a simple displacement control technique for pushover analysis

8/10/2019 A simple displacement control technique for pushover Analysis

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EARTHQUAKE ENGINEERING AND STRUCTURAL DYNAMICSEarthquake Engng Struct. Dyn. 2006; 35:851866

Published online 14 February 2006 in Wiley InterScience (www.interscience.wiley.com). DOI: 10.1002/eqe.560

A simple displacement control technique forpushover analyses

Armelle Anthoine;

ELSA; IPSC; Joint Research Centre; European Commission; I-21020 Ispra (VA); Italy

SUMMARY

This paper explains how to control in displacement any force proportional loading. Such a proceduremakes it possible to derive the complete (i.e. including the possible softening branch) response curveof a structure along any radial loading path in the force space. This is exactly what is required inthe so-called pushover analysis used in the seismic assessment of structures. The proposed procedureis simple in the sense that it can be easily implemented in any classical (displacement-based) niteelement code through a standard displacement control loading process. Furthermore, it leads to aninteresting denition of the controlled degree-of-freedom, which, in the case of the pushover analysis,could substitute the classical roof displacement. Copyright ? 2006 John Wiley & Sons, Ltd.

KEY WORDS: proportional loading; nite element method; pushover analysis

1. INTRODUCTION

In the pushover analysis, a non-linear nite element model of a given structure (e.g. a build-ing frame) subjected to gravity loads, is laterally loaded until either a predened targetdisplacement is met, or the model collapses. The lateral load has a prescribed distribution(e.g. uniform, inverse triangular, modal, etc.) and the control node is a particular point ofthe structure (usually the centre of mass at the roof level). The result of the analysis can bethus expressed in terms of a base shear versus control node displacement relationship, i.e. aso-called pushover curve. Depending on the mechanical behaviour of the structure and also onthe controlled degree of freedom, this pushover curve may exhibit limit points and softening

branches or even snap-backs and bifurcation points.Owing to the prescribed pattern of the lateral forces, a force control solution scheme is

straightforward but operates only along the ascending branch of the pushover curve. At alimit point, the applied forces cannot anymore increase either because of second-order eects

Correspondence to: Armelle Anthoine, ELSA, IPSC, Joint Research Centre, European Commission, I-21020Ispra (VA), Italy.

E-mail: [email protected]

Received 11 May 2005

Revised 12 October 2005Copyright ? 2006 John Wiley & Sons, Ltd. Accepted 14 November 2005


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852 A. ANTHOINE

(e.g. buckling, P-delta eect) or because a negative-sloped portion of a behaviour curve hasbeen reached somewhere in the structure. The lateral forces must therefore be decreased but thestructure should not undergo a simple elastic unloading. During this phase, the displacementof the control node will either increase (softening branch) or decrease (snap-back). To follow

a softening branch, a displacement control technique should be used (e.g. Reference [1])while only more powerful schemes (e.g. arc-length or work control methods) will be able tofollow also snap-back branches and detect bifurcation points. Actually, these latter schemeshave been developed appositely for non-linear analyses involving unstable responses due tosoftening and=or buckling for example.

It is worth mentioning that force control strategies based on a partial=total unloading ofthe structure (or a part of it) followed by a reloading with damaged characteristics, as themethods proposed in SAP 2000 NL [2], are not always able to pass a limit point and, ifsuccessful, do not necessarily produce the same result. In fact, the solution found beyondthe limit point depends on the whole path followed previously. Since each method followsa dierent unloading=reloading path and not a monotonic push, the obtained solutions mightdier from one another and from the monotonic pushover curve. Furthermore, they usuallylead to a stepwise decreasing pushover curve, each vertical drop corresponding to a sin-gle unloadingreloading path aiming at bypassing a softening or a snap-back portion of themonotonic pushover curve.

The simple displacement control method presented here is able to follow the monotonicpushover curve also along a softening branch. It does not resort to any artice such as theaddition of ctitious springs as proposed by Archer [1] but simply relies on an appropriatedenition of the displacement variable to be used for controlling the loading process.

2. PRINCIPLE OF THE PROPOSED METHOD

The method will be introduced through a simple example, a cantilever beam subjected to xedgravity loads QG and increasing lateral load Fwith an inverse triangular pattern (Figure 1(a)).The beam is composed of three elements of equal length and the same lumped mass m

3

2

1

2F

3F

6F

0

mg

mg

mg

3

2

1

0

F

23

23

23

5

4

mg

mg

mg

(a) (b)

Figure 1. (a) Pushover analysis on a cantilever beam; and (b) equivalent isostatic loading system.

