Reconciling Elastic and Equilibrium Methods for Static ...people.csail.mit.edu/hishin/projects/fem_vs_equilibrium/paper.pdf · Reconciling Elastic and Equilibrium Methods for Static

13

Reconciling Elastic and Equilibrium Methods for Static Analysis

HIJUNG V SHIN and CHRISTOPHER F PORSTMassachusetts Institute of TechnologyETIENNE VOUGAThe University of Texas at AustinandJOHN OCHSENDORF and FREDO DURANDMassachusetts Institute of Technology

We examine two widely used classes of methods for static analysis of ma-sonry buildings linear elasticity analysis using finite elements and equilib-rium methods It is often claimed in the literature that finite element analysisis less accurate than equilibrium analysis when it comes to masonry analy-sis we examine and qualify this claimed inaccuracy provide a systematicexplanation for the discrepancy observed between their results and presenta unified formulation of the two approaches to stability analysis We provethat both approaches can be viewed as equivalent dual methods for gettingthe same answer to the same problem We validate our observations withsimulations and physical tilt experiments of structures

Categories and Subject Descriptors I35 [Computer Graphics] Compu-tational Geometry and Object ModelingmdashPhysically based modeling

General Terms Algorithms Experimentation Theory

Additional Key Words and Phrases Static equilibrium analysis finiteelement method limit-state analysis masonry analysis self-supporting ma-sonry contact constraints

ACM Reference Format

Hijung V Shin Christopher F Porst Etienne Vouga John Ochsendorf andFredo Durand 2016 Reconciling elastic and equilibrium methods for staticanalysis ACM Trans Graph 35 2 Article 13 (February 2016) 16 pagesDOI httpdxdoiorg1011452835173

1 INTRODUCTION

The design of self-supporting masonry is an ancient and eleganttechnique that combines the form and function of geometry The

This project is partially funded by NSF DMS-1304211 and Royal DutchShell Hijung V Shin acknowledges Samsung Scholarship for their supportAuthorsrsquo addresses H V Shin email hishinmitedu C F Porstemail cporst14gmailcom E Vouga email evougacsutexasedu JOchsendorf email jaomitedu F Durand email fredomiteduPermission to make digital or hard copies of all or part of this work forpersonal or classroom use is granted without fee provided that copies are notmade or distributed for profit or commercial advantage and that copies bearthis notice and the full citation on the first page Copyrights for componentsof this work owned by others than the author(s) must be honored Abstractingwith credit is permitted To copy otherwise or republish to post on serversor to redistribute to lists requires prior specific permission andor a feeRequest permissions from permissionsacmorg2016 Copyright is held by the ownerauthor(s) Publication rights licensedto ACM0730-0301201602-ART13 $1500DOI httpdxdoiorg1011452835173

stability of a structure depends directly on its geometry rather thanrelying on the strength of reinforcement materials such as steelArchitects and engineers have explored the link between geome-try and mechanics since antiquity and recent computer graphicsapproaches have expanded this endeavor and created design toolsthat are informed by or optimized for structural goals (eg Deusset al [2014] Whiting et al [2012] Panozzo et al [2013] Liu et al[2013] Vouga et al [2012] and Block [2009]) This article consid-ers masonry structuresmdashthose made of unreinforced stone or bricklike the Aqueduct of Segovia or Gothic cathedralsmdashthat are stableunder their own weight Masonry has several well-accepted char-acteristics it is extremely strong in compression and for practicalpurposes its compressive strength can be assumed to be infinite[Heyman 1966] When a masonry structure fails usually it is notdue to material failure but to geometric failure blocks hinge apartfrom each other and topple Sometimes mortar is placed betweenblock interfaces but mortar is structurally weak and susceptibleto cracking and weathering so we assume that mortar does notcontribute to the stability of the structuremdashthat there is no tensilecapacity in between blocks

The best approach for assessing the structural stability of masonrydesign remains a subject of debate Given the success of the FiniteElement Method (FEM) for strength analysis in reinforced struc-tures and the wide availability of software tools [Autodesk 2014Computers and Structures Inc 2014 Dassault Systeme 2014] prac-titioners have attempted to apply it to traditional masonry designsThey use linear elastic analysis to predict whether a structure willstand under certain loads either by directly measuring displace-ments of blocks or by measuring indicators of instability such astension at block interfaces with little success Consider the Aque-duct of Segovia the most prominent Roman monument left onthe Iberian peninsula which has stood for nearly two millennia(see Figure 1 top) Bravo et al [1994] modeled the aqueduct as-suming a homogeneous and isotropic material and neglecting theeffect of masonry joints between blocks When they analyzed thestress state of the pillars using FEM they could not conclusivelydemonstrate that the aqueduct is stable under gravity without ten-sion forces Similarly Block [2009] and Whiting [2012] remarkedthat using conventional FEM on masonry arch modeled as a singlecontinuous structure they could not distinguish between stable andunstable circular arches (see Figure 1 bottom) Instead they advo-cated a different type of approach based on direct analysis of forceequilibrium (discussed later) which can reliably predict whether astructure is stable or not without computing the elastic deformationof materials

Although it is expected that naıvely modeling a structure as asingle continuum and applying linearized FEM is inadequate to an-alyze masonry this is the method employed by many practitioners

ACM Transactions on Graphics Vol 35 No 2 Article 13 Publication date February 2016

132 bull H V Shin et al

Fig 1 The Aqueduct of Segovia a masonry structure that has been standing since antiquity (top right) Finite element analysis of a section of the aqueductdoes not conclusively demonstrate its stability (top left image from Bravo et al [1994]) Similarly linear elastic FEM fails to differentiate between thestability of an infeasible arch that is too thin to stand on its own (bottom leftmost) and a thicker feasible arch (bottom left) (image from Block et al [2006])In contrast equilibrium methods correctly predict that the thin arch is infeasible (bottom right) and that the thick arch is feasible (bottom rightmost)

partly due to the lack of alternatives in commercial software Onthe other hand in the broader context FEM has been tremendouslysuccessful at numerically simulating a variety of physical systems inengineering biology mechanics and medicine So it seems plausi-ble that with proper assumptions and material models FEM shouldbe able to correctly analyze masonry structures Is there a better wayto apply FEM to masonry structures Certainly we would expecta full dynamics simulation of the masonry using complex nonlin-ear material and contact models and very fine spatial and temporaldiscretization to correctly predict the behavior of masonry albeitat a tremendous computational cost What are the necessary mod-ifications to continuum linearized FEM to obtain accurate resultsfor masonry In other words what is its primary source of inaccu-racy Several possible explanations are cited but to our knowledgeno systematic examination exists for which of these are importantsources of error and which are not

mdashIt is possible for masonry structures to hinge while still remainingstable Modeling the entire masonry as a single continuous struc-ture does not reflect the contact conditions at the block interfaces

mdashSimulating accurate stresses in the structure requires choosinga correct constitutive model for the stone blocks The simplestchoice of constitutive modelmdasha linear stress-strain relationshipand a linearized strain measuremdashis not correct for stone

mdashSimilarly stone is extremely stiff and its high Youngrsquos mod-ulus potentially introduces numerical stability issues duringsimulation

mdashMost analysis assumes that the shape and positions of the blocksspecified as input by the user are initial conditions and thensolve for the final deformation of the structure In other wordsstability is analyzed using a forward simulation The final de-formed geometry is not the same as the input geometry This canbe a problem when we are interested in the stability of the inputstructure as is without further deformation

Table I Comparison of the FEM VersusEquilibrium Methods

Method FEM EquilibriumSolution variable Deformation Force

Material Elastic RigidGeometry Deforms Fixed

mdashThe expected stresses in a building are impossible to determinefrom the geometry alone due to static indeterminacy Manypossible sets of internal stresses are consistent with the observedstate of the building and slight changes in the environment cancause large changes in measured stresses An everyday exampleof this indeterminacy is a four-legged stool resting on a rigid levelfloor the stool is stable with only three of the four legs carryingthe load and which legs carry how much load can change dras-tically for even infinitesimal changes in the lengths of the legsFinite element analysis necessarily solves for only one possibleset of equilibrium stresses and this solution is highly sensitive toinitial conditions

Equilibrium methods have been proposed as alternatives to finiteelement analysis to sidestep the aforementioned challenges Thisline of attack can be traced back to the seminal work of Heyman[1966] Equilibrium methods also called limit analysis recentlybecame popular in the graphics community as an efficient wayto guarantee stability of procedurally generated or interactivelydesigned architecture [Whiting et al 2009 Vouga et al 2012Panozzo et al 2013] The key idea in these methods is to ignorethe elastic deformation of the stone entirely and instead to treat thestructure as piecewise rigid The unknowns in these methods are theforces acting on these rigid pieces rather than their displacementsTable I summarizes the key differences between finite-element- andequilibrium-based methods (Equilibrium methods and FEM aresurveyed in more detail in Sections 3 and 4 respectively) Generally


Reconciling Elastic and Equilibrium Methods for Static Analysis bull 133

speaking equilibrium methods are known to predict the stabilityof structures more reliably but they are restricted to certain simpleclasses of masonry structure So how is it that equilibrium analysisobtains reliable results even without sophisticated modelingparameters What is the source of discrepancy between linearanalysis using FEM and equilibrium analysis

In this article we examine both classes of methods and seekan explanation for the discrepancy between their results In par-ticular we rigorously evaluate the claim that equilibrium methodsmore reliably predict stability than finite elements We expose theprimary source of difficulty in using linear elastic FEM to analyzemasonry namely that it does not correctly account for the geometricand topological nonlinearity of hinging and breaking contact at theblock interfaces and that moreover these problems can persist evenwhen the analysis is augmented with proper one-sided contact con-straints Finally we propose a small modification to the FEM thatavoids these problems and demonstrate that with this adjustmentand proper interpretation of the FEM both approaches to stabilityanalysis can be viewed as equivalent dual methods for getting thesame answer to the same problem Consequently both equilibriumand finite-element methods can be used to reliably predict the sta-bility of masonry even with very simple and efficient models ofthe structurersquos material and geometry We validate this observationby comparing the simulation results of both approaches to knownanalytic solutions the commercial finite-element Abaqus packageand data from physical tilt tests Both equilibrium and modifiedfinite-element methods correctly predict the maximum tilt angle ofstructures agreeing with analytical solutions and physical tests

2 RELATED WORK

21 Equilibrium Versus Elastic Approach

The analysis of masonry structures has been a topic of study fromas early as the 17th century [Huerta 2008] and can be divided intotwo schools of thought the equilibrium approach and the elasticapproach We refer the reader to Heyman [1995] Huerta [2001]and Roca [2010] for an extensive historical overview

The purpose of the equilibrium approach is to demonstratewhether a static state of equilibrium is possible That is can theforces and torques acting on each rigid block be balanced usingcompressive forces alone In 2D the interaction between gravityand the contact forces acting on each block can be elegantly summa-rized using graphical methods Force polygons translate equilibrium(force and torque balance) into the requirement that forces acting ona block link together to form a closed polygon Thrust lines connectthe resultant force at the interface between pairs of blocks and thecompression-only constraint requires that the thrust line go throughthe interior of the interface (Figure 2)

In 1748 Poleni used thrust line analysis to demonstrate theequilibrium of the cracked masonry dome of St Peterrsquos in Rome[Heyman 1995] The foundation for the modern equilibrium ap-proach was formally laid by Heyman [1966] Under certain physicalassumptions about masonry Heymanrsquos Safe Theorem of limitanalysis posits if it is possible to find an internal system of forces inequilibrium with the loads that does not violate the yield conditionof the material the structure will not collapse and it is ldquosaferdquo ThisSafe Theorem is the basis of several recent computational algo-rithms discussed later for determining or optimizing the stabilityof masonry structures The failure modes unveiled using equilib-rium methods tend to be hinging and correspond to failures of thegeometry of the structure rather than material failures (Figure 3)

Fig 2 The static equilibrium of an arch and a voussoir is demonstratedusing a thrust line and a force polygon The thrust line represents the pathconnecting the resultant compressive forces acting at the block interfacesFor the structure to be in equilibrium there must be a line of thrust containedentirely within the section The force polygon is constructed by linkingthe forces acting on a single block A closed force polygon indicates thatthe block is in equilibrium (Figure modified from Block and Ochsendorf[2007])

Fig 3 A four-hinge collapse mechanism of an arch subject to a point loadThe voussoirs rotate about their edge forming hinges and eventually causinggeometric failure (left) Cracks developing on a wall causing material failure(right) Equilibrium methods focus on geometric failures whereas elasticmethods concern material failures

In particular equilibrium analysis is independent of the density orelastic modulus of the material (forces simply scale linearly)

