulae. The finite element model exhibits good correlations when compared with the fiber model results as well as previous accurate models. The analysis results indicate that the new model is accurate and computational efficient. 2012 Elsevier Ltd. All rights reserved.
1. Introduction
In recent years, there were numerous researches on the simulation of the nonlinear nonlinear behavior behavior of beam–col beam–columns umns in space space steel steel frames [1–6]. [1–6]. In general, general, the nonlinear nonlinear behavior behavior of steel steel frame can be predicted by using finite element method in which frame membe members rs are modele modeled d by using using solid, solid, plate plate or shell shell eleme elements nts [7,8].. This method could successfully capture the nonlinear behav[7,8] ior of the structure but it is too time-consuming because of the great number of elements required for this type of analysis. Moreover, the model processing of this analysis type is not easy at all. In the other direction of nonlinear analysis of steel frames, a ‘‘line elements’’ approach is widely used. These studies may be categorized into two main types: plastic hinge analysis and spread of plasticity analysis. The plastic hinge formulation is the most direct approach for representing inelasticity in a beam–column element [9–12] [9–12].. In plastic hinge approach, the effect of material yielding is lumped into a dimensionless plastic hinge. Generally, this type of analysis is limited by its ability to provide the correct strength assessment of beam–columns that fail by inelastic buckling. This is because the plastic hinge analysis assumes that the cross-section behaves as either elastic or fully plastic, and the element is fully elastic between the member ends [13–15] ends [13–15].. In this model, the effect of residual ual stress stresses es betwe between en hinges hinges is not accoun accounted ted for either either.. The advantages of this method are its simplicity in formulation as well
E-mail address: Zubydan@gm
[email protected] ail.com
0141-0296/$ - see front matter 2012 Elsevier Ltd. All rights reserved. http://dx.doi.org/10.1016/j.eng http://dx.d oi.org/10.1016/j.engstruct.2012.09.0 struct.2012.09.024 24
as implemen implementation tation and the least elements elements needed for member member modeling. The stability functions may be introduced to consider geometric geometric nonlinea nonlinearitie ritiess using only one beam–co beam–column lumn element element to define the second-order effect of an individual member so it is an economical method for frame analysis [16,17] analysis [16,17].. This method accounts for inelasticity but not the spread of yielding through the section section or between between the plastic plastic hinges. For slender slender members in which failure mode is dominated by elastic instability, the plastic hinge method compares well with spread of plasticity solutions. However, for stocky members that suffer significant yielding, it overestimates the capacity of members due to neglect of gradual reduction reduction of stiffness stiffness as yielding yielding progresses progresses through through and along the member. member. The so-called so-called refined plastic-h plastic-hinge inge analysis, analysis, based based on simple refinements of the plastic hinge model, was proposed for frames analysis in order to overcome disadvantages of the elastic–plastic hinge method [18–20] method [18–20].. On the other hand, the spread of plasticity method uses the highest refinement for predicting the inelastic behavior of framed structures. In the spread of plasticity method, the gradual spread of yielding is allowed through the volume of the members. In this method, method, a frame member member is divided divided into subeleme subelements, nts, and the crosscross-sec sectio tion n of each each eleme element nt is subdiv subdivide ided d into into many many fibers fibers [21–25].. The internal forces are calculated by integrating the cross [21–25] sectional subelement forces. In such case, residual stress in each fiber can be explicitly considered, so, the gradual spread of yielding can be traced [26–31] traced [26–31].. Because of considering the spread of plasticity and residu residual al stress stresses es in a direct direct way, a spread spread of plasti plasticit city y solution is considered an exact method. Although the spread of
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A.H. Zubydan/ Engineering Structures 48 (2013) 155–165
plas plasti tici city ty solu soluti tion on may may be cons consid ider ered ed ‘exa ‘exact ct’, ’, it is stil stilll too computationally intensive and too costly. Among these two types of analysis the co-called fiber hinge method was developed in an attempt to take the advantages of the two methods [32] [32].. In which, the element is divided into three segments, two end-fiber hinge segments and an interior elastic segment, to simulate the inelastic behavior of the material according to the concentrated inelastic approximation. The mid-length of end hinge segment is divided into fibers so that the uniaxial stress–strain relationships of the fibers can be monitored during the analysis process. Recently, a new simplified model was proposed by the Author based on the spread of plasticity method in an attempt to eliminate the need of cross section discretization [33,34] discretization [33,34].. In this model, closed form formulae were derived to predict the tangent modulus of steel steel cross cross sectio sections ns subjec subjected ted to combin combined ed axial axial force force