Copyright ? 2006 John Wiley & Sons, Ltd. Earthquake Engng Struct. Dyn.2006; 35:851866


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DISPLACEMENT CONTROL TECHNIQUE FOR PUSHOVER ANALYSES 853

is aected to the three nodes above the ground, numbered 1, 2 and 3 (the base node isnumbered 0). The inverse triangular pattern consists in horizontal nodal forces proportionalto the mass and to the perpendicular height of the node considered. In the case considered, if

F is the total base shear, this pattern gives F=6 at node 1, F=3 at node 2 and F=2 at node 3.

No assumption is made on the mechanical behaviour of the beams elements and thus, thefailure mode is not known and can be dierent from the one expected when the three beamelements are homogeneous and mechanically identical (failure at the base).

The aim of the pushover analysis is to compute the response of the structure when thelateral loads are increasing (ascending branch of the pushover curve) and, if a limit point isreached, to pursue the computation until complete failure (descending branch of the pushovercurve). During the second phase, the lateral loads are decreasing but this is not a simpleunloading since the damage caused to the structure is still increasing.

Let us imagine how such a proportional loading could be imposed experimentally.A possible solution consists in superposing two simply supported rigid beams (Figure 1(b)).The rst beam of length rests on nodes 1 and 2 of the structure while the second beamof length 4=3 rests on node 3 of the structure and on the rst beam at a distance 5=3above the ground (node 4). Finally, the lateral force Fis applied at the middle of the second

beam (node 5). This loading system is statically determinate and the forces transmitted to thestructure are exactly F=6 at node 1, F=3 at node 2 and F=2 at node 3. This simply resultsfrom the equilibrium of each of the rigid beams. The applied force F gives rise to equalforces F=2 at nodes 3 and 4. On its turn, the force F=2 transmitted at node 4, gives rise to aforce F=6 at node 1 and a force F=3 at node 2.

In this experimental set-up,Fhappens to be the reaction force associated to the horizontaldisplacement u5 of node 5. Thus, if u5, instead of F, is made increasing monotonously, the

pushover analysis can be pursued also beyond the possible limit point independently of thefailure mode, even in case of local mechanisms leading to partial unloading of the structure. Inother words, u5 is the suitable control displacement and the pushover curve should therefore

be expressed in terms of F versus u5, and not in terms of F versus a particular horizontaldisplacement (e.g. u3). A posteriori, the Fu5 pushover curve can always be expressed interms of any horizontal displacement by simply replacing the monotonic evolution of u5 bythe computed evolution of u3 (roof displacement), u1 or u2.

In a nite element code environment, this experimental set-up could be numerically repro-duced as it is, together with the structure. The Fu5 pushover curve would thus result from astandard displacement control loading process (imposed horizontal displacement of node 5).However, the numerical modelling of many perfectly rigid beams may quickly become cum-

bersome. Mechanically speaking, each rigid beam is nothing else than a linear relationshipbetween the horizontal displacements of the three nodes involved. For the two rigid beams inFigure 1(b), one has:

3u4= u1+ 2u2

2u5= u3+u4(1)

Eliminating u4 in Equation (1), one obtains:

u5=u1

6 +

u23

+u3

2 (2)



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854 A. ANTHOINE

The quantity u5 happens to depend exclusively on the lateral load distribution, which appearsexplicitly in Equation (2), and not on the chosen loading set-up, which is not unique. Here,two rigid beams can be installed in three dierent ways, depending on which structural nodesthe rst beam rests on. As a matter of fact, Equation (2) can be derived directly on the original

structure (Figure 1(a)), through the expression of the virtual work done by the external forces(horizontal Hi and vertical Vi) in any cinematically admissible displacement eld, i.e.