Whereas equilibrium methods assume rigid geometry and focuson the balance of the external forces acting on each block elasticanalysis considers the internal stress and strain and relates forces(and stress) to the displacement or deformation of the structureEarly noncomputational work pioneering this idea in architectureincludes the photo-elastic method of Mark [1982] which estimatesthe stress distributions on plane sections of cathedrals

In elastic analysis the behavior of the material is modeled math-ematically often using a linearized elastic approximation and theelastic energy of the structure is expressed as a function of the dis-placement of each point of the structure The configurations in staticequilibrium are those that minimize the energy (with external forcestaken into account via the principle of virtual work) or equivalentlythose for which all forces balance at every point Elastic analysisis often used to identify potential material failures when stress istoo high as opposed to the geometric failures found by equilibriummethods Because of the difficulty in analytically solving the equi-librium equations elastic analysis was initially limited to simplegeometry [Huerta 2001] In the 1970s with the development of theFEM and increase in computational capabilities elastic analysis viaFEM became one of the most powerful and ubiquitous techniquesin engineering It is widely used for analyzing structures with bothtensile and compressive forces such as steel or reinforced concretebuildings While FEM refers to a wide family of methods for solv-ing PDEs over discretized function spaces and need not involveelastic analysis specifically in this article we use the term ldquoFEMrdquoin contrast to ldquoequilibriumrdquo methods as a shorthand for the classof primal methods that analyze masonry stability by solving for



displacements of elements in the structure using a linear stress-strain relationship

22 Physics-Based Simulation

Many techniques [Martin et al 2010 Xu et al 2009 Terzopouloset al 1987 Terzopoulos and Fleischer 1988] use simulation tomodel geometry with physical constraints but these approachesseek to compute realistic deformations rather than answer questionsabout static stability

Modeling contact and friction accurately is especially importantto simulating objects at rest Contact and friction resolution hasbeen recognized as a difficult problem For instance it is wellknown that rigid body dynamics with both contact and Coulombfriction is inconsistent [Painleve 1895] Recent strides in graphicstoward incorporating realistic frictional contact into rigid bodydynamics include a staggered algorithm for solving contact andfriction together [Kaufman et al 2008] with the goal of robustlysimulating complex stacks of rigid and deformable bodies addingcontact constraints and friction to continuum-based models [Narainet al 2010] to convincingly capture the complex behaviors ofgranular flow or using shock propagation with an impulse-based [Guendelman et al 2003] or velocity-based [Erleben 2007]contact solver to simulate stable stacking However these worksfocus primarily on dynamic simulations whereas we are interestedin static analysis Hsu et al [2012] use static equilibrium analysisto simulate large-scale stacking behavior but they are interested increating plausible animation rather than studying physical stability

FEM and its variations are widely used in computer modelingand animation Most relevant to our work are recent techniques byChen et al [2014] and Coros et al [2012] on computing the elasticinverse problem finding the rest shape of an elastic object so thatthe object deforms into a desired input shape by the given externalforces However the former is targeted at 3D printing with elasticmaterials and the latter deals with dynamic simulations rather thanstatics

23 Structural Optimization Using Static Analysis

Static analysis has been applied to a wide range of areas with the goalof designing physically realistic objects For example static equi-librium has been used to determine plausible character poses [Shiet al 2007] and realistic tree structures [Hart et al 2003]

Several applications have proposed structural optimization tar-geted at fabrication Umetani et al [2012] developed a guidedmodeling interface for interactive design of wood furniture Pre-vost et al [2013] implemented stability objectives in order to print3D models that stand upright without requiring an oversized baseor additional stilts FEM has been applied to strengthen 3D inputobjects [Stava et al 2012] and to find worst-case loading scenar-ios [Zhou et al 2013] Umetani et al [2010] presented real-timeFEM analysis for interactive geometric modeling of objects such asmetallophones

In the context of building structures Smith et al [2002] appliedstatics to automatic truss design that optimizes for minimal materialconsumption Recently a number of works appeared in the graph-ics community focusing specifically on the design of structurallysound masonry for instance integrating static analysis with proce-dural modeling [Whiting et al 2009] a method that was extendedby Whiting et al [2012] to generic quad meshes Several meth-ods [Vouga et al 2012 Panozzo et al 2013 de Goes et al 2013Liu et al 2013] optimize structures based on the Thrust NetworkAnalysis technique introduced by Block et al [2007] All of theseworks are in the tradition of equilibrium analysis

Table II NotationSymbol Means Size

lowast

x Known (ldquoinputrdquo) quantitiesxaft Geometry after deformation n times 1xbfr Geometry before deformation n times 1K Stiffness matrix n times n

fin Internal forces from elastic deformationDagger dagger

fw Self-weight amp external applied forces dagger

fct Contact forces between adjacent blocks dagger

ni Normal direction of interface 3 times 1r Torque arm 3 times 1

lowastn = number of element nodesdaggerForces are discretized at different points of applicationindicated bysuperscripts n (element nodes) i (block interfaces) b (block) Vectorsize depends likewise on the superscriptDaggerFinite element method only

Fig 4 Variables used in the block equilibrium (Section 3) and finite el-ement methods (Section 4) In the former force and torque equilibrium iscomputed per block (b) and contact forces fi

ct act at discrete contact nodeson the interfaces (i) between blocks In the latter deformation occurs atelement nodes (n) where each element can be a subpart of a block The de-formation in turn induces elastic forces fn

in at the nodes Contact constraintsare formulated as functions of nodal positions (x) and consequently contactforces (fn

ct ) act also at the nodes

Somewhat related to our goal of reconciling the equilibrium andFEM traditions Fraternali [2010] and de Goes et al [2013] derivedcorrespondences between Thrust Networks and finite element dis-cretization of stress functions in the case of masonry with shelltopology We discuss in detail the relationship between Thrust Net-work Analysis equilibrium of blocks and FEM in the appendix

3 EQUILIBRIUM METHODS

In what follows we describe the equilibrium method and the FEMformulation and discuss their similarities and differences We be-gin with the equilibrium method which is known to predict thestability of masonry structures with high reliability and is favoredby researchers as a powerful yet simple alternative to the FEM formasonry analysis

Please refer to Table II and Figure 4 for our notation and variables

Assumptions In contrast to elastic approaches equilibriummethods assume that stones are infinitely stiff and strong In ad-dition the method makes Heymanrsquos three assumptions about thephysics of masonry [1995]

(1) Masonry has no tensile strength(2) It can resist infinite compression(3) Sliding failure does not occur within the masonry



Equilibrium methods rely on Heymanrsquos Safe Theorem if thereexists an internal system of forces that are in equilibrium with theexternal loads (including self-weight) and that satisfy the previousassumptions the structure will stand Notice that the Safe Theoremguarantees stability if we can find any equilibrium statemdashwe donrsquothave to find or prove anything about the ldquoactualrdquo state of the stresseswithin the structure (which in any case can change discontinuouslyfor small perturbations of the structurersquos geometry or loads due tostatic indeterminacy) Equilibrium methods therefore check com-putationally for the existence of any possible equilibrium state

In what follows we describe one simple example of an equilib-rium method introduced by Whiting et al [2012] for analyzing thestability of masonry structures built from polyhedral blocks (whichwe will call the ldquoblock equilibrium methodrdquo) Other equilibriumapproaches such as Block et alrsquos Thrust Network Analysis [2009]and the improvements described in recent graphics papers (see Sec-tion 23) also rely on the Safe Theorem but are tailored to ana-lyzing structures with the geometry of a thin surface rather thana block network Here we will focus on the simple and generalblock equilibrium method rather than these surface-based methodsas the connection between it and finite elements will turn out to beparticularly simple

Static Equilibrium Condition In order for a structure to standin static equilibrium it is required that the net force and net torqueacting on the center of mass of each block b equal zero includingthe contact forces (fb

ct) and any forces arising from self-weight andexternally applied loads (fb

w)sum

iisinb

fict + fb

w = fbct + fb

w = 0 forall b (1)

sum

iisinb

(fict times ri

ct

) + (fbw times rb

w

) = 0 forall b (2)

where rbw is the torque arm with respect to the blockrsquos center of

mass A three-dimensional force fict is positioned at a finite subset

of vertices on the interface Assuming a linear force distributionacross the surface three contact points per interface are sufficient torepresent the force distribution In our implementation we followWhiting et al [2009] and place the forces on each vertex of thepolygon that defines the interface between two blocks (overlappingregion between the adjacent block faces) The assumption that ma-sonry can resist infinite compression but no tension means that thecontact forces at the interfaces between blocks must be compres-sive This is expressed as an inequality constraint for each interfacecontact force

fict middot ni ge 0 (3)

where ni is the interface normal pointing into the block (ie thedirection of compression)

In addition to the equilibrium constraints friction constraints canbe approximated by constraining the tangential component of theinterface forces (I minus niniᵀ) middot fi

ct to be within a conservative frictionpyramid of their normal component (niniᵀ) middot fi

ct

|(I minus niniᵀ) middot fict| le α|fi

ct middot ni | (4)

where α is the coefficient of static friction In our implementationwe assume infinite friction (α = infin) so that all failure is caused byhinging and not sliding

Force Solution According to the Safe Theorem if a feasiblesolution fct exists that satisfies the aforementioned constraints thestructure will stand Due to static indeterminacy if a solution existsthat solution is typically far from unique (Consider the case of the

four-legged stool mentioned earlier) To facilitate numerically solv-ing for a solution or to choose a particularly intuitive set of forcesfor visualization often a linear or quadratic objective function (egWhiting et al [2009] uses weighted sum-of squares) is included andthe problem is solved as a linear or quadratic program with linearconstraints Note that the solution fct is not intended to represent theactual state of the forces in the structure The key to stability is theexistence of a feasible solution and not the solution itself

To summarize in the equilibrium method we take an input geom-etry of a structure and formulate a linear program (LP) or a quadraticprogram (QP) that enforces equilibrium constraints [Livesley 1978]The requirement that forces must be compressive is a first-classcitizen included in the constraints If a solution exists the struc-ture is stable One beauty of the method is that it handles contactconstraints and hyperstatic equilibrium naturally by treating theblocks as rigid contact constraintsmdashthat blocks can hinge but can-not interpenetratemdashare implicitly satisfied and by relying on theSafe Theorem hyperstatic equilibrium is also taken into account

4 THE FINITE ELEMENT METHOD

We now quickly survey the Finite Element Method particularlyfocusing on the analysis of masonry structures Excellent text-books [Bathe 2006] cover the method in detail and we refer to othersurveys [Muller et al 2008 Sifakis and Barbic 2012] for more ex-tensive discussion of the application of the FEM to computer graph-ics simulations While there are numerous FEM formulations withdifferent assumptions we first introduce the simplest most naıveversion of the algorithm with a linear material model and continuumcontact constraints Although this may seem like a simplistic modelit is the version that is most commonly implemented in commer-cial software and widely employed by practitioners [Autodesk 2014Computers and Structures Inc 2014 Dassault Systeme 2014] Thisis the case even for masonry analysis although it has been demon-strated that the results are unreliable in some cases (eg Block et al[2006]) Later we also consider different variations that improveon this method

Linear Elastic Material The linear FEM treats structures as a fi-nite set of elements of finite size Each element is defined by a set ofnodes and it is assumed that a linear stress-strain relationship holdsacross the element encoded as a per-element stiffness matrix Thestiffness matrix depends on the physical behavior of the materialand for homogeneous and isotropic elastic materials it is usuallycharacterized by the Lame parameters or Youngrsquos modulus andPoissonrsquos ratio A global stiffness matrix K can be constructed forthe entire physical system by assembling the element stiffness ma-trices K relates nodal displacements (xaft minus xbfr) to internal forcesat element nodes (fn

in) where xbfr and xaft refer respectively to nodalpositions before and after the structure undergoes deformation

fnin = K(xaft minus xbfr) (5)

For most physical materials the simplified linear stress-strain rela-tionship and multilinear strain measure is a good approximation forsmall displacements

Contact Constraints Contact constraints limit the space of al-lowable nodal displacements The simplest FEM models the struc-ture as a continuum (ie a monolith)mdashdisplacements (xaft minus xbfr)across the structure are continuous and fracture is disallowed Thiscondition is expressed as a constraint that overlapping nodes onadjacent elements have identical position

xj

aft = xkaft forallj k st

(x

j

bfr = xkbfr

) (6)



Fig 5 A classic example of a structure that is stable despite its blocks not resting flush against each other is the three-hinged arch Begin with a stablecircular arch (left) and pull the base apart (center) The two halves of the arch hinge inward with the top two blocks remaining in contact only along their topboundary t Thrust line analysis demonstrates that the hinged configuration is still stable since a thrust line exists embedded within the volume of the archMany real-world bridges such as the Salginatobel Bridge in the Swiss Alps (right) stand by this principle