and uniaxial bending moment about major or minor axis considering the residual stresses. Due to eliminating the integration of internal forces on the cross section level, a lot of consumed computational time time could could be saved. saved. In the presen presentt paper paper tangen tangentt modulu moduluss of wide-flan wide-flange ge steel steel cross sections subjected subjected to axial axial compressio compression n force and biaxial bending is derived. New formulae are derived by simulation of the results obtained from the fiber model. Prior to the derivation of tangent modulus, new plastic strength surfaces for H-shaped H-shaped cross sections sections subjected subjected to axial axial force and biaxial biaxial bending moment are proposed. The model achieves the accuracy of the spread of plasticity method but in an easy and a direct way. The research aims to eliminate complex calculations and so mini minimi mizi zing ng the the cons consum umed ed runn runnin ing g time time and and the the cost cost.. The The updated-Lagrangian method is applied in the formulation of the increm increment ental al matrix matrix equil equilibr ibrium ium equat equation ionss of the propos proposed ed beam–element beam–element model [35,36]. [35,36]. The minimum minimum residual residual displacedisplacement ment comb combin ined ed with with Newt Newton on–R –Rap aphs hson on meth method od is used used to satisfy the convergence when solving the nonlinear equilibrium equations. 2. Numerical model 2.1. Basic assumptions
The following assumptions are made in the formulation of the beam–column element: (1) A plane cross section remains plane after deformation. (2) (2) Loc Local al buck buckli ling ng and and late latera rall tors torsio iona nall buck buckli ling ng are are not not considered. (3) Small Small strain strainss but large large displa displacem cement entss and and rotatio rotations ns 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 surface as well as the tangent modulus. 2.2. Cross-section plastic strength
The determination of cross-section plastic strength surface is very essential in order to predict the nonlinear behavior of structural members. The most common formulae that describe the full plastifica plastification tion surface for cross sections sections are those proposed proposed by AISC–LRFD and Orbison. Recently, analytical plastic interaction criteria for steel I-sections under biaxial moment and axial force were developed by Baptista [37]. [37]. Although the method requires many calculations, it is considered as an exact method. For cross sections subjected to axial force and biaxial bending moments about both axes, Orbison’s formula is given as [38]
z (minor axis)
y (major axis)
Fig. 1. Cross sectional fiber model.
2
1:15 pr
4 2 2 m4ry ¼ a þ m2ry þ mrz þ 3:67 pr 2 m2ry þ 3 pr 6 mrz þ 4:65mrz
ð1Þ
while the AISC–LRFD plastic surface formula is given as [39] 1
pr þ mrz þ mry ¼ a
2
8
8
9
9
pr þ mrz þ mry ¼ a
for p r 6
2 9
for p r >
2
ð2:aÞ
mrz þ mry
2 9
9
2
ð2:bÞ
mrz þ mry 9
where a is a factor that equal unity at full plastification surface, pr is the ratio of the applied normal force P to the yield value P y y at the plastic strength envelope ( pr = P /P y), and m rz and m ry are the ratios of the applied applied bending moments M z (about (about minor axis) and M y (about major axis) to the corresponding plastic moments M pz and M py , respectively, at the plastic strength envelope. In the present paper, a new formula is derived to describe the wide flange cross section plastic strength surface based on the results obtained from the analysis of many cross sections. The cross sections are analyzed using the fiber model in which the cross section is discretized into small fibers as shown in Fig. in Fig. 1. 1. The analyzed cross sections are selected to cover all popular universal column cross cross sectio sections. ns. Twenty Twenty univer universal sal column column sectio sections ns (H-sha (H-shape ped d section) are analyzed in which the ratios B /T = = 5.5–22.4, D /t = = 10– 34.2 and D/B = 0.97–1.13, where D, B are the cross section depth and the flanges breadth, respectively, and t and T are the thicknesses of cross section web and flanges, respectively. The cross sections are analyzed using linear strain distribution along their axes. For each cross section, curvatures with different ratios are gradually increased until reaching the maximum possible bending moments at a fixed value of axial force. The internal forces (P , M y and M z ) are evaluated by accumulation of uniaxial stresses for all cross section discrete as follows:
m rz
p r1
* m rz
(m ryL, m rzL )
p r2
limiting points m ry * m ry (0,0)
1.0
Fig. 2. Proposed cross sectional plastic strength.
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A.H. Zubydan / Engineering Structures 48 (2013) 155–165 n
P ¼ ¼
X X X
ð3:aÞ
ai f i
i¼1
n
M y ¼
ai f i z i
ð3:bÞ
ai f i yi
ð3:cÞ
i¼1 n
M z ¼
i¼1
where ai is the steel fiber area, y i and z i are coordinates of each fiber, f i is the uniaxial stress at the steel fiber and n is the number of steel fibers. It is observed that, the plastic strength surface is not affected by the presence of residual stresses. The proposed formula is graphically illustrated in Fig. 2 for 2 for two different values of P r r. As shown in the figure the formula consists of two groups of curves that intersect at limiting points of coordinates ( mryL, mrzL). The proposed plastic strength strength surface surface formu formula la for a cross cross sectio section n can be given given as follow:
Fig. 3. Plastic strength for UC steel sections.