Wext=i=03

(Hiui+Vivi) =F

u1

6 +

u23

+u3

2

mg(v1+v2+v3) (3)

where ui (resp. vi) stands for the horizontal (resp. vertical) virtual displacement of node iand g is the acceleration due to gravity. The gravity loads being constant, Equation (3) provesthat the structure is subjected to a one-parameter loading mode according to the terminologyintroduced by Halphen and Salenon [3] and that F is actually the load parameter, whereas thecorresponding displacement parameterueq is, by duality, the same quantity as in Equation (2),i.e.

ueq= u1

6 + u2

3 + u3

2 (4)

Therefore, the loading set-up can be removed provided that the displacement control is shiftedfrom a single nodal displacement of the loading set-up (u5) to a linear combination of thenodal displacements of the structure (u1, u2 and u3). In practice, if the original force controlelastic problem reads

K:u =Fp+M:g (5)

where p is the spatial distribution of force, M the mass matrix and F a scalar increasingfrom 0, it can be replaced by the following displacement control problem:

K:u = M:g submitted to p

T

:u = ueq (6)where ueq is now the scalar increasing from 0. General-purpose nite element codes usuallyoer the possibility of imposing linear constraints on nodal displacements such as Equation (4),

by resorting to dierent numerical methods (e.g. elimination, penalty, Lagrange multiplier).Among them, the Lagrange multiplier method is one of the most ecient and is brieyrecalled in Appendix A. Other methods may be found in Reference [4].

3. GENERALIZATION OF THE PROPOSED METHOD

The proposed method can be generalized to the case of any structure having N degreesof freedom (DOFs) generically denoted (ui)i=1

N. The term ui may stand for a 2D or 3D

translation or rotation DOF. The kinematical admissibility (boundary conditions) imposes Mlinear constraints on the DOFs

(ckiui= hk)k=1M (7)

which, in most cases, reduce to (uk= 0)k=1M. In the case of a pushover analysis, the loadingis composed of two parts, the xed loads (Qi)i=1N (usually gravity loads) and the varying



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loads (Fi)i=1N (lateral seismic equivalent loads), applying on the N rst nodes (N6N)according a given pattern:

i= 1N ; Fi=Fpi (8)

In Equation (8), F is a scalar value that may coincide with the resultant load if the pattern(pi)i=1N is normalized appositely. For example, in the case of the cantilever beam presentedearlier, the scalar F coincides with the lateral resultant (and the base shear) because the

pattern (pi)i=13= (16

; 13

; 12

) is such that

i=13pi= 1.The key-point is to express the virtual work of the external forces in any cinematically

admissible displacement eld, Wext. With the adopted notations, one obtains:

Wext=i=1N

(Fiui+Qiui) =Fi=1N

piui+i=1N

Qiui (9)

Equation (9) proves that the structure is subjected to a one-parameter loading mode and thatFis the load parameter, whereas the corresponding displacement parameterueq is, by duality,

the following linear combination:ueq=

i=1N

piui (10)

Therefore, the Fueq pushover curve can be obtained by increasing monotonically the linearcombination of the DOFs as shown on the right-hand side of Equation (10). The Lagrangemultiplier method described in Appendix A may be applied. However, one should keep inmind that the corresponding computing time and=or memory might be substantially increasedif the force pattern involves most of the DOFs (e.g. in fully distributed loading). The numberof additional unknowns is only 1 (or 2 if the Lagrange multiplier is duplicated) but the

bandwidth of the augmented stiness matrix is at least N.Again, it is possible to build a statically determined loading system able to impose

Equation (10). Generalizing what has been done in paragraph 2, (

N 1) new DOFs canbe introduced through the following (N 1) relations:

uN+1= p1u1+p2u2

uN+i= uN+i1+pi+1ui+1 if 26 i6 N 1(11)

Here, each relation stands for a rigid link between the three DOFs involved (e.g. simplysupported rigid beam in the case of three parallel displacements) and the last added DOFuN+N1 coincides with the control DOF ueq dened in Equation (10). It is worth noting thatEquation (11) can be used as a numerical strategy for implementing Equation (10). Insteadof considering one linear relation involving N existing DOFs, one can introduce (N 1)new DOFs and consider (N 1) relations involving only three DOFs at a time. With the

Lagrange multiplier method, the number of additional unknowns is 2(N 1) (or 3(N 1) ifthe Lagrange multipliers are duplicated) but the bandwidth of the augmented stiness matrixremains roughly unchanged.

Independently of the numerical method chosen for imposing Equation (10), it is importantto check the consistency of the problem to be solved, i.e. the orthogonality between the force

pattern and the boundary conditions. For example, any force (resp. moment) applied on ablocked translation (resp. rotation) DOF should be removed. More generally speaking, the



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856 A. ANTHOINE

loading pattern should aect only free DOFs and=or free combinations of DOFs. In the forcecontrol formulation, incompatibilities between imposed forces and imposed displacements arenot an issue because the displacement conditions naturally prevail on the force conditions.Conversely, in the displacement control formulation, such incompatibilities cause the problem

to have no solution because the displacement constraint equation (10), which derives fromthe loading pattern equation (8), is incompatible with the boundary conditions equation (7).It is therefore fundamental to start from a well-dened problem, after having eliminating any

possible inconsistency.