Over the entire structure this constraint can be expressed as a sparselinear system of equalities

A(xaft minus xbfr) = 0 (7)

where A is a sparse matrix of coefficientsThis constraint models the entire structure as a continuum In

traditional FEM the element boundaries are artificial boundariesinside a continuum object that have both compressive and tensilecapacity The FEM computes the deformation of the object whichcauses internal stress distributed throughout the elements When anode is subject to stress that exceeds its capacity the object under-goes fracture or material failure at this location

As mentioned earlier while the compressive strength of stonesis extremely high and practically infinite compared to the averageworking stress the same is not true for tensile strength particularlyfor masonry structure built of weakly connected blocks Mortarbetween blocks appears to connect the structure but in practice it isvery weak and provides close to zero tensile capacity Tensile stressat the interfaces of blocks has thus been interpreted as an indicationthat the masonry structure is not safe (eg Brune and Perucchio[2012])

Here lies the important difference between masonry structuresand other continuum structures modeled in traditional FEM Blockinterfaces in masonry are physical boundaries that have practicallyzero tensile capacity Tensile stresses at the interface mean that theadjacent elements or blocks will separate from each other How-ever in case of masonry this does not automatically mean that thestructure will collapse Blocks can hinge without completely sep-arating from each other and in fact some stable configurations ofstructures involve such hinging

The arch in Figure 5 is an example of a stable hinged structureThe three-hinged arch in the center is formed by taking a stablecircular arch and pulling the bases apart At the top part t ofthe crown interface the blocks are pushing against each other sothe adjacency constraint applies On the other hand as the base ispulled apart at the bottom part of the same interface b and at theouter parts of the base interfaces the blocks separate away forminga hinge The three-hinged arch is a well-known stable configura-tion [Heyman 1996] and many real-world arch bridges stand by thisprinciple Notice that in the final stable configuration there is notensile stress at the opening of the hinges The blocks simply sepa-rate from each other at these locations In order for the FEM to findsuch configurations contact constraints must be carefully specifiedto express the physical reality allowing separation between blocksand disallowing interpenetration Naıve continuum FEM does notdo this

Many different strategies have been developed to address thisissue For example Brasile et al [2007] and Lourenco [1998] useanisotropic continuum models for separate brick and mortar layersDolbow and Belytschko [1999] developed the eXtended FiniteElement Method (XFEM) to model fractures and discontinuitiesby enriching the solution space with discontinuous functions andCundall et al [1971] introduced the Discrete Element Method(DEM) which simulates discrete particles and their contact throughtime These methods are computationally expensive and requiretuning parameters [DeJong 2009] Perhaps for these reasons theyare rarely used by practitioners

In graphics simulations it is a well-known technique to add in-equality constraints to the contact constraints to express unilateralcontacts [Baraff 1993] Do correct contact constraints reconcile thedifferences between equilibrium methods and FEM when analyz-ing masonry stability That is can the FEM with correct contactconstraints predict equilibrium of a masonry structure As we willsee in Section 5 the answer is ldquonot quiterdquo But before we examinea failure case in detail we will finish describing the FEM solutionand make some remarks about how the FEM equations relate tothose of the block equilibrium method

Deformation Solution Once the material properties and con-tact constraints are appropriately assigned the FEM solves forthe deformation of the structure (xaft) at static equilibrium Theprinciple of virtual work states that the equilibrium configurationminimizes the sum of elastic strain energy stored in the structure12 (xaft minus xbfr)ᵀK(xaft minus xbfr) the gravitational potential energy andthe work done by external forces fnᵀ

w (xaft minus xbfr)

U(xbfr xaft) = 1

2(xaft minus xbfr)

ᵀK(xaft minus xbfr) + fnᵀw (xaft minus xbfr) (8)

where fnw is the force applied by self-weight and external loads

The FEM is then a quadratic program that given an input un-deformed geometry (xbfr) solves for the deformed geometry (xaft)

xlowastaft = argmin

xaft

U(xaft xbfr)

st Contact forces at block interfaces are compressive(9)

Inequality constraints express the compression-only constraint onall adjacent nodes at the interfaces

(xj

aft minus xkaft

)N (xjk) le 0 forallj k st

(xj

bfr = xkbfr

) (10)

where N (xjk) is the normal direction of the interface where nodesxj and xk meet pointing away from xj If it is known a prioriwhich nodes are in compression and which are in tension thecompression-only constraint can be expressed instead as a linear



Fig 6 The three-legged π geometry is a simple model that illustrates the problem with the forward FEM A horizontal block rests on three supports ofequal height but varying width (a) The undeformed geometry of the three-legged π (b) The deformed settled state of the π model using finite elements withnaıve (continuum) contact constraints between elements The method predicts tension at the indicated location on the right block (c) An Abaqus simulationof the tilted π model with inequality contact constraints still incorrectly predicts toppling of the structure at 30 (d) Inverse FEM (Section 6) and (e) blockequilibrium method (Section 3) both predict that the tilted π is stable up to 191 The solved-for rest shape of the blocks and one possible set of contact forcescertifying equilibrium of the structure are shown for the two methods respectively (f) We performed an experimental tilt test of the three-legged π (g) The π

stands even when tilted to 136 well beyond the point that failure was predicted by forward FEM simulation

constraint on the active set by adjusting A the contact constraintin Equation (7) to apply only to adjacent vertices in compressionIn either case the blocks are allowed to hinge and prevented frominterpenetrating

Note that the output of the previous optimization (xlowastaft) is a de-

formed structuremdashthe input geometry (xbfr) is not preserved butsettles into a minimum energy state This does not answer quitethe same question as the inverse problem given the geometry ofthe masonry ldquoas it lies (already settled or deformed)rdquo does thereexist a physically plausible undeformed state for which the currentconfiguration is stable Since masonry is stiff and displacementsare very small it seems reasonable to assume that this difference isnegligible We will return to this point in Section 6

Comparison to the Equilibrium Method The equations ofthe FEM presented previously turn out to be quite similar to that ofthe equilibrium methods in Section 3 This is clearer if we look atthe dual formulation of the FEM The Lagrangian form of the FEMEquation (9) is

L(xaft λ) = U(xaft) + λᵀA(xaft minus xbfr) (11)

where U is the potential energy as a function of the deformedgeometry (xaft) A(xaft minus xbfr) is the contact constraint and λ isthe vector of Lagrange multipliers Taking the derivative of theLagrangian with respect to xaft we get

dL

dxaft= K(xaft minus xbfr) + fn

w + Aᵀλ (12)

At xlowastaft the minimum potential energy state the derivative is equal

to zero

K(xlowastaft minus xbfr) + fn

w + Aᵀλ = fn

in + fnw + fn

ct = 0 (13)

where K(xlowastaft minus xbfr) = fn

in is the elastic force fnw is the force from

self-weight and external loads and Aᵀλ = fn

ct is the contact forceinduced by the contact constraints all defined at the element nodesEquation (13) the condition that the derivative of the Lagrangian iszero at the solution turns out to be the condition that the net forceper element node equals zero Compare this to the equilibriumEquations (1) and (2) in the block equilibrium method Equilibriummethods state that the net force and net torque are zero per blockAlso whereas the compression-only constraint is directly specifiedas an essential part of the equilibrium method in the FEM it is ap-plied to the contact constraints in an implicit manner (by specifyingthe active set where the contact constraints apply)

5 THE PROBLEM WITH FORWARD FEM

Despite the similarity of the FEM formulation to the equilibriummethods the FEM even with correct one-sided contact constraintsstill fails to predict a no-tension equilibrium of certain structuresConsider the three-legged π model in Figure 6 which consists ofa long horizontal block balanced on top of three vertical columnsof different thicknesses The structure is clearly stable the centerof mass of the top block falls inside the support region of thecolumns defined by the convex hull of the parts of the structure



Table III Comparison of Naıve FEM Inverse FEM and Equilibrium MethodsMethod Naıve FEM Inverse FEM Equilibrium

Solution variable Deformed geometry (xaft) Undeformed geometry (xbfr) Contact forces at interfaces (fict)

Constraints Nodal displacements Lagrangian derivative Sum force and torqueObjective function Minimum potential energy None User defined None User defined

Equilibrium condition Sum force per element node Sum force and torque per block

touching the ground Let us examine closely what happens to thestructure according to the FEM described in the previous sectionDue to the force of gravity and the weight of the top block pushingagainst them the three columns deform and shorten Since theleftmost column is thinner and carries more of the weight it shortensslightly more than the middle column As a result the top blocktilts leftward leaning against the left and the middle column If thestructure is treated as a monolith with the naıve contact constraint inEquation (6) artificial tensile stresses appear at the interface of therightmost column and the top block (Figure 6 top left) If we followthe practice of interpreting tension at block interfaces as an indicatorof structural failure we would make an incorrect conclusion thatthe structure cannot stand

Replacing the continuum contact constraints by inequality con-tact constraints allows hinging at the block interfaces Howeverthis still does not guarantee a correct stability prediction Considerwhat happens if we start tilting the ground counterclockwise (iethe structure is standing on a slope) According to the inequality-constrained FEM soon a hinge starts to form between the leftmostcolumn and the top block The top block tilts leftward lifting upfrom the rightmost block Very soon the center of mass of therightmost block falls outside its support region and the columntopples over causing the structure to fail Indeed when we run theexperiment on Abaqus with discontinuous boundaries at the blockinterfaces it failed to find an equilibrium solution for the structureat 30 However when we analyze the structurersquos stability usingthe block equilibrium method and perform a physical experimentof the same model we see that the structure can tilt well beyond30 (see Figure 6 bottom) What is causing the discrepancy

Naturally it is possible to improve the FEM further by addingmore sophisticated features For instance we can model nonlinearstress-strain relationships or use a nonlinear strain measure In factnumerous variations of the FEM allow such modifications at theexpense of increasing complexity and computational cost [Lourencoet al 1998 Cundall 1971 Dolbow and Belytschko 1999] Howeverit is unclear exactly which of the assumptions mentioned in theintroduction are responsible for the FEMrsquos error and hence whichmodifications are worth their cost when it comes to answering thequestion ldquowill the structure standrdquo

In the next section we turn to the key advantage of equilibriummethods and apply it to FEM to make it more suitable for masonryanalysis We show that with only a simple modification to the origi-nal FEM formulation and without adding more complexity we canobtain the same predictions as equilibrium methods

6 INVERSE FINITE ELEMENT METHOD

One of the main advantages of equilibrium methods is that theytreat the blocks as rigid bodies Since the blocks do not displacehandling of the contact constraints is simple we need not worryabout contacts forming or breaking over time (due to eg hing-ing) Since the material properties are already built into the modelequilibrium methods also avoid the need for specifying other mate-rial parameters This greatly simplifies the problem formulation

We can avoid the problem of contact resolution in the FEMif we know the final geometry So instead of solving for the

deformation of the input geometry consider the following modi-fication let the input be the deformed state Now given the input-deformed geometry which is the final minimum energy state (xlowast

aft)solve for the original undeformed geometry (xbfr) Hence the orig-inal FEM Equation (9) becomes

xaft = argminxaft

U(xaft xbfr)


where xbfr is the unknown and U is the potential energy of thestructure from the original FEM formulation Equation (8) Sincethe final geometry (and hence the normal direction of each inter-face N (xjk) in the compression-only constraint) is known we canexpress it as a linear inequality constraint

AN(xaft minus xbfr) le 0 (15)

where AN is a sparse matrix of coefficients encoding the normaldirection and adjacency of element nodes

To solve for xbfr take the Lagrangian form of the new FEMEquation (14)

L(xaft xbfr λ) = U(xaft xbfr) + λᵀAN(xaft minus xbfr) (16)

Since the minimum solution is given as the input (xaft) we can write

dL

dxaft

∣∣∣∣xaft

= K(xaft minus xbfr) + fnw + Aᵀ

Nλ

= fnin + fn

w + fnct

= 0

(17)

where the unknowns are the undeformed geometry (xbfr) and theLagrangian multiplier (λ) The stiffness matrix depends nonlinearlyon the nodal rest positions xbfr but since displacements will be smallits dependence on the rest positions is negligible (see appendixfor derivation) Similar to Equation (13) which we get from theLagrangian of the forward FEM Equation (17) is an equilibriumcondition that states that for each node the sum of the forces equalszero The difference is that in the inverse FEM the final geometryand hence the solution to the optimization x

aft is known whereas inthe forward FEM only the undeformed geometry (xbfr) is knownFixing the final geometry resolves the contact problem naturallyThe nodes that are adjacent in the input geometry are the ones thatare also in contact after the deformation