mrz mrzL mry þ mrz mrzL mryL
R1
mry mryL m mry ryL
R2
! mrz þ mrzL
¼ 1 for M rz rz
mrzL =mryL
ð4:aÞ
¼ 1 for M rz rz y < m rzL =mryL
ð4:bÞ
y P
where M rz rz _ y = M rz rz /M ry ry , M rz rz = M z /M pz , M ry ry = M y/M py and R 1 and R 2 are factors that depend on the axial force ratio ( P r r ) and they are given as
R1 ¼ 0:58P r r þ 2:33 for P r 6 0:4 R1 ¼ 2 for P r r > 0 :4
ð5:aÞ ð5:bÞ
R2 ¼ 3:25P r 2 þ 3:24P r r þ 1
ð5:cÞ
mryL and mrzL are coordinates of a limiting point that can be given as
mryL ¼ 0:2P r 2 0:8P r r þ 1
ð6:aÞ
mrzL ¼ 1:06P r 3 0:41P r 2 þ 1:47P r r
ð6:bÞ
The values mrz and mry are the bending moment capacity ratios about z - and y -axis, respectively, for cross sections subjected to axial force and uniaxial bending moment. These values are used as proposed by Zubydan for cross sections subjected to a bending moment about major or minor axis in addition to an axial force [33,34] [33,34].. For a cross section subjected to bending moment about major axis, mry is evaluated using the following formulae:
Fig. 4. Plastic strength for HD 400 1086 steel section.
pr 1:5 þ mry ¼ 1 for p r 6 0:2
7
pr þ mry ¼ 1 for p r > 0 :2
8
Fig. 5. ECCS residual stress for hot-rolled H and I sections.
ð7:aÞ ð7:bÞ
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A.H. Zubydan/ Engineering Structures 48 (2013) 155–165
On the other hand, when the cross section is subjected to bending moment moment about minor axis, mrz is evaluated evaluated from the followin following g relationships: 1
2 pr þ mrz ¼ 1 for p r 6 0:05C H þ 0:43C H H 0:37
8:55
2 H H pC for p r > 0:05C H þ 0:43C H H 0:37 r þ 0:95mrz ¼ 1
ð8:aÞ ð8:bÞ
In which C H and A is the cross sec 2 T )t and H = 4.9( Aw/ A) + 1.33, A w = ( D 2 tional area. The correlation of the proposed plastic surface formulae in Eq. (4) with the fiber model results are represented in Fig. 3. 3. Since AISC–LRFD plastic formula is very conservative especially for sections subjected to bending moment about minor axis, only Orbison’s formula is compared to the derived one. As shown in Fig. 3, 3, it is observed that the proposed formula correlates very well with the fiber model results. On the other hand, the Orbison’s formula may deviate from the fiber model results at various axial force ratios. The proposed formula for plastic strength is also compared to the the resu result ltss obta obtain ined ed from from Bapt Baptis ista ta’s ’s meth method od [37] for for HD 400 1086 steel section as shown in Fig. 4. 4. It is clearly observed that the proposed formula gives an excellent correlation with the results obtained from Baptisa’s method. It should be mentioned that the proposed formula is simpler and more direct than Baptista’s method. 2.3. Tangent modulus for cross sections
It is known that the cross sectional tangent modulus exhibits a premature degradation when it is subjected to forces due to the presence of residual stresses. In order to account for the effect of residual stresses, researchers researchers [40–42] use an empirical tangential modulus ratio depending on the formula proposed in AISC–LRFD [39] to [39] to describe the tangent modulus of columns under axial force. This formula is given as follows:
E tr for a tr ¼ 1
6 0:5
E tr tr ¼ 4 að1 aÞ for a > 0 :5
ð9:aÞ ð9:bÞ
where E tr ), E is the elastic tr is the tangent modulus ratio ( E tr tr = E tang tang /E ), modulus and a is a force-state parameter that measures the magnitude of axial force and bending moments which may be calculated from Eq. (1) Eq. (1) or Eq. (2) Eq. (2).. In the present paper, a new tangent modulus is determined for H-shaped H-shaped sections sections considerin considering g the effect effect the residual residual stresses. stresses. The residual stresses adopted in the present paper are based on
the recommend recommendatio ation n by European European Conventi Convention on for Constructio Construction n Steelwork (ECCS) [43] (ECCS) [43].. The magnitude of residual stress (rr ) is assumed to be dependent on depth/breadth ratio as shown in Fig. 5. 5. For cross sections subjected to axial compression force and biaxial bending moments, the flexural tangent modulus is evaluated by applying the concepts of fiber model. The selected cross sections are analyzed using the fiber model in which the bending moments with different ratios are incrementally applied to the cross section at constant values of axial compression forces. Through each load increment, the flexural tangent modulus ratio for each direction is calculated as follows:
dM y y =du y EI y y dM z z =du z E trz trz ¼ EI z E try try ¼
ð10:aÞ ð10:bÞ
where EI y y and EI z z are the cross sectional elastic rigidities about y and z -axis, -axis, respectively, and u y and u z are the curvatures about the same axes. By plotting the relationships of the tangent modulus ratios E try try and E trz trz versus the moment ratios M ry ry and M rz rz , respectively, as shown in Fig. in Fig. 6, 6, two possible multilinear paths may be obtain tained ed for for each each mome moment nt dire direct ctio ion. n. When When the the valu value e of axia axiall compression force ratio P r r is less than P r r 0, (P r r 0 = 1 rr /r y [33,34] [33,34])) , the relationships follow the path abcde for E try for E trz try and abcdef for trz . For such case, the tangent modulus modulus ratios remain constant constant and equal to the elastic values until reaching M ry0, ry 0,rz rz 0 (point b ) after they decrease linearly with the shown slopes until they vanish when reaching the plastic plastic surface. On the other hand, when the axial compression force exceeds P r r0 the values E try try and E trz trz follow the paths a0 b0 c 0 d0 e0 and a 0 b0 c 0 d0 e0 f 0 , respectively. In such case, the tangent modulus ratios start with E try1 try1 and E trz trz 1 that are less than unity due to the effect of high axial force. force. The tangent tangent modulus modulus ratios decrease decrease again with the increase increase of bending bending moments until reaching the plastic surface. The values M ry0 found to be relat related ed to each each other other by ry0 and M rz rz 0 are found the following relationship:
M rz M ry rz 0 ry0 þ ¼ 1 :0 M rz M 0 ry0
ð11Þ
where M ry0 ¼ ðP r r 0 P r r Þ Z y y =S y ; M rz [33,34], Z y, r 0 P r r Þ Z z =S z [33,34], y , z z and 0 ¼ ðP r S y, y , z z are elastic and plastic modulus of cross section about y- and z -axis, -axis, respectively. The slopes S y1, y 1, z z 1, S y2, y 2, z z 2 and S y3, y3, z z 3 and the values E ty0, ty0,tz tz 0, E ty1, ty 1,tz tz 1, E tz depend on the biaxial moment ratio M rz tz 2, E tz tz 3 and M ry2, ry 2,rz rz 2 depend rz _ y
Fig. 6. Inelastic tangent modulus ratios for cross section under biaxial bending moment and axial compression force.
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Fig. 7. Inela Inelast stic ic flexur flexural al modulu moduluss ratios ratios for cross cross sectio sections ns about about major major axis. axis.
Fiber Fiber model model;;
prese present nt model. model.
Fig. 8. Inelas Inelastic tic flexur flexural al modulu moduluss ratio ratioss for cross cross sectio sections ns about about minor minor axis. axis.
Fiber Fiber model; model;
presen presentt model. model.
(where M rz rz _ y = M rz rz /M ry ry) and the axial force ratio P r r as illustrated in the Appendix the Appendix A. A. The proposed tangent modulus ratios are compared to those obtained from the fiber model as shown in Figs. 7 and 8 for 8 for biaxial moment ratios (M rz = 0.25, 0.5, 1 and 2). Fig. 2). Fig. 7 7 illustrates illustrates E try rz _ y try versus M ry while versus are illustrated in Fig. in Fig. 8 8. . For each biaxial E M ry trz trz rz rz moment ratio, the axial force ratio varies from 0.2 to 0.8. It is clearly observed that the proposed formulae give very good correlations with the fiber model results for various values of P r r and M rz rz _ y . It should be mentioned that the proposed model is valid for all values of P P r r . 2.4. Cross section incremental secant modulus
For nonlinear analysis, the accuracy of results essentially depends on the accuracy of the internal forces calculation. The use of tangent stiffness for calculating the internal forces may slightly overestim overestimate ate the structure structure strength, strength, so the increment incremental al secant secant stiffness is required in order to evaluate accurate values of internal forces. The same sequence followed by the Author [33] [33] is is used in the present research research to evaluate evaluate the secant stiffness stiffness for a load increment. For a load increment from steps j to j + 1, the curvature
Fig. 9. Moment-curvature curve for cross section.