4. PROPERTIES OF THE EQUIVALENT DISPLACEMENT ueq AND OF THEASSOCIATED CURVE Fueq

This denition of the displacement parameter ueq coincides with the energy-based displace-ment proposed by Hernandez-Montes et al. [5]. As a matter of fact, from Equations (9)

and (10), the incremental work done by the lateral force F is

WF=Fueq (12)

Equation (12) means that the area beneath the Fueq pushover curve coincides with the energyprovided to the structure by the lateral force. This last property has precisely been used byHernandez-Monteset al. to compute ueq from the results of a conventional pushover analysis,through a step-by-step integration based on Equation (12). For being a xed linear combinationof structural displacements, Equation (10) is undeniably a simpler way of computing ueq, butits main advantage is to allow the pushover analysis to be controlled directly by ueq and thusto obtain directly the Fueq curve.

Compared to other displacement control methods based on a particular displacement (e.g.Reference [1]), the proposed method is much simpler since it relies on a standard displacementcontrol loading process where the displacement parameter is a linear combination of severalnode displacements. In particular, it is potentially compatible with any numerical solver with-out requiring the addition of ctitious springs.

However, the proposed method is still unable to follow a snap-back branch. To do so,a more advanced numerical strategy is required (e.g. arc-length method). The question thusarises as whether the Fueq curve exhibits more or less frequently snap-backs than a particular

Fui0 curve. Since the original Fueq curve is nothing else than a particular weighted averageof all possible pushover Fui curves, the following property can be easily established: Ifthe Fueq curve exhibits a snap-back, then at least one of the Fui curve does the same.However, a snap-back in at least one Fui curve does not necessarily imply a snap-back in

the Fueq curve. Therefore all four combinations are possible: snap-back in Fueq but not inFui0 , snap-back in Fui0 but not in Fueq, snap-back in both or in none. The most frequentcombination depends on the structural typology and only an extensive numerical investigationon a large class of structures could give a statistical answer.

A snap-back in the Fueq pushover curve may be given an interesting energetic interpreta-tion. For a while, let us assume that the gravity loads are zero. Then, the energy WF provided

by the lateral load coincides with the sum of the stored elastic energyWe and of the dissipated



7/16


energy Wd. Incrementally, this equality reads:

WF= We+Wd (13)

Here, only Wd should be positive (or zero) whereas no restriction holds on the other twoquantities. A snap-back is characterized by the fact that the release of elastic energy due tothe decrease of the lateral force is higher than the energy dissipated in the structure:

We0 and |We| Wd WF0 (14)

The energy provided by the lateral load is therefore negative, which means that the excessof energy is restored. In particular, the extreme case of a perfectly brittle structure (Wd= 0)appears equivalent to a simple elastic unloading (WF= We 0). This last property is pre-cisely at the basis of the rst method (Unload Entire Structure) in SAP 2000 NL but holdstrue only for drops of strength (strength degradation without dissipation). Snap-backs are in-herent to the static character of the pushover curve and do not appear in a dynamic analysis

where the excess of released energy is mostly transformed into kinematical energy.Now, if the gravity loads are also present, their incremental workWg should be added on

the left-hand side of Equation (13) or, equivalently, on the right-hand side with a negativesign. Then, Wg is the variation of the gravitational potential energy Wg, which can beadded to the elastic energy We, so as to form the total potential energy of the structure. Thus,a snap-back is characterized by the fact that the release of the total potential energy due tothe decrease of the lateral force, i.e. (We Wg), is higher than the energy dissipated inthe structure. The gravity loads will therefore increase (resp. decrease) the snap-back eect iftheir incremental work Wg happens to be positive (resp. negative).

It is worth mentioning also the eventuality for the pushover curve to be early terminated(without snap-back) for a non-zero lateral force because the gravity loads cannot be equili-

brated anymore. Last but not least, the pushover curve may also exhibit bifurcation points.

These two latter cases as well as the softening and snap-back cases will now be illustratedon an analytical example.