Similar to the equilibrium methods the inverse FEM searchesfor one of possibly many solutions (xbfr and λ) that satisfy theequilibrium condition in Equation (17) Likewise in practice a user-defined objective function (H) is added to obtain a unique solutionFor instance H can be set to minimize the amount of deformation(xaftminusxbfr) or the internal force induced by the deformation K(xaftminusxbfr) Finally the solution is obtained by solving the optimizationproblem

[xlowastbfr λ

lowast] = argminxbfrλ

H(xbfr λ)

st K(xaft minus xbfr) + fnw + Aᵀ

Nλ

(18)



So both the inverse FEM and the equilibrium approach keep theinput geometry as the final geometry of the structure Both constrainthe contact forces to act in compression only and both solve for apossible solution state that satisfies some equilibrium condition Incase of the inverse FEM the equilibrium condition is in terms ofnet force per element node and in case of the equilibrium methodsit is net force and torque per block Next we prove that in factthese equilibrium conditions are equivalent

Equivalence of the Inverse FEM and Equilibrium Methods

I Forces induced by the displacements solved for using theinverse FEM also satisfy the constraints of the block equi-librium method In other words every solution of the inverseFEM problem corresponds to a set of forces at the block inter-faces that are solutions of the block equilibrium method

The equilibrium condition of inverse FEM Equation (17)states that the sum of the forces acting on each element node iszero It easily follows that the sum of the forces acting on eachblock (b) is also zero

iisinb

(fiin + fi

w + fict

) = 0 forallb (19)

Since the elastic energy of each block is invariant under uni-form translation of the blockrsquos nodes by Noetherrsquos theorem thegradient of elastic energy has no component of rigid translationand

iisinb

fiin = 0 forallb (20)

That is the sum of internal forces per block is zero SubtractingEquation (20) from Equation (19) it follows that

iisinb

(f iw + fi

ct

) = 0 forallb (21)

That is the sum of contact forces external forces and self-weight is zero This is the exactly the per-block force balancein Equation (1) An identical argument holds for infinitesimalrotation of the blocks and torque balance at the center of mass(see appendix for full proof)

II For a given input geometry if the block equilibrium methodfinds a set of forces that puts the geometry in static equilib-rium then there exists an undeformed (rest) configurationof the input geometry that is in static equilibrium when de-formed to the input positions In other words every set offorces at the block interfaces that are solutions of the blockequilibrium method correspond to some rest configuration thatcertifies stability of the input geometry as judged by the inverseFEM

Let fiin = minusfi

ct minus fiw By definition these forces satisfy the

inverse FEM equilibrium equation

fiin + fi

w + fict = 0 foralli (22)

It remains only to show that there exists an undeformed con-figuration that generates these elastic forcesmdasha deformationu = xaft minus xbfr for which fin = Ku It suffices to show this oneach block b separately The Fredholm alternative states thatexactly one of the following cases holds(a) fb

in = Kbx has a solution In this case x is the desiredundeformed positions of the nodes of b

(b) (Kb)ᵀy = 0 has a solution y with yᵀfbin = 0 But Kb

is symmetric and the only vectors in its kernel are theinfinitesimal rigid translations and rotations of the blockTherefore y can be decomposed into pure translation (t)

and infinitesimal rotation (r) components

yᵀfbin = (t + r)ᵀfb

in

= t middot fbin + [

ω times (xb

aft minus cbaft

)] middot fbin = 0 (23)

where cbaft is the blockrsquos center of mass and ω is the rotation

vector of r and is understood to act per node on the torquearm xi

aft minus ciaft The last equality (= 0) follows since

[ω times (

xbaft minus cb

aft

)] middot fbin = [(

xbaft minus cb

aft

) times fbin

] middot ω = 0

(24)

and by the block equilibrium methodrsquos constraints fbin does

not have a force or torque component Equation (23) con-tradicts (b) Therefore (a) is true and fb

in = Kbx has asolution

We have shown that the inverse FEM obtains results equivalentto those of the block equilibrium method and vice versa Bothapproaches to stability analysis can be viewed as equivalent dualmethods for getting the same answer to the same problem with theFEM formulating the problem in terms of displacement variablesand equilibrium methods in terms of force variables By making thefinal geometry a known input both formulations avoid the difficultyof specifying the contact constraints The block equilibrium methodhas the additional simplicity of expressing the compression-onlyconstraint directly in terms of forces Inverse FEM expresses thesame constraint but indirectly as a linear inequality constraint interms of displacements (see Equation (15))

7 EXPERIMENTAL RESULTS AND VALIDATION

For several different example structures both didactic and com-plex we compared the stability of the structure as predicted by thefollowing methods

mdashnaıve FEM using continuum linear elastic finite elements themost widely used computational analysis used by practitioners(We expect it to give overconservative results far from the ana-lytical or the physical experiment results)

mdashFEM with inequality contact constraints still using forward sim-ulation This one-sided handling of contact constraints improvesupon the previous model but small deformations in the elementscan translate to nonlinear changes in the geometry and topologyof the block network that lead to inaccurate results

mdashthe block equilibrium methodmdashinverse FEM (as we saw in the previous section this method

ought to give identical results as the block equilibrium method)mdashexact analysis (for the simple examples where the analytical so-

lution is tractable)mdashfor five of the models physical experiment

To most effectively evaluate the predictions of these methodswe subjected each structure to a tilt test we continuously tilted theground plane on which the structure rests and computed the crit-ical angle at which the structure topples In the simulations thiscorresponds to increasing the horizontal component of the gravita-tional acceleration This tilt analysis is commonly used as an initialassessment of the lateral stability of structures [Zessin 2012]

We use the commercial software package AbaqusStandard[Dassault Systeme 2014] for the forward FEM implementationsFor all examples we used linear elastic material with Youngrsquosmodulus 1 MPa Poissonrsquos ratio 02 and uniform density Detailedsettings for Abaqus parameters are attached in the appendix aswell as displacement scaling factors used to aid visualization



Table IV Tilt TestModel Critical Tilt Angle ()

( of blocks) Naıve FEM FEM w Inequality Inverse FEM Equilibrium Analytic ExperimentCube (1) 184 448 450 450 450 410 plusmn 03

Arch infeasible (36) infeasible infeasible infeasible infeasible infeasiblelowast -Arch thin (36) infeasible infeasible 01 01 01lowast -Arch thick (36) infeasible 72 82 82 82lowast 47 plusmn 02Gothic arch (7) 01 168 174 174 - 157 plusmn 023-legged π (4) infeasible 30 191 191 - 143 plusmn 02

Flying Buttress (22) infeasible 00 57 57 - 44 plusmn 07Aqueduct (490) infeasible infeasible 65 65 - -

lowastAnalytical angles for circular arch [Ochsendorf 2002]

For the physical experiments we created small-scale models byassembling 3D printed blocks The blocks were printed using aZCorp 400 Series powder printer with clear binder After printingblocks were infiltrated with 3D Systems ColorBond also commonlyreferred to as Z-Bond 90 in order to reduce material deteriora-tion from multiple tests The material density was measured to be15gcm3 with a friction angle of 43 We used a tilt table consistingof a stiff platform attached to a hinge along one edge and a stringalong the opposite edge To minimize dynamic effects the table wastilted by pulling the string at a very low speed The models weretilted to collapse three times All of the experiments were filmedwith a high-speed video camera and the angle of tilt was monitoredwith a digital protractor

For each method we determined the failure of a structure bythe following criteria The inverse FEM and equilibrium methodspredict failure when a feasible solution that satisfies their respectiveequilibrium conditions do not exist For naıve FEM we interpretedtensile forces at the block interfaces as failure since we are assum-ing they have zero tensile capacity FEM with inequality contactconstraints cannot predict tensile forces at block interfaces sinceblocks are allowed to separate from each other Instead we mea-sured the critical tilt angle when Abaqus failed to converge to anequilibrium solution This can happen for example when the dis-placements of the blocks become too large (eg when the blockstopple) Admittedly the Abaqus results can be more or less accuratedepending on numerous variables such as the exact way blocks arediscretized into mesh elements the maximum number of iterationattempts allowed for the solver or the numerical precisiontolerancesettings Finding the right parameters that work for each model isnot straightforward and is a major source of difficulty in correctlyanalyzing masonry stability using elastic analysis We have exper-imented with changing the parameters (see appendix on Abaqussettings for detail) and show the least conservative critical angle

Table IV shows the result of the experiments The models wetested are in order of increasing complexity as follows

Simple Cube A lone cube should hinge and topple when tiltedpast the point where its center of mass lies over its support at 45With the exception of continuum FEM1 all methods predict thecorrect critical angle for this trivial example

Circular Arches The stability of a circular arch can be charac-terized by the dimensionless thickness-to-radius ratio tr any archwith tr lt 01075 is unstable and the arch becomes increasinglystable as tr increases Analytic solutions for the critical angle are

1A cube is a single continuum block For this example only to distinguishbetween naıve FEM and the formulation with inequality constraints thenaıve FEM models the cube and the ground as a single continuum (ie theblock is glued to the ground) Failure is predicted when tension first appearsat the interface between the ground and the block

Fig 7 Results of the tilt-test applied to circular arches (a) Inverse FEMand (b) the block equilibrium method both correctly predict that an archwith thickness-to-radius ratio 01075 the thinnest possible stable circulararch is stable For a thicker arch (tr = 0150) both the (c) inverse FEMand (d) block equilibrium method correctly predict that the arch is stableup to a tilt angle of 82 (e) The naıve continuum FEM predicts tension atblock interfaces (circled) even at the rest position (00) (f) With correctedcontact constraints that allow hinging between blocks forward FEM predictsfailure at a more accurate but still overly conservative angle of 72 (gh) In a physical tilt test the model fails at 47 likely due to imperfectionsin the fabrication process of the blocks corresponding to an effectivenessthickness of 87

available [Ochsendorf 2002] Even at the horizontal rest positionthe naıve continuum FEM predicts failure due to finding tension atthe interfaces Note that this does not mean that the structure willfail Rather it means that hinges are trying to form between blocksbut they are prevented from doing so because of incorrect contact



Fig 8 The Gothic arch at the critical tilt angle as predicted by (a) theinverse FEM and (b) the block equilibrium method where the one possibleset of contact forces is shown with arrows (c) Continuum FEM predictstension immediately after tilt (e f) The physical model fails at a criticalangle slightly less than that predicted by (d) the inequality-constrainedforward FEM inverse FEM and equilibrium method Red dots indicatelocations where hinges develop between blocks

constraints The FEM with inequality contact constraints performsbetter but fails to predict the stability of the minimum thicknessstable arch and for a thicker arch (tR = 0150) predicts a criticalangle over 10 smaller than the analytic solution (see Figure 7)Both the block equilibrium method and inverse FEM predict a crit-ical angle equal to the analytic solution The physical structure canfail the tilt test before the ldquotheoretical critical anglerdquo for severalreasons including slight variations in block size imperfections inthe constructed geometry due to fabrication error wearing off ofblock corners from repeated experiments or insufficient friction atthe block interfaces The printed arch failed at a surprisingly lowangle of 47 which corresponds to an effective thickness of 87or tR = 0130 most likely due to these factors

Gothic Arches Similar to the circular arch the naıve continuumFEM predicts failure as soon as the Gothic arch is tilted beyondthe horizontal rest position The FEM with inequality contact con-straints predicts a higher critical angle but slightly less than thatpredicted by the inverse FEM and the block equilibrium methodAn analytical ground-truth value is not known but we compare withresults from the physical test Similar to the circular arch example

Fig 9 The flying buttress at the critical tilt angle as predicted by (a) theinverse FEM and (b) the block equilibrium method (c) Continuum FEMand (d) inequality-constrained FEM The latter two methods are overcon-servative as validated by (e f) a physical tilt test

the printed structure fails at a critical angle slightly below that pre-dicted by the inverse FEM and equilibrium methods

The Three-Legged π As discussed in Section 5 the three-legged π is a good didactic example as it is a simple structure withstatic indeterminacy The printed structure fails at a critical anglefar above that predicted by the forward FEM methods but slightlybelow that predicted by the inverse FEM and equilibrium methods(see Figure 6)

Flying Buttress We compared critical angle predictions of thefour numerical methods against a 3D-printed flying buttress model(see Figure 9) As with the π model the inverse FEM and blockequilibrium predict the same critical angle which was close tothe experimental results As expected the naıve FEM is overlyconservative and predicts tension even at the rest position Moresurprisingly as in three-legged π FEM with inequality contactconstraints also fails significantly below the experimental or equi-libriuminverse FEM results We observe that small deformationscause hinges to appear even at the rest orientation and this geomet-ric nonlinearity in the topology of the contacts likely leads to earlyfailure