changes from u j to u j+1 j+1 and the bending moment changes from M j 0 to M jþ when the initial tangent modulus is followed as shown in jþ1 0 Fig. 9. 9. The value of M can be calculated as M jþ jþ1
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A.H. Zubydan/ Engineering Structures 48 (2013) 155–165
0 M jþ tr j EI jþ1 ¼ M j j þ d uE tr
ð12Þ
where d u is the change in the curvature of the cross section from step j to step j + 1 and EI is the elastic flexural flexural rigidity rigidity of the cross j which can be calsection. E tr is the tangent modulus ratio at step tr j culated from Fig. from Fig. 6 using 6 using the forces values at point j . As E tr tr changes linearly linearly with M r r from from point point j– j + 1, i.e. the the followin following g relations relationship hip can be deduced: E tr ð13Þ tr ¼ c 1 M r r þ c 2 where c 1 ¼
0 E tr
M r0
E jþ1 tr j jþ1
M r
j
0 0 and c 2 = E tr tr j c 1M r r j . M jþ1 and E trjþ1 are the
bending moment and the corresponding tangent modulus ratio at point j 0 + 1 which is very close to point j + 1. Based on Eq. (12) (12),, the value of bending moment at j + 1( M j+1 j+1 ) and the increment incremental al secant secant modulus modulus ratio (E sr derived d sr ) can be derive and given as follow [33] follow [33]::
M jþ jþ1 ¼
1
k
ð14Þ
M p p ðe c 2 Þ c 1 1 M jþ jþ1 M j j E sr sr ¼ EI u jþ jþ1 u j
ð15Þ
where M p is the cross section plastic moment and u u k ¼ c 1 jþu1 j þ lnðc 1 M r j þ c 2 Þ. Eq. (15) Eq. (15) can can be applied for the bend y
ing moments about y - and z -axis -axis to evaluate the corresponding secant modulus E sry sry and E srz srz , respectively. 2.5. Finite element model
A stiffness method for the analysis of space frames is developed considering both geometric and material nonlinearities. The equation of equilibrium in terms of geometry of the deformed system is given as follows [44] follows [44]:: ½K þ K g fDDg ¼ fDFg ð16Þ where {DF} and {DD} are the incremental force and displacement vectors, respectively. [ K] is the stiffness matrix of the structure considering material nonlinearity and [ K g ] is the geometric stiffness matrix matrix which which represent representss the change in the stiffness stiffness that results from deformation effects. Consider a prismatic element of a symmetric cross section about y and z axes. This element is subjected to axial force N , torsional moment M x and bending moments M y and M z while the corresponding displacements are u,h x,h y and h z , respectively as shown in Fig. 10. 10. The shape functions for the axial displacement (u), the axial rotation (h x), the displacement in x y plan (w) and the displacement in x z plan plan (v ) are introduced considering the shear effect as follow:
u ¼ ð1 nÞu1 þ nu2 h x ¼ ð1 nÞh x1 þ nh x2
ð17:aÞ ð17:bÞ
wð1 þ U y =2Þ ¼ ½ð 1 þ U y Þ U y n 3n2 þ 2n3 w 1 þ ½ð 1 þ U y =2Þ xh z 1 þ ½ U y n þ 3n2 2n3 w2 ð2 þ U y =2Þn þ n2 x xh z 2 þ ½U y =2 ð1 U y =2Þn þ n2 x
ð17:cÞ
ð1 þ U z =2Þ ¼ ½ð1 þ U z Þ U z n 3n2 þ 2n3 v 1 þ ½ð1 þ U z =2Þ
v
xh y1 þ ½ U z n þ 3n2 2n3 v 2 þ ð2 þ U z =2Þn n2 x xh y2 þ ½U z =2 þ ð1 U z =2Þn n2 x
ð17:dÞ
In which n = x/L and L is the member length, Uy = 12 EI z /GAQyL2,Uz = 12EI y/GAQz L2, G is the shear modulus and A Qy and AQz are the shear areas corresponding to y - and z -axis, -axis, respectively. Since the matrices [K] and [K g ] are displacement dependent, Eq. (16) (16) cannot cannot be directl directly y solved. solved. Various Various procedure proceduress can be used used to solve solve the equiequilibrium equations. Generally, members are subdivided into subelements to produce satisfactory results. The modulus ratios E try, try ,trz or E sry, evaluated at each member ends and then average values sry ,srz are evaluated at each direction are applied for each modulus ratio in the stiffness matrix [k]. 2.6. Internal force recovery
The employing of equilibrium equation in conjunction with the incremen incremental tal analysis analysis requires requires that the structural structural geometry includes cludes all accumula accumulated ted deformati deformations. ons. For the current current analysis analysis,, the node coordinates are updated after each iteration. That is, the coordinates of each node are modified or updated to include the translatio translational nal displacem displacement ent components components that occur during during iteraiterations. In updating the coordinates of nodes or element ends, the deformed geometry of the structure is achieved by changing the position and hence the orientation of each element with respect to the global coordinates system. For all the elements of a structure, the element stiffness equation, as given in Eq. (16) (16),, can be assembled to yield the stiffness equation of the structure for an incremental step. For an incremental–iterative nonlinear analysis, the element incremental displacement vector { Dd} is used to calculate the incremental axial strain d e, torsional strain dv and flexural strains du y and du z based on the assumed assumed shape functions functions and so the secant modulus at element ends can be evaluated as illustrated in Section 2.4 Section 2.4.. The material stiffness matrix [ k] for each element is reformulated using the average secant flexural modulus ratios E sry sry and E srz srz instead of the tangent modulus. The secant modulus for axial force and torsional moment can be modified by using the average of E E sry sry and E srz srz . On the other hand, the geometric stiffness matrix [k g ] is also formulated again using a new axial force P which is calculated by adding the incremental axial force to the total previous value as follows:
P ¼ ¼ P pre pre
ious þ d
v
eE sr sr EA
ð18Þ
The increments of internal force vector { D f int int } for an element can be calculated as
fD f int int g ¼ ½k þ k g fDdg
ð19Þ
By summing the element forces at at the structural nodes and comparing them them with with the applie applied d loads loads,, the unbala unbalanc nced ed forces forces for the structure can be obtained. Finally, by treating the unbalanced forces as applied loads, other iterations can be repeated. 3. Numerical solution
Fig. 10. Space frame element.