5. AN ILLUSTRATIVE EXAMPLE

The irregular frame sketched in Figure 2 is not intended to be realistic since it has beendesigned appositely to exhibit all the possible diculties that can be encountered in real cases,namely limit point, softening branch, snap-back, early termination and even bifurcation. Threehinges (white disks in Figure 2) have been introduced to keep it as simple as possible. Twoequal masses m are lumped at the middle of the storey beams. Besides gravity, an inverse

triangular lateral load is considered. Therefore, the equivalent displacement corresponding tothe load pattern is simply:

ueq= uI=3 + 2uII=3 (15)

where uI (resp. uII) is the horizontal displacement of the rst (resp. second) oor. All beamsare linear elastic (same Young modulus E and inertia I) until the bending moment reaches



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858 A. ANTHOINE

2

1

4 5 23

F

mg

3

6

mg

3F

Figure 2. Pushover analysis on an irregular frame.

a given limit 0 in some section i. Then a plastic hinge forms, which is characterized by alinear isotropic softening

i= 0(1 +i|pi |) (16)

where i is the reduced yielding moment of the section, pi is the plastic rotation and i0 is

a constant softening coecient, which depends on the section. A section that has completelysoften (i.e. i= 0) is thus replaced by a free hinge. Given the loading, any section coincidingwith a beam extremity and=or a point of load application, is one of the six potential plastichinges (numbered black points in Figure 2) that might form during the loading because the

bending moment varies linearly between these critical sections.The structure is twice hyperstatic. Ifmi denotes the bending moment in the critical section i ,

the elastic solution can be found as the minimum of the potential bending energy of thestructure We:

We= (2m21+ 2m1m2+ 3m

22+m2m3+ 2m

23+ 3m

24+ 5m

25+ 4m5m6+ 4m

26)l=12EI (17)

under the following conditions of equilibrium:

m1+m2 m4=Fl

3

2m4 m5+m6=4Fl3

m2+ 2m3=mgl

2

m4 m5=mgl

2

(18)



9/16


Each equilibrium condition is here derived from a particular virtual displacement eld featuringa frame or beam mechanism. The solution is

m1=

67

728mgl

16

39Fl

m2=237

1820mgl+

16

65Fl

m3= 673

3640mgl+

8

65Fl

m4=809

3640mgl+

21

65Fl

m5=1011

3640mgl

21

65Fl

m6=

607

3640 mgl+

71

195Fl

(19)

When F is increasing (or decreasing) from 0, Equation (19) proves that the location of therst plastic hinge depends on the value of the ratio mgl=0. In the following, the structurewill be assumed elastic for F =0, which is true if mgl=0

36401011

3:60, and three particularcases will be considered. In case 1 (F0 and mgl=0= 0:25) the rst plastic hinge appearsin section 1 while, in case 2 (F0 and mgl=0= 3), it appears in section 4. Finally, in case 3(F0 and mgl=0=

24530:453), two plastic hinges appear simultaneously in sections 1 and 6.

This example can be solved analytically: at each step, the elasto-plastic solution minimizes

the function G= We+6

i=1mipi under the equilibrium conditions (18) and the hypothesis of

yielding (|mi|= i and mipi 0) or non-yielding (|mi|6i and

pi = 0) in each section. Moregenerally, this method can be applied to any beam system in which plasticity is conned to

pre-selected critical sections and is characterized by a piece-wise linear constitutive law. Thereader is referred to Reference [6] who applied recent results of mathematical programmingto this particular class of problems.

5.1. Case 1: F0 andmgl=0= 0:25

The elasto-plastic response is conditioned by the ratio l0i=EI of the activated hinge(s).Assuming l01=EI= 0:85, l06=EI= 1:5 and l02=EI= 0:75, the pushover analysisis composed of eight successive steps (Table I). Each step is characterized by a dierent

combination of damage (Di= 1 i=0= i|pi |) and damage evolution (Di= i|

pi |)

in the three activated hinges (1, 6 and 2) as shown in the last three columns of Table I:

D =0 (and D = 0): no hinge. 0D1 and D0: yielding hinge. D =1 (and D = 0): destroyed hinge. 0D1 and D = 0: elastic unloadingreloading of a damaged hinge.