Fig 10 Tilt-test results for our aqueduct model similar in spirit to the Roman Aqueduct of Segovia (a) The stress distributiondisplacement at the restconfiguration predicted by naıve continuum FEM Colors indicate magnitude of stress (red is higher and blue is lower) and circled regions indicate areasof tension (b) Inequality-constrained forward FEM fails to find an equilibrium solution even at the rest position (c) The solved-for rest shape of the blockspredicted by the inverse FEM and (d) one possible set of contact forces output by the block equilibrium method at the critical angle (65) In (b) Abaqus fails to converge to a solution at 00 but here we show the deformation at an intermediate increment stage

Aqueduct Scanned geometry for the Aqueduct of Segovia isunfortunately not available so we analyzed the stability of a similartwo-tier aqueduct we modeled ourselves (see Figure 10) Similar tothe flying buttress the naıve FEM outputs tension immediately andthe forward FEM with inequality contact constraints also fails toconverge even at the rest position The aqueduct is made of numer-ous blocks that constitute a complex topology of interfaces exacer-bating the problem of geometric nonlinearity The inverse FEM andblock equilibrium methods predict the same critical angle of 65

8 DISCUSSION AND CONCLUSIONS

This article examines two different traditions of methods used forstatic analysis of masonry linear elastic analysis using FEM andequilibrium methods The former is widely employed by practition-ers across numerous fields of engineering but when naıvely appliedto masonry it gives incorrect results The latter approach is favoredby many researchers as a simple but more reliable alternative forpredicting whether or not a building stands After examining thesetwo classes of methods in detail we draw several conclusions

First linear elastic finite element analysis with continuum equal-ity constraints used to model interblock contacts (as is often done inpractice) is wholly inadequate for predicting stability as it is far tooconservativemdashit cannot differentiate between safe and unsafe hing-ing modes crucial to the analysis of all of the examples presentedpreviously

Replacing the contact constraints with one-sided constraints toallow hinging gives good results in some cases particularly where

the contact geometry of the blocks is simple (eg the arch andGothic arch) However in others replacing the contact constraintsalone still fails to give accurate results (the three-legged pi fly-ing buttress and aqueduct) This is because interblock contact isgeometrically nonlinear even geometrically discontinuous smalldeformations of the blocks can completely change the set of pos-sible hinging modes Elastic FEM resolves the problem of staticindeterminacy by way of choosing one particular material modelDifferent ways of resolving the indeterminacy would lead to differ-ent deformations of the block which leads to different admissiblehinging modes

Other culprits suggested in the literature for the error of forwardFEM have included an insufficiently sophisticated material modelthe extreme stiffness of stone size of the elements and so forthIt is true that an increasingly complex implementation of the FEMmust eventually converge to a correct analysis of stability Howeverthe results are sensitive to the choice of parameters (which are notstraightforward to tune even in commercial packages) and compu-tationally expensive to obtain On the other hand inverse FEM andequilibrium methods are reliable even with a simple material modeland elements as coarse as the structurersquos blocks By fixing the de-formed geometry and consequently also the contact geometry theinverse FEM and equilibrium methods both sidestep all of these dif-ficulties without incurring the complexity and computational costof FEM with sophisticated contact and material models

Indeed we have shown that linear elastic analysis using finiteelements with a minor modification to solve the inverse ratherthan the forward problem becomes a dual formulation of the block



equilibrium method and gives the same equally reliable resultsreconciling the two traditions The two classes of methods turn outto be two different ways of expressing an equivalent problem Thatbeing said the block equilibrium method is particularly elegantSince it is formulated using force variables it does not introducecounterintuitive (inverse) displacements and it also includes thecompression-only constraints directly (instead of formulating themindirectly in terms of displacements)

Limitations We have made several assumptions about the be-havior and modeling of masonry structures that could be relaxedto increase the sophistication of the stability analysis Perhaps thebiggest assumption we have made is that of infinite frictionmdashafully satisfactory theory combining equilibrium analysis with evena simple Coulomb friction model remains an elusive and extremelyimportant goal in computational architecture Infinitely thin hingesoccurring at fixed locations are also an idealistic assumption vi-olated in reality For example as we see in the arch and Gothicarch physical experiments blocksrsquo corners become rounded withwear which affects the location of the hinges and the stability ofthe structures We have also assumed that the blocks in the structurecan support arbitrarily high internal tensile stress This assumptionis reasonable for small blocks in simple contact configurations butnot for large horizontal slabs like the one in our three-legged πmodel Finally both the equilibrium and FEM results depend on thediscretization of a structure into blocks especially the configurationof interfaces Finding the optimal level of discretization between thetwo extremesmdashmodeling the structure as a monolith and modelingeach block individuallymdashis a difficult problem that must take intoaccount practical cost and desired accuracy

APPENDIX

Relationship to Thrust Network Analysis (TNA)

For a full explanation of Thrust Network Analysis we refer to Block[2009] Here we focus on illustrating the relationship between TNAand the block equilibrium method and how FEM has been usedwithin equilibrium analysis in the past

TNA Versus Block Equilibrium Method Every thrust networkin equilibrium corresponds to a set of blocks in frictionless equilib-rium (The reverse is also true but only if the topology of the set ofblocks is sufficiently simple ie surface-like) To see this take thethrust network and create a set of blocks so that (1) every vertex ofthe thrust network is the center of mass of one block (2) the mass ofeach block is equal to the mass of the vertex (3) two blocks meet atan interface if and only if the corresponding vertices share an edgeand (4) the interface between blocks is perpendicular to the thrustnetwork edge

One construction is to first compute the intrinsic dual of thethrust network mesh The dual cells will satisfy properties (3) and(4) automatically and can be ldquofattenedrdquo in the normal direction ofthe faces to create blocks Fattening arbitrarily will make the blockshave the wrong total mass and center of mass To fix this add orremove material from each block (while keeping it in touch with itsneighbors) until (1) and (2) are satisfied we can always do this bymaking some parts of the block thinner and other parts thicker Nowfor each edge eij of the thrust network with weight wij add a contactforce wijeij at the point where eij intersects the interface betweenblocks i and j (We can also distribute this force to the interfacecorners) Since this force is parallel to the torque arm it exerts notorque on block i or j Moreover the total force on the center ofmass of block i is Fg + jsimiwijeij where Fg is the gravitational

force and the sum is over neighboring blocks of i This is equal tozero since this is also the equation of thrust network equilibrium

So our proof in Section 6 also holds for thrust networks everythrust network in equilibrium corresponds to some set of blocks(constructed as described earlier) that are stable under inverse FEM

TNA Versus FEM Fraternali [2010] has shown that the edgeweights in a thrust network can be interpreted as gradient jumps in afinite element discretization of the polyhedral Airy stress functiondiscretized using piecewise linear elements However there aretwo key differences between this approach and the FEM (directstiffness method) we consider in our work (1) Analyzing stabilityvia the Airy stress function still falls into the equilibrium schoolof analysis since it treats the geometry of the structure as staticand then computes whether a stress function exists that satisfies theequilibrium There is no deformation involved While the unknownAiry stress function is found using FEM in the broader sense ofthe term (as a numerical method to solve arbitrary PDEs) this isdifferent from the direct stiffness method that we consider in ourwork (2) The work based on discretizing Airy stress analyzes onlyinfinitesimally thin shells and doesnrsquot apply to volumetric blocks

Linearization of Inverse FEM

In the linear formulations of elasticity used earlier for both forwardand inverse FEM we assume that the block deformation u = xaft minusxbfr is small (but not negligible) but that u2 is even smallerand can be neglected [Mase et al 2009] Here we show that K isthe correct linearization of the forces arising in the inverse FEMformulation described in Equation (17) We first review how theforces are linearized in forward FEM and then extend the analysisto inverse FEM

Consider the full nonlinear elastic energy U (xaft xbfr) and Taylorexpand it about xaft = xbfr

U(xbfr + u xbfr) = U(xbfr xbfr) + partU

partxaft(xbfr xbfr)u

+ 1

2uᵀ part2U

partx2aft

(xbfr xbfr)u + o(u3)(25)

The first two terms vanish since U is minimized when xaft = xbfrThe force expressed in displacement degrees of freedom is then

dU(xbfr + u xbfr)

du= part2U

partx2aft

(xbfr xbfr)u + o(u2) (26)

and K = part2Upartx2aft(xbfr xbfr) is constant with respect to xaft This

linearization is used in the elastic energy term in Equation (8)In the case of inverse FEM xbfr is the unknown variable We thus

take the Taylor expansion about xbfr = xaft

U(xaft xaft minus u) = U(xaft xaft) minus partU

partxbfr(xaft xaft)u

+ 1

2uᵀ part2U

partx2bfr

(xaft xaft)u + o(u3)(27)

Once again the first two terms are zero since U has a local minimumat xbfr = xaft The force with respect to displacement is

dU(xaft xaft minus u)

du= part2U

partx2bfr

(xaft xaft)u + o(u2) (28)

Finally we show that part2U

partx2bfr

(xaft xaft) = part2U

partx2aft

(xaft xaft) = K the

stiffness matrix in Equation (17) To see this notice that partU

partxaft(x x) is



zero (thus constant) for any x since U has a minimum at xbfr = xaftTherefore

0 = d

dxpartU

partxaft(x x) = part2U

partx2aft

(x x) + part2U

partxbfrpartxaft(x x) (29)

and by an identical argument

0 = d

dxpartU

partxbfr(x x) = part2U

partx2bfr

(x x) + part2U

partxaftpartxbfr(x x) (30)

Combining the previous two equations and using equality of mixedpartials yields

part2U

partx2bfr

(x x) = part2U

partx2aft

(x x) (31)

for any x (and in particular x = xaft)

Equivalence of the Inverse FEM and EquilibriumMethods (Proof Continued from Section 6)

Torques induced by the displacements solved for using the in-verse FEM also satisfy the constraints of the block equilibriummethod Consider a single block and its full nonlinear elastic energyU(xaft xbfr) where xaft and xbfr are expressed in coordinates withthe origin at the blockrsquos center of mass Let w be a unit vector and[w] be the cross-product matrix of w then the matrix Rw(t) = e[w]t

represents rotation about the center of mass by t radians counter-clockwise around the w axis Since elastic energy is invariant undertransformation of xaft by rigid motions

U(Rw(t)xaft xbfr) = U(xaft xbfr) (32)

for any t Differentiating with respect to t gives

0 = dRw

dt(t)

sum

i

partU

partxiaft

(Rw(t)xaft xbfr) (33)

Since dRwdt

(0) = [w] evaluating the previous at t = 0 gives

0 = [w]sum

i

partU

partxiaft

(xaft xbfr) (34)

or since fiin = minus partU

partxiaft

0 = [w]sum

i

fiin(xaft xbfr) (35)

Since w is arbitrary the full net nonlinear internal force exerts notorque on the block We can see that the same must hold for thelinearized forces Taylor expand U about xbfr to get

U(xbfr + u xbfr) =sum

i

partU

partxiaft

(xbfr xbfr)ui + o(u2) (36)

and

0 = minus[W]K(xbfr)u + o(u2) (37)

where u = xaftminusxbfr and [W] is a block-diagonal matrix with blocks[w] The equation holds for all u so differentiating and setting u = 0yields 0 = [W]K and so

0 = [W]Ku =sum

i

[w]fiin (38)

for any u where fiin are now linearized forces It follows from

Equation (19) that 0 = [w]sum

i(fiw + fi

ct) which is the torque balanceof the block

Experiment Parameters and Data

Table V describes the parameter setting for Abaqus simulationsand Table VI shows the complete results from the physical tilt test

Table V AbaqusStandard SettingsYoungrsquos modulus 10 times 106

Poissonrsquos ratio 02Density 100

Meshelement type CPS4RContact relationship Hard

Friction penaltylowast coeff = 100lowastWe experimented with larger values of coeff as well as theldquoRoughrdquo contact formulation We also experimented withdifferent increment sizes and toggling NLGEOM option Wereport the least conservative result

Table VI Physical Tilt-Test ResultsModel 1 2 3 AverageCube 413 408 409 410

Arch thick 48 48 45 47Gothic arch 158 157 155 1573-legged π 143 145 142 143

Flying buttress 52 43 38 44

In the naıve FEM we model a structure as a single continuouspart In contrast in the FEM with inequality contact constraintseach block composing a structure is modeled as a separate part andthe interaction between the blocks is defined through tangential andnormal contact properties a high coefficient of friction to preventsliding and a hard contact to disallow interpenetration but allowseparation

Scaling Factors for Figures

Table VII lists the scaling factors used in the figures to visualizedisplacements

Table VII Displacements Scale FactorTo aid visualization the displacements in thefigures of this article have been exaggeratedFigure Factor