The nonlinear analysis algorithm consists of four basic steps; the formulation of the initial stiffness matrix, the solution of the equilibrium equations for the displacement increments, the determination of the new updated stiffness and member forces using the cross sectional model, and the check of conversion. Since the global stiffness matrix of the structure depends on the displacement increments, increments, the solution solution of the equilibrium equilibrium equations equations is typically accompanied by an iterative method through the convergence check. In the present model, the Newton Raphson method is used by updating updating the tangent tangent stiffness stiffness matrix at each iteration iteration
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A.H. Zubydan / Engineering Structures 48 (2013) 155–165
[45]. Also, the minimum residual displacement method is used in [45]. order to trace the post-peak path path [46]. [46]. A full nonlinear iterative solution solution discussed discussed by Zubydan Zubydan [33] is follow followed ed in the present present paper. 4. Numerical analysis and results
A comput computer er progra program m is develo develope ped d to predic predictt the nonlin nonlinear ear behavior of space frames using the derived model. The fiber model as well as the proposed simplified model is first compared to wellestablished benchmark results. After that the analysis results obtained from the proposed and the previous simplified models are compared to those obtained from the fiber model.
Table 1
Out-of-plumbness imperfection of two-storey space frame.
Leve Levell
Roof Second floor Base
Imp Imperf erfecti ect io on (mm) Column 1
Column 2
Column 3
Column 4
x
y
x
y
x
y
x
y
4.51 1.39 1.39
11.08 6.88 6.88
5.49 0.68 0.68
11.41 6.77 6.77
8.17 8.17 5.11 5.11
6.58 6.58 2.11 2.11
4.31 4.31 3.96 3.96
12.0 12.04 4 6.19 6.19
0
0
0
0
0
0
0
0
4.1. Two-storey space frame
The two storey space steel frame shown in Fig. 11 11 was previously analyzed by Kim and Lee [7] [7] using finite element package ABAQUS [47] ABAQUS [47].. For this study the frame was only subjected to the gravit gravity y loads loads and and latera laterall loads loads in X -direction -direction (i.e. the value value of 0). All All fram frame e memb member erss are are comp compri rise sed d of H150 H150 160 r = 0). 10 6.5mm section. The frame has 2.5 m wide in Y -direction -direction and 3 m long in X -direction. -direction. The height of the second floor is 1.76 m from the column base while the roof elevates 2.2 m from the second floor. The yield stress, Young’s modulus, and shear modulus of material are 320 MPa, 221 GPa, and 85 GPa, respectively. The frame frame columns columns have out-of-plumbne out-of-plumbness ss imperfecti imperfection on as listed listed in Table 1. 1. The ABAQUS shell element S4R was employed to model frame members. The total numbers of shell elements used to model the structure by ABAQUS are 49,840. The shown applied loads were proportionall proportionally y increased increased until failure failure of the structure. structure. For such such case case of loadin loading g (r = = 0), the frame frame is reanal reanalyze yzed d using using the present fiber model and also using the proposed simplified model. For the analyzed frame, each frame member is discretized into 10 equal equal elements elements.. The comparisons comparisons of the obtained and the published analysis results are illustrated in Fig. 12. 12. It is observed that the results obtained from the proposed model correlates well with the fiber model model result resultss as well well as the results results obtain obtained ed from from ABAQUS. The frame is also analyzed using the both the simplified and the fiber models with different values of r ( r = = 0.25, 0.5, 1.0). Comparison of the analysis results are shown in Fig. 13 which 13 which illustrates
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 and 1.0).
the resultant displacement at point A versus the load P. It is clearly obse observ rved ed that that the the resu result ltss obta obtain ined ed from from simp simpli lifie fied d mode modell correlates very well the results obtained from the fiber model. It is observed that the frame capacity is greatly reduced with the increase of lateral loads in Y -direction -direction due to the increase of bending moments produced about cross sectional minor axes. 4.2. Six storey space frame
Fig. 11. Two storey space frame.
The six-story space steel frame shown in Fig. in Fig. 14 was 14 was first analyze lyzed d by Orbi Orbiso son n et al. al. [38] and later later by many many resea research rchers ers
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Fig. 14. Six storey space frame.