Once the three hinges are destroyed, a free mechanism forms (Figure 3(a)) and the displace-ments are indeterminate (last row of Table I). Nonetheless, the equilibrium equations (18) can



10/16

860 A. ANTHOINE

Ta

bleI.Case

1ana

lys

is.

n

F=

lF=

0

uIEI=l

2

0

uIIEI=l

2

0

ueq

EI=l

2

0

D1

D6

D2

1

0+

dF()

7

3120

+

56

585

dF

29

3120

+

158

585

dF

17

3120

+

124

585

dF

0

0

0

2

8937

3584

+

dF(

)

227

960

20216

4815

dF

18353

26880

9038

4815

dF

7177

13440

12764

4815

dF

1904

321

dF

0

0

3

127557

52192

+

dF(

)

4973

11184

+

98

1503

dF

15185

19572

38

501

dF

52097

78288

130

4509

dF

2193

7456

497

501

dF

0

4

5321091

3705632

+

dF()

100297

264688

+

49

432

dF

118438

3

138961

2+

305

648

dF

11581301

16675344

+

2284

7101

d

F

2193

7456

1

0

5

15027

7456

+

dF()

4973

11184

2527

864

dF

9437

8388

2015

1296

dF

90415

100656

15641

7776

dF

2193

7456

595

144

dF

1

0

6

10107

5152

+

dF()

3985

6624

763

2160

dF

84053

69552

+

517

3240

dF

419897

417312

221

19440

dF

1139

2208

119

360

dF

1

161

360

dF

7

1893

3808

+

dF(

)

10957

9792

14

27

dF

10022

3

10281

6+

4 81

dF

37117

36288

34

243

dF

1

1

1

069

1

632

7 9d

F

8

3 56

()

259

192

+

3

1921

2016

+

2

13123

12096

+

7 3

1

1

1



11/16


0

1

2

3

0.0 0.5 1.0 1.5 2.0

Displacement

Force

3

equ

IIu

Iu

( )equ S AP

(a)

(b)

Figure 3. Case 1: (a) failure mechanism; and (b) analytical pushover curve in terms of ueq, uII and uI(continuous lines) compared with SAP 2000 NL Restart Using Secant Stiness (dotted line).

still be satised provided that the lateral force does not change and is thus able to equilibratethe gravity loads, but an unloading is impossible.

The displacements being normalized to 0l2=EI and the lateral force to 0=l, the complete

Fueq pushover curve is shown in Figure 3(b) together with its two other possible formsFuII and FuI (continuous lines). These curves do not begin at the origin because the chosenreference state is the totally unloaded structure so that, for F= 0, the displacements arethose induced by the gravity loads. Furthermore, the nal horizontal branches do not coincide

with thex-axis because the gravity loads are involved in the failure mechanism. Here, the workof the gravity loads in the mechanism is opposed to the work of the lateral force. Ifit were of the same sign, the nal horizontal branch would fall below the x-axis. Only whenthe work of the gravity loads in the failure mechanism is zero, does the nal horizontal branchof the pushover curve coincide with the x-axis.

The slope of the pushover curve depends on the ratio l0i=EI of the active hinge(s) andalso on the chosen control DOF (ueq, uI oruII). In the present case, the Fueq curve does not



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862 A. ANTHOINE

present any snap-back whereas the two other curves do. The proposed method is thereforeable to follow completely the pushover curve whereas a displacement control method basedeither on uII or on uI fails at some stage. Of course, the proposed method would also failin case of a snap-back of the Fueq curve. By varying the softening ratios, it is possible to

derive cases where the best control DOF would be uI, other cases where it would be uII, butif the hinges soften quickly enough, all pushover curves exhibits a snap-back. In particularfor l0i=EI= , i.e. for perfectly brittle hinges, the snap-back branch coincides with theelastic unloading branch.

For comparison purposes, the example has also been run with SAP 2000 NL. The numericalsolution turned out to depend not only on the member unloading method adopted but alsoon the monitored DOF (uII or uI) and on the control method (use of conjugate displacementor not). The rst method (Unload Entire Structure) always stops prematurely when thesecond plastic hinge appears in section 6. The second method (Apply Local Redistribution)is able to give a complete solution only if the conjugate displacement control is not used.However, the solution then depends on the monitored DOF. Only the third method (RestartUsing Secant Stiness) gives the same and complete solution, independently of the monitoredDOF and on the control method. The solution is however quite dierent from the analyticalone (dotted line in Figure 3(b)). In particular, although the collapse mechanism is right, thecorresponding ultimate displacements are largely overestimated because the order of failureof the three sections is incorrect. This is obviously due to the SAP 2000 NL third method,which forces any hinge that has become non-linear to reach directly the bottom end of anegatively sloping segment of its moment-rotation curve. Since the hinges are rigid plastic witha unique softening segment down to zero, the method gives the same solution independentlyof the i values. Unfortunately, the present example proves that such a solution is not alwaysconservative.