6(c) 64 times 102

7(e) 75 times 102

7(f) 21 times 108(c) 10 times 102

8(d) 50 times 101

9(c) 51 times 102

9(d) 21 times 102

10(a) 80 times 102

10(b) 20 times 105

ACKNOWLEDGMENTS

We would like to thank William Plunkett for his assistance with thephysical experiments We are grateful to Sylvain Paris David LevinShinjiro Sueda Desai Chen and Danny Kaufman for their insightfulfeedback and discussions We also thank Zhandos Orazalin for hishelp with Abaqus software

REFERENCES

Autodesk 2014 Robot Structural Analysis Professional (November 2014)httpwwwautodeskcomproductsrobot-structural-analysisoverview



David Baraff 1993 Non-penetrating rigid body simulation In Eurographics93 State of the Art Report J Fagerberg D C Mowery and R R Nelson(Eds) Eurographics Association

Klaus-Jurgen Bathe 2006 Finite Element Procedures Klaus-Jurgen Bathe

Philippe Block Thierry Ciblac and John Ochsendorf 2006 Real-time limitanalysis of vaulted masonry buildings Computers amp Structures 84 29(2006) 1841ndash1852

Philippe Block and John Ochsendorf 2007 Thrust network analysis A newmethodology for three-dimensional equilibrium Journal of the Interna-tional Association for Shell and Spatial Structures 48 3 (2007) 167ndash173

Philippe Philippe Camille Vincent Block 2009 Thrust Network Anal-ysis Exploring Three-Dimensional Equilibrium PhD DissertationMassachusetts Institute of Technology

Sandro Brasile Raffaele Casciaro and Giovanni Formica 2007 Multilevelapproach for brick masonry wallsndashPart I A numerical strategy for thenonlinear analysis Computer Methods in Applied Mechanics and Engi-neering 196 49 (2007) 4934ndash4951

Jaime Cervera Bravo Jose Ignacio Hernando Garcia and Juan Francisco DeLa Torre Calvo 1994 Acueducto de Segovia Comportamiento MecanicoTechnical Report Universidad Politecnica de Madrid

Philip Brune and Renato Perucchio 2012 Roman concrete vaulting in thegreat hall of Trajans markets Structural evaluation Journal of Architec-tural Engineering 18 4 (2012) 332ndash340

Xiang Chen Changxi Zheng Weiwei Xu and Kun Zhou 2014 An asymp-totic numerical method for inverse elastic shape design ACM Transactionson Graphics (In Proceedings of SIGGRAPH 2014) 33 4 (Aug 2014)

Computers and Structures Inc 2014 SAP2000 (June 2014) httpwwwcsiamericacomproductssap2000

Stelian Coros Sebastian Martin Bernhard Thomaszewski ChristianSchumacher Robert Sumner and Markus Gross 2012 Deformable ob-jects alive ACM Transactions on Graphics 31 4 Article 69 (July 2012)9 pages DOIhttpdxdoiorg10114521855202185565

P A Cundall 1971 A computer model for simulating progressive large-scale movements in blocky rock systems In Proceedings of the Sympo-sium of the International Society of Rock Mechanics Vol 2 Nancy

Dassault Systeme 2014 Abaqus Unified FEA (June 2014) httpwww3dscomproducts-servicessimuliaportfolioabaqusoverview

Fernando de Goes Pierre Alliez Houman Owhadi Mathieu Desbrun andothers 2013 On the equilibrium of simplicial masonry structures ACMTransactions on Graphics 32 4 (2013) 931ndash9310

Matthew J DeJong 2009 Seismic Assessment Strategies for Masonry Struc-tures PhD Dissertation Massachusetts Institute of Technology Depart-ment of Architecture

Mario Deuss Daniele Panozzo Emily Whiting Yang Liu PhilippeBlock Olga Sorkine-Hornung and Mark Pauly 2014 Assembling self-supporting structures ACM Transactions on Graphics (SIGGRAPH ASIA)(2014)

John Dolbow and T Belytschko 1999 A finite element method for crackgrowth without remeshing International Journal of Numerical Methodsin Engineering 46 1 (1999) 131ndash150

Kenny Erleben 2007 Velocity-based shock propagation for multibodydynamics animation ACM Transactions on Graphics (TOG) 26 2(2007) 12

Fernando Fraternali 2010 A thrust network approach to the equilibriumproblem of unreinforced masonry vaults via polyhedral stress functionsMechanics Research Communications 37 2 (2010) 198ndash204

Eran Guendelman Robert Bridson and Ronald Fedkiw 2003 Nonconvexrigid bodies with stacking In ACM Transactions on Graphics (TOG)Vol 22 ACM 871ndash878

John C Hart Brent Baker and Jeyprakash Michaelraj 2003 Structuralsimulation of tree growth and response The Visual Computer 19 2ndash3(2003) 151ndash163

Jacques Heyman 1966 The stone skeleton International Journal of Solidsand Structures 2 (1966) 249ndash279

Jacques Heyman 1995 The Stone Skeleton Structural Engineering of Ma-sonry Architecture Cambridge University Press

Jacques Heyman 1996 Elements of the Theory of Structures CambridgeUniversity Press

Shu-Wei Hsu and John Keyser 2012 Automated constraint placementto maintain pile shape ACM Transactions on Graphics (TOG) 31 6(2012) 150

Santiago Huerta 2001 Mechanics of masonry vaults The equilibrium ap-proach Universidade do Minho (2001)

Santiago Huerta 2008 The analysis of masonry architecture A historicalapproach To the memory of professor Henry J Cowan ArchitecturalScience Review 51 4 (2008) 297ndash328

Danny M Kaufman Shinjiro Sueda Doug L James and Dinesh K Pai2008 Staggered projections for frictional contact in multibody systemsIn ACM Transactions on Graphics (TOG) Vol 27 ACM 164

Yang Liu Hao Pan John Snyder Wenping Wang and Baining Guo 2013Computing self-supporting surfaces by regular triangulation ACM Trans-actions on Graphics 32 4 (July 2013) 92

R K Livesley 1978 Limit analysis of structures formed from rigid blocksInternational Journal of Numerical Methods in Engineering 12 (1978)1853ndash1871

Paulo B Lourenco Jan G Rots and Johan Blaauwendraad 1998 Contin-uum model for masonry Parameter estimation and validation Journal ofStructural Engineering 124 6 (1998) 642ndash652

Robert Mark 1982 Experiments in Gothic Structure Vol 11 MIT PressCambridge MA

Sebastian Martin Peter Kaufmann Mario Botsch Eitan Grinspun andMarkus Gross 2010 Unified simulation of elastic rods shells and solidsACM Transactions on Graphics 29 4 (2010) 391ndash3910

G Thomas Mase Ronald E Smelser and George E Mase 2009 ContinuumMechanics for Engineers CRC Press

Matthias Muller Jos Stam Doug James and Nils Thurey 2008 Real timephysics Class notes In ACM SIGGRAPH 2008 classes ACM 88

Rahul Narain Abhinav Golas and Ming C Lin 2010 Free-flowing granularmaterials with two-way solid coupling In ACM Transactions on Graphics(TOG) Vol 29 ACM 173

John Ochsendorf 2002 Collapse of Masonry Structures PhD DissertationUniversity of Cambridge

Paul Painleve 1895 Sur les lois du frottement de glissement ComptesRendus Academie des Sciences Paris 121 112ndash115

Daniele Panozzo Philippe Block and Olga Sorkine-Hornung 2013 De-signing unreinforced masonry models ACM Transactions on Graphics(In Proceedings of ACM SIGGRAPH) 32 4 (2013) 911ndash9112

Romain Prevost Emily Whiting Sylvain Lefebvre and Olga Sorkine-Hornung 2013 Make it stand Balancing shapes for 3D fabrication ACMTransactions on Graphics (In Proceedings of ACM SIGGRAPH) 32 4(2013) 811ndash8110

Pere Roca Miguel Cervera Giuseppe Gariup and others 2010 Struc-tural analysis of masonry historical constructions Classical and advancedapproaches Archives of Computational Methods in Engineering 17 3(2010) 299ndash325

Xiaohan Shi Kun Zhou Yiying Tong Mathieu Desbrun Hujun Bao andBaining Guo 2007 Mesh puppetry Cascading optimization of meshdeformation with inverse kinematics ACM Transactions on Graphics 263 (2007) 81 httpdoiacmorg10114512763771276479



Eftychios Sifakis and Jernej Barbic 2012 FEM simulation of 3D deformablesolids A practitionerrsquos guide to theory discretization and model reductionIn ACM SIGGRAPH 2012 Courses ACM 20

Jeffrey Smith Jessica K Hodgins Irving Oppenheim and Andrew Witkin2002 Creating models of truss structures with optimization In Pro-ceedings of SIGGRAPH 2002 (Computer Graphics Proceedings AnnualConference Series) John Hughes (Ed) ACM ACM Press ACM SIG-GRAPH 295ndash301

Ondrej Stava Juraj Vanek Bedrich Benes Nathan Carr and Radomır Mech2012 Stress relief Improving structural strength of 3D printable objectsACM Transactions on Graphics 31 4 Article 48 (July 2012) 11 pagesDOIhttpdxdoiorg10114521855202185544

Demetri Terzopoulos and Kurt Fleischer 1988 Modeling inelastic de-formation Viscolelasticity plasticity fracture SIGGRAPH ComputerGraphics 22 4 (June 1988) 269ndash278 DOIhttpdxdoiorg101145378456378522

Demetri Terzopoulos John Platt Alan Barr and Kurt Fleischer 1987 Elas-tically deformable models SIGGRAPH Computer Graphics 21 4 (Aug1987) 205ndash214 DOIhttpdxdoiorg1011453740237427

Nobuyuki Umetani Takeo Igarashi and Niloy J Mitra 2012 Guided explo-ration of physically valid shapes for furniture design ACM Transactionson Graphics 31 4 (2012) 86

Nobuyuki Umetani Kenshi Takayama Jun Mitani and Takeo Igarashi2010 Responsive FEM for aiding interactive geometric modeling IEEEComputer Graphics and Applications 99 (2010) 43ndash53

E Vouga Mathias Hobinger Johannes Wallner and Helmut Pottmann 2012Design of self-supporting surfaces ACM Transactions on Graphics 31 3(2012) 871ndash8711

Emily Whiting John Ochsendorf and Fredo Durand 2009 Proceduralmodeling of structurally-sound masonry buildings ACM Transactionson Graphics 28 5 (2009) 112

Emily Whiting Hijung Shin Robert Wang John Ochsendorf and FredoDurand 2012 Structural optimization of 3D masonry buildings ACMTransactions on Graphics 31 6 (2012) 1591ndash15911

Emily Jing Wei Whiting 2012 Design of Structurally-Sound MasonryBuildings Using 3D Static Analysis PhD Dissertation MassachusettsInstitute of Technology

Weiwei Xu Jun Wang KangKang Yin Kun Zhou Michiel van de PanneFalai Chen and Baining Guo 2009 Joint-aware manipulation of de-formable models ACM Transactions on Graphics 28 3 Article 35 (July2009) 9 pages DOIhttpdxdoiorg10114515313261531341

Jennifer Furstenau Zessin 2012 Collapse Analysis of Unreinforced Ma-sonry Domes and Curving Walls PhD Dissertation Massachusetts In-stitute of Technology

Qingnan Zhou Julian Panetta and Denis Zorin 2013 Worst-case structuralanalysis ACM Transactions on Graphics 32 4 Article 137 (July 2013)12 pages DOIhttpdxdoiorg10114524619122461967

Received November 2014 revised August 2015 accepted October 2015



Fig 1 The Aqueduct of Segovia a masonry structure that has been standing since antiquity (top right) Finite element analysis of a section of the aqueductdoes not conclusively demonstrate its stability (top left image from Bravo et al [1994]) Similarly linear elastic FEM fails to differentiate between thestability of an infeasible arch that is too thin to stand on its own (bottom leftmost) and a thicker feasible arch (bottom left) (image from Block et al [2006])In contrast equilibrium methods correctly predict that the thin arch is infeasible (bottom right) and that the thick arch is feasible (bottom rightmost)

partly due to the lack of alternatives in commercial software Onthe other hand in the broader context FEM has been tremendouslysuccessful at numerically simulating a variety of physical systems inengineering biology mechanics and medicine So it seems plausi-ble that with proper assumptions and material models FEM shouldbe able to correctly analyze masonry structures Is there a better wayto apply FEM to masonry structures Certainly we would expecta full dynamics simulation of the masonry using complex nonlin-ear material and contact models and very fine spatial and temporaldiscretization to correctly predict the behavior of masonry albeitat a tremendous computational cost What are the necessary mod-ifications to continuum linearized FEM to obtain accurate resultsfor masonry In other words what is its primary source of inaccu-racy Several possible explanations are cited but to our knowledgeno systematic examination exists for which of these are importantsources of error and which are not

mdashIt is possible for masonry structures to hinge while still remainingstable Modeling the entire masonry as a single continuous struc-ture does not reflect the contact conditions at the block interfaces

mdashSimulating accurate stresses in the structure requires choosinga correct constitutive model for the stone blocks The simplestchoice of constitutive modelmdasha linear stress-strain relationshipand a linearized strain measuremdashis not correct for stone

mdashSimilarly stone is extremely stiff and its high Youngrsquos mod-ulus potentially introduces numerical stability issues duringsimulation

mdashMost analysis assumes that the shape and positions of the blocksspecified as input by the user are initial conditions and thensolve for the final deformation of the structure In other wordsstability is analyzed using a forward simulation The final de-formed geometry is not the same as the input geometry This canbe a problem when we are interested in the stability of the inputstructure as is without further deformation