[11,31,48,49].. The yield strength of frame members is 250 MPa, [11,31,48,49] Youn Young’ g’ss modu modulu luss E = 206, 206,85 850 0 MPa MPa and and shea shearr modu modulu luss 79,293 MPa. The frame is subjecte subjected d to proportiona proportionall gravity gravity G = 79,293 and lateral loads. Uniform floor pressure of 9.6 kN/m2 is converted into equivalent concentrated loads on the top of columns. Wind loads are simulated by point loads of 53.376 kN in the Y -direction -direction at every beam–column joints. Jiang et al. analyzed the frame using fiber plastic zone model in which the cross section flange is discredited credited into fibers and the frame frame member member into nine elements elements [31].. The frame is reanalyzed using the present model with/with[31] out the shear effect. Since most plastic deformation is concentrated at member edges, each frame member is discretized into three elements with two short edge elements of length = 0.1 L. The tangent modulus for I-shaped cross sections (as in case of frame girders) is calculated as proposed by Zubydan [33] [33].. Comparisons of the proposed simplified model with Jiang’s analysis results are shown in
Fig. 16. Single storey space frame.
Fig. 15. 15. It is clearly observed that the present model results with shear effect correlate very well the Jiang results. On the other hand, when the shear effect is neglected, the model slightly overestimates the frame capacity.
4.3. Single storey space frame
Fig. 15. Load-displacement relationships for the six storey frame.
The frame shown in Fig. in Fig. 16 is 16 is assumed to be made of steel with yield strength r y = 250 Mpa and elastic modulus E = = 200,000 MPa. The frame is subjected to incremental lateral loads H in in X -direction -direction and rH in in Y -direction -direction in addition to constant value of vertical loads analyzed ed using the fiber fiber model and also by using using the P . The frame is analyz derived model neglecting the shear effect. For all analysis types, each frame member is divided divided into 10 elements elements to accumula accumulate te the plastic deformation through the whole member. The tangent modulus proposed by researchers (Eq. (9) (9))) accompanied with the Orbison’s plastic surface formula is also used to analyze the frame. The The cros crosss sect sectio ion n of the the fram frame e memb member erss is assu assume med d to be UC203 203 60 and the frame is analyzed using different values of r (r = 0.25, 0.5, 1.0 and 2.0). For each value of r , vertical load
A.H. Zubydan / Engineering Structures 48 (2013) 155–165
163
Fig. 17. Load-displacement relationships for the single storey frame.
Table 2
Normalized ultimate loads for single storey space frame. P r r = 0.2
P r r = 0.4
P r r = 0.6
r = = 0.25 Present model Eqs. (9) Eqs. (9) & (1)
0.9977 1 .0823
0.9917 1.0198
1.0523 1.1377
r = = 0.5 Present model Eqs. (9) Eqs. (9) & (1)
0.9958 1 .1233
1.02160 1.1904
0.9950 1.5412
r = = 1.0 Present model Eqs. (9) Eqs. (9) & (1)
0.9976 1 .1708
0.9902 1.2822
1.0264 1.9594
r = = 2.0 Present model Eqs. (9) Eqs. (9) & (1)
0.9884 1 .1088
1.0223 1.1088
1.0638 2.1503
ratios P r r = 0.2, 0.4 and 0.6 are used. Fig. 17 shows 17 shows comparisons between the results obtained from different models. The figure illustrates the relationships between the lateral load H and the lateral resultant displacement at A. It is clearly clearly observed observed that that the proposed proposed model gives excellent correlations with the fiber model for all values of r r and P r r. On the other hand, the previously used formula in Eq. (9) Eq. (9) mostly overestimates the stiffness and the capacity of the frame compared to the fiber model. Table model. Table 2 shows 2 shows the normalized
capacities of the frame using the proposed model and that from Eqs. (9) and (1) with (1) with respect to the fiber model results. As shown in Table Table 2, the diversion diversion of the analysis analysis results obtained obtained from Eqs. (9) and (1) from (1) from the fiber model increases with the increase of r and P r r. The overall capacity of the frame obtained from the pre r and vious model more (Eqs. (9) (Eqs. (9) and (1)) (1)) may be than twice the capacity obtained from the fiber model. 5. Conclusions