5.2. Case 2: F0 andmgl=0= 3

Assuming l04=EI= 0:2 and 50, the pushover analysis is composed of two successivesteps (see Table II) but is interrupted without snap-back when the second plastic hinge appearsin section 5. This is because the gravity loads cannot be equilibrated anymore. On the onehand, as soon as yielding begins in section 5, dp50 and m50. On the other hand,owing to the previous yielding in section 4, m464= 0=2. Under these two conditions, the

Table II. Case 2 analysis.

n F= lF=0 uIEI=l20 uIIEI=l

20 ueqEI=l20 D4

1 0 + dF() 7

260+

56

585d F

29

260+

158

585d F

17

260+

124

585d F 0

2 1213

1176+ dF()

95

756+

287

1935d F

221

1323

2734

1935d F

811

5292

1727

1935d F

147

215d F

3 353

1176

13

756

1588

1323

4265

5292

1

2



13/16


fourth equilibrium equation in Equation (18), which reads

m4 m5=30

2 (20)

cannot be veried. This means that the pushover curve terminates at this point (third rowof Table II) and the reader will easily check that even an unloading is impossible. Moregenerally, the early termination of the pushover curve occurs when the work of the lateralforce in the failure mechanism is zero. Here, the failure mechanism is a beam mechanism atthe second storey (Figure 4(a)). The complete Fueq pushover curve is shown in Figure 4(b)together with its two other possible forms FuII and FuI. Again, the proposed method isable to follow completely the pushover curve, as a displacement control method based onuII would do. However, nothing would indicate that the nal point is not a snap-back point,unless an advanced numerical strategy is used (e.g. arc-length).

For case 2, the rst method of SAP 2000 NL gives the right answer. However, the programsends the same warning message as in case 1 so that nothing indicates that no solution exists

beyond. The second method either stops prematurely or goes beyond the nal point, which is

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

-1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2

Displacement

Force

equ

IIu

Iu

(b)

(a)

Figure 4. Case 2: (a) failure mechanism; and (b) pushover curve in terms of ueq; uII and uI.



14/16

864 A. ANTHOINE

Table III. Case 3 analysis.

n F= lF=0 uIEI=l20 uIIEI=l

20 ueqEI=l20 D1 D6

1 0 + dF()

14

3445+

56

585d F

58

3445+

158

585d F

34

3445+

124

585d F 0 0

2a 942

371+ dF()

38

159

8344

20 835d F

782

1113+

458

20 835d F

610

1113

2476

20 835d F

1456

1389d F 0

2b 942

371+ dF()

38

159+

49

1710d F

782

1113

419

855d F

610

1113

1627

5130d F 0

497

285d F

2c 942

371+ dF()

38

159

1043

6030d F

782

1113

1367

3015d F

610

1113

6511

18 090d F

91

201d F

1421

1005d F

0

0.5

1

1.5

2

2.5

3

0 0.2 0.4 0.6 0.8 1.2

Displacement

Force

ueq

uII

uI

ueq(SAP)

1

Figure 5. Case 3: bifurcation point of the analytical pushover curve expressed in termsof ueq, uII and uI (continuous lines) and comparison with SAP 2000 NL Restart

Using Secant Stiness (dotted line).

obviously wrong. The third method gives an absurd answer, which was foreseeable becausethe gravity load is preponderant.

5.3. Case 3: F0 andmgl=0= 2453

Assuming l01=EI= 1:3 and l06=EI= 1:1, three solutions are possible for the second

step of the pushover analysis, namely yielding in section 1 and elastic unloading in section 6,yielding in section 6 and elastic unloading in section 1 or yielding in both sections 1 and 6.The end of the elastic segment of the pushover curve is therefore a bifurcation point. Thethree possible solutions are given in Table III and shown in Figure 5 (continuous lines).For each hypothesis, the pushover curve has been computed only until the complete failureof the activated plastic hinge. In this case, the proposed method will follow one among the

possible solutions that do not exhibit a snap-back if any, otherwise it will fail. If at least two



15/16


possible solutions do not exhibit a snap-back, the selection of the solution will probably berandom because of the limited numerical precision. In particular, the bifurcation will not bedetected.