Table I Comparison of the FEM VersusEquilibrium Methods

Method FEM EquilibriumSolution variable Deformation Force

Material Elastic RigidGeometry Deforms Fixed

mdashThe expected stresses in a building are impossible to determinefrom the geometry alone due to static indeterminacy Manypossible sets of internal stresses are consistent with the observedstate of the building and slight changes in the environment cancause large changes in measured stresses An everyday exampleof this indeterminacy is a four-legged stool resting on a rigid levelfloor the stool is stable with only three of the four legs carryingthe load and which legs carry how much load can change dras-tically for even infinitesimal changes in the lengths of the legsFinite element analysis necessarily solves for only one possibleset of equilibrium stresses and this solution is highly sensitive toinitial conditions

Equilibrium methods have been proposed as alternatives to finiteelement analysis to sidestep the aforementioned challenges Thisline of attack can be traced back to the seminal work of Heyman[1966] Equilibrium methods also called limit analysis recentlybecame popular in the graphics community as an efficient wayto guarantee stability of procedurally generated or interactivelydesigned architecture [Whiting et al 2009 Vouga et al 2012Panozzo et al 2013] The key idea in these methods is to ignorethe elastic deformation of the stone entirely and instead to treat thestructure as piecewise rigid The unknowns in these methods are theforces acting on these rigid pieces rather than their displacementsTable I summarizes the key differences between finite-element- andequilibrium-based methods (Equilibrium methods and FEM aresurveyed in more detail in Sections 3 and 4 respectively) Generally





2 RELATED WORK






















lowast























w)sum

iisinb

fict + fb

w = fbct + fb

w = 0 forall b (1)

sum

iisinb

(fict times ri

ct

) + (fbw times rb

w

) = 0 forall b (2)








ct


ct middot ni | (4)












xj


(x

j

bfr = xkbfr

) (6)














w (xaft minus xbfr)

U(xbfr xaft) = 1

2(xaft minus xbfr)




xlowastaft = argmin

xaft

U(xaft xbfr)



(xj

aft minus xkaft


(xj

bfr = xkbfr

) (10)












dL


w + Aᵀλ (12)


to zero


w + Aᵀλ = fn

in + fnw + fn

ct = 0 (13)






















xaft = argminxaft

U(xaft xbfr)








dL

dxaft

∣∣∣∣xaft


Nλ

= fnin + fn

w + fnct

= 0

(17)




[xlowastbfr λ


H(xbfr λ)


Nλ

(18)







iisinb

(fiin + fi

w + fict

) = 0 forallb (19)


iisinb



iisinb

(f iw + fi

ct

) = 0 forallb (21)



Let fiin = minusfi



fiin + fi








in

= t middot fbin + [

ω times (xb

aft minus cbaft





[ω times (

xbaft minus cb

aft

)] middot fbin = [(

xbaft minus cb

aft

) times fbin

] middot ω = 0

(24)





















































APPENDIX













+ 1

2uᵀ part2U

partx2aft



dU(xbfr + u xbfr)

du= part2U

partx2aft







+ 1

2uᵀ part2U

partx2bfr




du= part2U

partx2bfr



partx2bfr


partx2aft

(xaft xaft) = K the


partxaft(x x) is




0 = d

dxpartU


partx2aft

(x x) + part2U



0 = d

dxpartU


partx2bfr

(x x) + part2U



part2U

partx2bfr

(x x) = part2U

partx2aft

(x x) (31)







0 = dRw

dt(t)

sum

i

partU

partxiaft


Since dRwdt


0 = [w]sum

i

partU

partxiaft

(xaft xbfr) (34)


partxiaft

0 = [w]sum

i




i

partU

partxiaft


and



0 = [W]Ku =sum

i

[w]fiin (38)



i(fiw + fi















6(c) 64 times 102

7(e) 75 times 102

7(f) 21 times 108(c) 10 times 102

8(d) 50 times 101

9(c) 51 times 102

9(d) 21 times 102

10(a) 80 times 102

10(b) 20 times 105

ACKNOWLEDGMENTS


REFERENCES



































































2 RELATED WORK






















lowast























w)sum

iisinb

fict + fb

w = fbct + fb

w = 0 forall b (1)

sum

iisinb

(fict times ri

ct

) + (fbw times rb

w

) = 0 forall b (2)








ct


ct middot ni | (4)












xj


(x

j

bfr = xkbfr

) (6)














w (xaft minus xbfr)

U(xbfr xaft) = 1

2(xaft minus xbfr)




xlowastaft = argmin

xaft

U(xaft xbfr)



(xj

aft minus xkaft


(xj

bfr = xkbfr

) (10)












dL


w + Aᵀλ (12)


to zero


w + Aᵀλ = fn

in + fnw + fn

ct = 0 (13)






















xaft = argminxaft

U(xaft xbfr)








dL

dxaft

∣∣∣∣xaft


Nλ

= fnin + fn

w + fnct

= 0

(17)




[xlowastbfr λ


H(xbfr λ)


Nλ

(18)







iisinb

(fiin + fi

w + fict

) = 0 forallb (19)


iisinb



iisinb

(f iw + fi

ct

) = 0 forallb (21)



Let fiin = minusfi



fiin + fi








in

= t middot fbin + [

ω times (xb

aft minus cbaft





[ω times (

xbaft minus cb

aft

)] middot fbin = [(

xbaft minus cb

aft

) times fbin

] middot ω = 0

(24)





















































APPENDIX













+ 1

2uᵀ part2U

partx2aft



dU(xbfr + u xbfr)

du= part2U

partx2aft







+ 1

2uᵀ part2U

partx2bfr




du= part2U

partx2bfr



partx2bfr


partx2aft

(xaft xaft) = K the


partxaft(x x) is




0 = d

dxpartU


partx2aft

(x x) + part2U



0 = d

dxpartU


partx2bfr

(x x) + part2U



part2U

partx2bfr

(x x) = part2U

partx2aft

(x x) (31)







0 = dRw

dt(t)

sum

i

partU

partxiaft


Since dRwdt


0 = [w]sum

i

partU

partxiaft

(xaft xbfr) (34)


partxiaft

0 = [w]sum

i




i

partU

partxiaft


and



0 = [W]Ku =sum

i

[w]fiin (38)



i(fiw + fi















6(c) 64 times 102

7(e) 75 times 102

7(f) 21 times 108(c) 10 times 102

8(d) 50 times 101

9(c) 51 times 102

9(d) 21 times 102

10(a) 80 times 102

10(b) 20 times 105

ACKNOWLEDGMENTS


REFERENCES











































































lowast























w)sum

iisinb

fict + fb

w = fbct + fb

w = 0 forall b (1)

sum

iisinb

(fict times ri

ct

) + (fbw times rb

w

) = 0 forall b (2)








ct


ct middot ni | (4)












xj


(x

j

bfr = xkbfr

) (6)














w (xaft minus xbfr)

U(xbfr xaft) = 1

2(xaft minus xbfr)




xlowastaft = argmin

xaft

U(xaft xbfr)



(xj

aft minus xkaft


(xj

bfr = xkbfr

) (10)












dL


w + Aᵀλ (12)


to zero


w + Aᵀλ = fn

in + fnw + fn

ct = 0 (13)






















xaft = argminxaft

U(xaft xbfr)








dL

dxaft

∣∣∣∣xaft


Nλ

= fnin + fn

w + fnct

= 0

(17)




[xlowastbfr λ


H(xbfr λ)


Nλ

(18)







iisinb

(fiin + fi

w + fict

) = 0 forallb (19)


iisinb



iisinb

(f iw + fi

ct

) = 0 forallb (21)



Let fiin = minusfi



fiin + fi








in

= t middot fbin + [

ω times (xb

aft minus cbaft





[ω times (

xbaft minus cb

aft

)] middot fbin = [(

xbaft minus cb

aft

) times fbin

] middot ω = 0

(24)





















































APPENDIX













+ 1

2uᵀ part2U

partx2aft



dU(xbfr + u xbfr)

du= part2U

partx2aft







+ 1

2uᵀ part2U

partx2bfr




du= part2U

partx2bfr



partx2bfr


partx2aft

(xaft xaft) = K the


partxaft(x x) is




0 = d

dxpartU


partx2aft

(x x) + part2U



0 = d

dxpartU


partx2bfr

(x x) + part2U



part2U

partx2bfr

(x x) = part2U

partx2aft

(x x) (31)







0 = dRw

dt(t)

sum

i

partU

partxiaft


Since dRwdt


0 = [w]sum

i

partU

partxiaft

(xaft xbfr) (34)


partxiaft

0 = [w]sum

i




i

partU

partxiaft


and



0 = [W]Ku =sum

i

[w]fiin (38)



i(fiw + fi















6(c) 64 times 102

7(e) 75 times 102

7(f) 21 times 108(c) 10 times 102

8(d) 50 times 101

9(c) 51 times 102

9(d) 21 times 102

10(a) 80 times 102

10(b) 20 times 105

ACKNOWLEDGMENTS


REFERENCES





































































w)sum

iisinb

fict + fb

w = fbct + fb

w = 0 forall b (1)

sum

iisinb

(fict times ri

ct

) + (fbw times rb

w

) = 0 forall b (2)








ct


ct middot ni | (4)












xj


(x

j

bfr = xkbfr

) (6)














w (xaft minus xbfr)

U(xbfr xaft) = 1

2(xaft minus xbfr)




xlowastaft = argmin

xaft

U(xaft xbfr)



(xj

aft minus xkaft


(xj

bfr = xkbfr

) (10)












dL


w + Aᵀλ (12)


to zero


w + Aᵀλ = fn

in + fnw + fn

ct = 0 (13)






















xaft = argminxaft

U(xaft xbfr)








dL

dxaft

∣∣∣∣xaft


Nλ

= fnin + fn

w + fnct

= 0

(17)




[xlowastbfr λ


H(xbfr λ)


Nλ

(18)







iisinb

(fiin + fi

w + fict

) = 0 forallb (19)


iisinb



iisinb

(f iw + fi

ct

) = 0 forallb (21)



Let fiin = minusfi



fiin + fi








in

= t middot fbin + [

ω times (xb

aft minus cbaft





[ω times (

xbaft minus cb

aft

)] middot fbin = [(

xbaft minus cb

aft

) times fbin

] middot ω = 0

(24)





















































APPENDIX













+ 1

2uᵀ part2U

partx2aft



dU(xbfr + u xbfr)

du= part2U

partx2aft







+ 1

2uᵀ part2U

partx2bfr




du= part2U

partx2bfr



partx2bfr


partx2aft

(xaft xaft) = K the


partxaft(x x) is




0 = d

dxpartU


partx2aft

(x x) + part2U



0 = d

dxpartU


partx2bfr

(x x) + part2U



part2U

partx2bfr

(x x) = part2U

partx2aft

(x x) (31)







0 = dRw

dt(t)

sum

i

partU

partxiaft


Since dRwdt


0 = [w]sum

i

partU

partxiaft

(xaft xbfr) (34)


partxiaft

0 = [w]sum

i




i

partU

partxiaft


and



0 = [W]Ku =sum

i

[w]fiin (38)



i(fiw + fi















6(c) 64 times 102

7(e) 75 times 102

7(f) 21 times 108(c) 10 times 102

8(d) 50 times 101

9(c) 51 times 102

9(d) 21 times 102

10(a) 80 times 102

10(b) 20 times 105

ACKNOWLEDGMENTS


REFERENCES












































































w (xaft minus xbfr)

U(xbfr xaft) = 1

2(xaft minus xbfr)




xlowastaft = argmin

xaft

U(xaft xbfr)



(xj

aft minus xkaft


(xj

bfr = xkbfr

) (10)












dL


w + Aᵀλ (12)


to zero


w + Aᵀλ = fn

in + fnw + fn

ct = 0 (13)






















xaft = argminxaft

U(xaft xbfr)








dL

dxaft

∣∣∣∣xaft


Nλ

= fnin + fn

w + fnct

= 0

(17)