A new simplified simplified model was developed developed to predict predict the large deflection deflection inelastic inelastic behavior behavior of space space steel frames. New plastic plastic strength surfaces for H-shaped cross sections under axial forces and biaxial bending moment were derived. Moreover, the tangent modulus of cross sections subjected to the combined forces was predicted. The residual stress distributions were considered as recommended ommended by European European Conventio Convention n for Constructi Construction on Steelwork Steelwork (ECCS). The internal forces were recovered by using derived incremental mental secant secant stiffness stiffness.. The derived cross sectional sectional model was implemen implemented ted into a finite element element program program based based on stiffness stiffness method to predict the full nonlinear behavior of steel space frames. The simplified model correlated very well the fiber model without the need of cross section discretization. The proposed model could successfully simplify the plastic zone analysis and save a lot of comput computati ationa onall time time and data data storag storage e by the elimin eliminati ation on of
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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 fiber model ranged from 8 to 10 times the run time required using the simplified model. These run times included the formulation and the solving of nonlinear equilibrium equations and they also included the time consumed to equilibrate the fibers on the cross sectional level which was saved by using the proposed simplified model. Appendix A
Values of S S y1 y1 , S y2 y 2, S y3 y3 , E ty0 ty0, E ty1 ty1 and M ry1 ry 1: 2 S y1 ¼ 0 :38M rz rz y þ 0:74 þ f 1 y þ 0:05M rz
S y1 ¼ 0 :88M rz rz y þ 0:26 þ f 1
for M rz rz y 6 1:0
for M rz rz y > 1 :0
ðA:1aÞ ðA:1bÞ
where f 1 = 0 for P r r 6 0.5 and f 1 ¼ 7: 7 :81P 2r 6:84P r r þ 1:47 for P r r > 0.5 For M rz rz _ y 6 1.5: 2 S y2 ¼ 10:82M rz for P r 6 0:2 rz y þ 4:03 y þ 17:87M rz
ðA:2aÞ
2 S y2 ¼ 11:45M rz for 0:3 6 P r r 6 0:5 ðA:2bÞ rz y þ 7:37 y þ 15:65M rz 2 S y2 ¼ 17:57M rz for P r P 0:6 rz y þ 13:67 y þ 21:43M rz
ðA:2cÞ
For M rz rz _ y > 1.5:
S y2 ¼ 1 :4M rz rz y þ 5:12
ðA:2dÞ
2 S y3 ¼ 0 :05M rz þ 0:36M rz rz y þ 0:96 y
ðA:3Þ
2 E ty ty0 ¼ 0:65P r þ 0:22P r r þ 1:05
ðA:4Þ
2
E ty ty1 ¼ 1:3P r 0:45P r r þ 1:1 M ry ry1 ¼
f 2 1 þ M rz rz y
6
ðA:5Þ ðA:6Þ
mry
where M ry1 1.015 5 for for ry1 6 1.05P r r + 1.07; M ry1 ry1 6 0.98, f 2 = 0.4P r r + 1.01 3 2 P r r 6 0.3 and f 2 ¼ 3:89P r þ 4:44P r 2:27P r r þ 1:53 for P r r > 0.3 Values of S S z z 1, S z z 2 , S z z 3 , E tz tz 0, E tz tz 1, E tz tz 2, E tz tz 3 and M rz rz 1
S z 1 ¼ 1 :37M 2ry z 0:94M ry ry z þ 1:52 þ f 3 S z 1 ¼ 2 :38M ry ry z 0:49 f 4 þ f 3
for M ry ry z 6 1:0
for M ry ry z > 1 :0
ðA:7aÞ ðA:7bÞ
3 2 S y1 P 0:9; M ry where ry z ¼ M ry ry =M rz rz ; f 3 ¼ 4:65P r þ 1:87P r þ 1:03P r r 0:19; 19; f 4 ¼ 0 for for P r r 6 0.5 0.5 and and f 4 = (10.74P r r 5.4) 5.4)M ry_ ry_ z for P r r > 0.5.
S z 2 ¼ 13 :59M 2ry z 6:11M ry for M ry ry z þ 2:54 ry S z 2 ¼ 12 :77M ry ry z 2:65
z 6
for M ry ry z > 1 :0
1:0
ðA:8aÞ ðA:8bÞ
S z 3 ¼ 0 :94M ry ry z þ 0:55 6 1:5 E tz tz 0 ¼ 1:9P r r þ 1:94
ðA:9Þ ðA:10Þ
2 E tz tz 1 ¼ 2 :5P r 5:45P r r þ 3:1
ðA:11Þ
For P r r 6 0.5:
E tz for M ry tz 2 ¼ 0:4M ry ry z þ 0:65 ry z 6 5:0 E tz for M ry tz 2 ¼ 0 :007M ry ry z þ 0:44 ry z > 5 :0
ðA:12aÞ ðA:12bÞ
For P r r > 0.6:
E tz tz 2 ¼ 0 :007M ry ry z þ 0:44 ðP r r 0:6Þ
ðA:12cÞ
E tz tz 3 ¼ ð0:11 0:35 f 5 ÞM ry ry z þ f 5 E tz for M ry tz 3 ¼ 0 :3 ry z > 3 :0
ðA:13aÞ ðA:13bÞ
for M ry ry z 6 3:0
where f 5 = = 0.53 0.53P r r + 0.54
M rz rz 1 ¼
f 6 1 þ M ry ry z
6
mrz
ðA:14Þ
where M rz 1 6 0:95P 3r 0:16P 2r þ 0:07P r r þ 0:95; 95; f 6 ¼ 0:75P 2r þ 0:4P r r þ 2 1:07 for P r r 6 0.3 and f 6 ¼ 1:86P r þ 0:79P r r þ 1:05 for P r r > 0.3.
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