For case 3, the two rst methods of SAP 2000 NL stop prematurely at the bifurcation point,

independently of the monitored DOF and of the control method. As in case 1, only the thirdmethod gives a unique and complete solution, which is again independent of the i values.In particular, it diers from any of the three analytical pushover curves, at least immediatelyafter the bifurcation point (dotted line in Figure 5).

6. CONCLUSIONS

The proposed method is very simple and fully operational in the case of pushover curves(monotonic or cyclic) exhibiting limit points and softening branches. It is based on an appro-

priate denition of the displacement variable ueq to be used for controlling the loading. The

proposed equivalent displacement ueq is related by duality to the base shear F in the sensethat, the energy WF provided to the structure by the lateral load is exactly the surface belowthe Fueq pushover curve, i.e. WF=Fueq.

The method is however unable to follow snap-back branches of the Fueq curve and todetect bifurcation points. Snap-backs and bifurcation are likely to occur as soon as the soft-ening phenomenon is fast enough, which, in the given example, corresponds to low valueof the parameter . In particular, a sudden drop of strength (i.e. = ) always impliesa snap-back. A possible solution would be to implement an arc-length method in which themaximum increment of plastic strain would play a preponderant role in the measure of thearc along the Fueq curve.

In its present state, the proposed method could therefore be added to the three methodsalready available in SAP 2000 NL, as the rst method to be tried.

Finally, the proposed method naturally suggests the use of ueq as the control displacementvariable to be used in seismic displacement-based design methods instead of the roof dis-

placement. The implications of such a choice are nevertheless beyond the scope of this paperand the reader is referred to Reference [5] who eectively used the Fueq pushover curve(modal load) for seismic design.

APPENDIX A: THE LAGRANGE MULTIPLIER METHOD

To keep it as simple as possible, the method is presented in the linear elastic framework. Forthe non-linear case, the reader is referred to Reference [4].

Assuming that the node displacements involved in the boundary conditions have been

eliminated, the force control pushover problem reads

K:u = M:g+Fp (A1)

whereas the displacement control pushover problem takes the form

K:u = M:g submitted to pT:u = ueq (A2)



16/16

866 A. ANTHOINE

The constraint in Equation (A2) may be appended to the classical potential function W witha Lagrange multiplier so that the solution (u; ) is the stationary point of

WL=12

uT:K:u uT:M:g+ (pT:u ueq) (A3)

Hence, the derivatives of WL with respect to u and must vanish:

@WL@u

= K:u M:g+ p = 0

@WL@

= pT:u ueq= 0

K p

pT 0

u

=

M:g

ueq

(A4)

Compared to the force control system equation (A1), the displacement control systemequation (A4) has one additional unknown , which is the reaction force associated to theconstraint, i.e. F. This clearly results from comparing the original problem equation (A1)to the rst derivative in Equation (A4). However, if the original stiness matrix K is positivedenite, the augmented matrix is only symmetric. If this is incompatible with the numeri-

cal solver in use, an alternative solution consists in duplicating the Lagrange multiplier byconsidering the function

WLL= 12

uT:K:u uT:M:g+ 1(pT:u ueq) + 2(p

T:u ueq) + 12

(1 2)2 (A5)

and the associated system

K p p

pT 1 1

pT 1 1

u

1

2

=

M:g

ueq

ueq

(A6)

The augmented matrix is now positive denite. At the stationary point, the two additionalunknowns are obviously equal and their sum coincides with the reaction force F.

It is worth mentioning that the Lagrange multiplier method can also handle the displacementboundary conditions, which are only particular restraints on the node displacements.

REFERENCES

1. Archer GC. A constant displacement iteration algorithm for nonlinear static push-over analyses. Electronic Journalof Structural Engineering 2001; 2:120134.

2. Computers & Structures Incorporated (CSI). SAP 2000 NL, Berkeley, CA, U.S.A., 2000.3. Halphen B, Salencon J. Elasto-plasticite. Presses des Ponts et chaussees: Paris, 1987.4. Belytschko T, Liu WK, Moran B. Nonlinear Finite Elements for Continua and Structures. Wiley: England,

2000.5. Hernandez-Montes E, Kwon OS, Aschheim MA. An energy-based formulation for rst and multiple-mode non-

linear static (pushover) analyses. Journal of Earthquake Engineering 2004; 8(1):6988.6. Cocchetti G, Maier G. Elasticplastic and limit-state analyses of frames with softening plastic-hinge models by

mathematical programming. International Journal of Solids and Structures 2003; 40:72197244.


a simple displacement control technique for pushover analysis

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