[xlowastbfr λ


H(xbfr λ)


Nλ

(18)







iisinb

(fiin + fi

w + fict

) = 0 forallb (19)


iisinb



iisinb

(f iw + fi

ct

) = 0 forallb (21)



Let fiin = minusfi



fiin + fi








in

= t middot fbin + [

ω times (xb

aft minus cbaft





[ω times (

xbaft minus cb

aft

)] middot fbin = [(

xbaft minus cb

aft

) times fbin

] middot ω = 0

(24)





















































APPENDIX













+ 1

2uᵀ part2U

partx2aft



dU(xbfr + u xbfr)

du= part2U

partx2aft







+ 1

2uᵀ part2U

partx2bfr




du= part2U

partx2bfr



partx2bfr


partx2aft

(xaft xaft) = K the


partxaft(x x) is




0 = d

dxpartU


partx2aft

(x x) + part2U



0 = d

dxpartU


partx2bfr

(x x) + part2U



part2U

partx2bfr

(x x) = part2U

partx2aft

(x x) (31)







0 = dRw

dt(t)

sum

i

partU

partxiaft


Since dRwdt


0 = [w]sum

i

partU

partxiaft

(xaft xbfr) (34)


partxiaft

0 = [w]sum

i




i

partU

partxiaft


and



0 = [W]Ku =sum

i

[w]fiin (38)



i(fiw + fi















6(c) 64 times 102

7(e) 75 times 102

7(f) 21 times 108(c) 10 times 102

8(d) 50 times 101

9(c) 51 times 102

9(d) 21 times 102

10(a) 80 times 102

10(b) 20 times 105

ACKNOWLEDGMENTS


REFERENCES









































































dL


w + Aᵀλ (12)


to zero


w + Aᵀλ = fn

in + fnw + fn

ct = 0 (13)






















xaft = argminxaft

U(xaft xbfr)








dL

dxaft

∣∣∣∣xaft


Nλ

= fnin + fn

w + fnct

= 0

(17)




[xlowastbfr λ


H(xbfr λ)


Nλ

(18)







iisinb

(fiin + fi

w + fict

) = 0 forallb (19)


iisinb



iisinb

(f iw + fi

ct

) = 0 forallb (21)



Let fiin = minusfi



fiin + fi








in

= t middot fbin + [

ω times (xb

aft minus cbaft





[ω times (

xbaft minus cb

aft

)] middot fbin = [(

xbaft minus cb

aft

) times fbin

] middot ω = 0

(24)





















































APPENDIX













+ 1

2uᵀ part2U

partx2aft



dU(xbfr + u xbfr)

du= part2U

partx2aft







+ 1

2uᵀ part2U

partx2bfr




du= part2U

partx2bfr



partx2bfr


partx2aft

(xaft xaft) = K the


partxaft(x x) is




0 = d

dxpartU


partx2aft

(x x) + part2U



0 = d

dxpartU


partx2bfr

(x x) + part2U



part2U

partx2bfr

(x x) = part2U

partx2aft

(x x) (31)







0 = dRw

dt(t)

sum

i

partU

partxiaft


Since dRwdt


0 = [w]sum

i

partU

partxiaft

(xaft xbfr) (34)


partxiaft

0 = [w]sum

i




i

partU

partxiaft


and



0 = [W]Ku =sum

i

[w]fiin (38)



i(fiw + fi















6(c) 64 times 102

7(e) 75 times 102

7(f) 21 times 108(c) 10 times 102

8(d) 50 times 101

9(c) 51 times 102

9(d) 21 times 102

10(a) 80 times 102

10(b) 20 times 105

ACKNOWLEDGMENTS


REFERENCES














































































xaft = argminxaft

U(xaft xbfr)








dL

dxaft

∣∣∣∣xaft


Nλ

= fnin + fn

w + fnct

= 0

(17)




[xlowastbfr λ


H(xbfr λ)


Nλ

(18)







iisinb

(fiin + fi

w + fict

) = 0 forallb (19)


iisinb



iisinb

(f iw + fi

ct

) = 0 forallb (21)



Let fiin = minusfi



fiin + fi








in

= t middot fbin + [

ω times (xb

aft minus cbaft





[ω times (

xbaft minus cb

aft

)] middot fbin = [(

xbaft minus cb

aft

) times fbin

] middot ω = 0

(24)





















































APPENDIX













+ 1

2uᵀ part2U

partx2aft



dU(xbfr + u xbfr)

du= part2U

partx2aft







+ 1

2uᵀ part2U

partx2bfr




du= part2U

partx2bfr



partx2bfr


partx2aft

(xaft xaft) = K the


partxaft(x x) is




0 = d

dxpartU


partx2aft

(x x) + part2U



0 = d

dxpartU


partx2bfr

(x x) + part2U



part2U

partx2bfr

(x x) = part2U

partx2aft

(x x) (31)







0 = dRw

dt(t)

sum

i

partU

partxiaft


Since dRwdt


0 = [w]sum

i

partU

partxiaft

(xaft xbfr) (34)


partxiaft

0 = [w]sum

i




i

partU

partxiaft


and



0 = [W]Ku =sum

i

[w]fiin (38)



i(fiw + fi















6(c) 64 times 102

7(e) 75 times 102

7(f) 21 times 108(c) 10 times 102

8(d) 50 times 101

9(c) 51 times 102

9(d) 21 times 102

10(a) 80 times 102

10(b) 20 times 105

ACKNOWLEDGMENTS


REFERENCES





































































iisinb

(fiin + fi

w + fict

) = 0 forallb (19)


iisinb



iisinb

(f iw + fi

ct

) = 0 forallb (21)



Let fiin = minusfi



fiin + fi








in

= t middot fbin + [

ω times (xb

aft minus cbaft





[ω times (

xbaft minus cb

aft

)] middot fbin = [(

xbaft minus cb

aft

) times fbin

] middot ω = 0

(24)





















































APPENDIX













+ 1

2uᵀ part2U

partx2aft



dU(xbfr + u xbfr)

du= part2U

partx2aft







+ 1

2uᵀ part2U

partx2bfr




du= part2U

partx2bfr



partx2bfr


partx2aft

(xaft xaft) = K the


partxaft(x x) is




0 = d

dxpartU


partx2aft

(x x) + part2U



0 = d

dxpartU


partx2bfr

(x x) + part2U



part2U

partx2bfr

(x x) = part2U

partx2aft

(x x) (31)







0 = dRw

dt(t)

sum

i

partU

partxiaft


Since dRwdt


0 = [w]sum

i

partU

partxiaft

(xaft xbfr) (34)


partxiaft

0 = [w]sum

i




i

partU

partxiaft


and



0 = [W]Ku =sum

i

[w]fiin (38)



i(fiw + fi















6(c) 64 times 102

7(e) 75 times 102

7(f) 21 times 108(c) 10 times 102

8(d) 50 times 101

9(c) 51 times 102

9(d) 21 times 102

10(a) 80 times 102

10(b) 20 times 105

ACKNOWLEDGMENTS


REFERENCES






































































































APPENDIX













+ 1

2uᵀ part2U

partx2aft



dU(xbfr + u xbfr)

du= part2U

partx2aft







+ 1

2uᵀ part2U

partx2bfr




du= part2U

partx2bfr



partx2bfr


partx2aft

(xaft xaft) = K the


partxaft(x x) is




0 = d

dxpartU


partx2aft

(x x) + part2U



0 = d

dxpartU


partx2bfr

(x x) + part2U



part2U

partx2bfr

(x x) = part2U

partx2aft

(x x) (31)







0 = dRw

dt(t)

sum

i

partU

partxiaft


Since dRwdt


0 = [w]sum

i

partU

partxiaft

(xaft xbfr) (34)


partxiaft

0 = [w]sum

i




i

partU

partxiaft


and



0 = [W]Ku =sum

i

[w]fiin (38)



i(fiw + fi















6(c) 64 times 102

7(e) 75 times 102

7(f) 21 times 108(c) 10 times 102

8(d) 50 times 101

9(c) 51 times 102

9(d) 21 times 102

10(a) 80 times 102

10(b) 20 times 105

ACKNOWLEDGMENTS


REFERENCES























































































APPENDIX













+ 1

2uᵀ part2U

partx2aft



dU(xbfr + u xbfr)

du= part2U

partx2aft







+ 1

2uᵀ part2U

partx2bfr




du= part2U

partx2bfr



partx2bfr


partx2aft

(xaft xaft) = K the


partxaft(x x) is




0 = d

dxpartU


partx2aft

(x x) + part2U



0 = d

dxpartU


partx2bfr

(x x) + part2U



part2U

partx2bfr

(x x) = part2U

partx2aft

(x x) (31)







0 = dRw

dt(t)

sum

i

partU

partxiaft


Since dRwdt


0 = [w]sum

i

partU

partxiaft

(xaft xbfr) (34)


partxiaft

0 = [w]sum

i




i

partU

partxiaft


and



0 = [W]Ku =sum

i

[w]fiin (38)



i(fiw + fi















6(c) 64 times 102

7(e) 75 times 102

7(f) 21 times 108(c) 10 times 102

8(d) 50 times 101

9(c) 51 times 102

9(d) 21 times 102

10(a) 80 times 102

10(b) 20 times 105

ACKNOWLEDGMENTS


REFERENCES














































































APPENDIX













+ 1

2uᵀ part2U

partx2aft



dU(xbfr + u xbfr)

du= part2U

partx2aft







+ 1

2uᵀ part2U

partx2bfr




du= part2U

partx2bfr



partx2bfr


partx2aft

(xaft xaft) = K the


partxaft(x x) is




0 = d

dxpartU


partx2aft

(x x) + part2U



0 = d

dxpartU


partx2bfr

(x x) + part2U



part2U

partx2bfr

(x x) = part2U

partx2aft

(x x) (31)







0 = dRw

dt(t)

sum

i

partU

partxiaft


Since dRwdt


0 = [w]sum

i

partU

partxiaft

(xaft xbfr) (34)


partxiaft

0 = [w]sum

i




i

partU

partxiaft


and



0 = [W]Ku =sum

i

[w]fiin (38)



i(fiw + fi















6(c) 64 times 102

7(e) 75 times 102

7(f) 21 times 108(c) 10 times 102

8(d) 50 times 101

9(c) 51 times 102

9(d) 21 times 102

10(a) 80 times 102

10(b) 20 times 105

ACKNOWLEDGMENTS


REFERENCES



































































APPENDIX













+ 1

2uᵀ part2U

partx2aft



dU(xbfr + u xbfr)

du= part2U

partx2aft







+ 1

2uᵀ part2U

partx2bfr




du= part2U

partx2bfr



partx2bfr


partx2aft

(xaft xaft) = K the


partxaft(x x) is




0 = d

dxpartU


partx2aft

(x x) + part2U



0 = d

dxpartU


partx2bfr

(x x) + part2U



part2U

partx2bfr

(x x) = part2U

partx2aft

(x x) (31)







0 = dRw

dt(t)

sum

i

partU

partxiaft


Since dRwdt


0 = [w]sum

i

partU

partxiaft

(xaft xbfr) (34)


partxiaft

0 = [w]sum

i




i

partU

partxiaft


and



0 = [W]Ku =sum

i

[w]fiin (38)



i(fiw + fi















6(c) 64 times 102

7(e) 75 times 102

7(f) 21 times 108(c) 10 times 102

8(d) 50 times 101

9(c) 51 times 102

9(d) 21 times 102

10(a) 80 times 102

10(b) 20 times 105

ACKNOWLEDGMENTS


REFERENCES


































































0 = d

dxpartU


partx2aft

(x x) + part2U



0 = d

dxpartU


partx2bfr

(x x) + part2U



part2U

partx2bfr

(x x) = part2U

partx2aft

(x x) (31)







0 = dRw

dt(t)

sum

i

partU

partxiaft


Since dRwdt


0 = [w]sum

i

partU

partxiaft

(xaft xbfr) (34)


partxiaft

0 = [w]sum

i




i

partU

partxiaft


and



0 = [W]Ku =sum

i

[w]fiin (38)



i(fiw + fi















6(c) 64 times 102

7(e) 75 times 102

7(f) 21 times 108(c) 10 times 102

8(d) 50 times 101

9(c) 51 times 102

9(d) 21 times 102

10(a) 80 times 102

10(b) 20 times 105

ACKNOWLEDGMENTS


REFERENCES














































































































































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Reconciling Elastic and Equilibrium Methods for Static ...people.csail.mit.edu/hishin/projects/fem_vs_equilibrium/paper.pdf · Reconciling Elastic and Equilibrium Methods for Static