Finite Elements from the early beginning to the very end

Transcription

1 Finit Elmnts from th arly bginning to th vry nd A(x), E(x) g b(x) h x =. x = L An Introduction to Elasticity and Hat Transfr Applications x Prliminary dition LiU-IEI-S--8/535--SE Bo Torstnflt

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3 Contnts 1 Prlud Background Th Big Pictur I Linar Static Elasticity 5 2 Introduction 7 3 Bars Th Bar Displacmnt Assumption Th Local Equations A Strong Formulation A Wak Formulation A Galrkin Formulation A Matrix Formulation A 2-Nod Elmnt Stiffnss Matrix A 3-Nod Elmnt Stiffnss Matrix An Elmnt Load Vctor Th Assmbly Opration Strss and Strain Calculations Multi-dimnsional Truss Fram Works Numrical Exampls A 1D bar problm A 2D truss problm A 3D truss problm Common Pitfalls and Mistaks Summary i

4 ii CONTENTS 4 Bams Th Bam Displacmnt Assumption Th Local Equations A Strong Formulation A Wak Formulation A Galrkin Formulation A Matrix Formulation A 2D 2-nod Bam Elmnt An Elmnt Load Vctor A 2D Bam Elmnt with Axial Stiffnss A 3D Spac Fram Elmnt Strss and Strain Calculations Numrical Exampls A 2D consol bam Summary Solids Displacmnt Assumptions Th 2D Mmbran Displacmnt Assumption Th Axisymmtric Displacmnt Assumption Th 3D Displacmnt Assumption Unknown Strss Componnts Th Local Equations Th Balanc law Th Constitutiv rlation Th Compatibility rlation A Summary of Local Equations A Strong Formulation A Wak Formulation Th Galrkin Formulation Th Matrix Problm Th Assmbly Opration Th Elmnt Load vctor Th 2D Constant Strain Triangl Th Elmnt Stiffnss matrix A Numrical Exampl Four-nod Rctangular Alignd Elmnts A Numrical Exampl Isoparamtric 2D Elmnts A 2D 4-nod Quadrilatral Elmnt A Numrical Exampl Numrical Intgration Th Numrical Work Flow D 8- and 9-nod Quadrilatral Elmnts

5 CONTENTS iii Sub- or Hyprparamtric Elmnt Formulations Isoparamtric 3D Elmnts Isoparamtric Axisymmtric Elmnts Distributd Loads Lin Loads Surfac Loads Volum Loads Raction Forcs Strss Evaluation An Elmnt Library D and Axisymmtric lmnts D lmnts Summary II Linar Hat Transfr D Stady-stat Hat Transfr Th Local Equations A Strong Formulation A Wak Formulation A Galrkin Formulation A Matrix Formulation Multi-dim Transint Hat Transfr Local Equations Th Balanc Law Constitutiv Rlation Strong Formulation Wak Formulation Galrkin Formulation Matrix Formulation

6 iv CONTENTS

7 Prfac Th writing of this book has arisn as a natural nxt stp in my profssion as a tachr, rsarchr, program dvlopr, and usr of th Finit Elmnt Mthod. Of cours on can wondr, why I am writing just anothr book in Finit Elmnts. Th answr is qually obvious as simpl. Aftr many yars in th fild I hav, as hav many othrs, discovrd a larg varity of pitfalls or mistak don by othrs and myslf. I hav now rachd a point whr I would lik to dscrib my viw of th topic. That is, how to undrstand it, how to tach it, how to implmnt it and how to us it; ths will b main goals for th discussion to com! Th discussion to com will b influncd by xprincs from all four of ths branchs. As a tachr I hav taught both basic and advancd courss in Finit Elmnts with focus on both Solid Mchanics and Hat Transfr applications. Ths courss hav bn givn at th Linköping Univrsity in th mchanical nginring programm. Th txt to com is writtn by an nginr for nginrs. On ovrall goal for th dscription is to try to covr vry stp from how a crtain mathmatical modl appars from basic considrations basd on fact from rality, to a classical formulation with its possibl analytical solutions, and finally ovr to a study of numrical solutions usd by a Finit Elmnt program basd on a crtain finit lmnt formulation. That is, discussing th finit lmnt mthod from th arly bginning to th vry nd. Th grat challng is to mak this as short and intrsting as possibl without loosing or braking th mathmatical chain. It is strongly blivd that for succss in larning Finit Elmnts it is an absolut prrquisit to b familiar with th local quations and thir availabl analytical solutions. I think most popl who hav trid to tach Finit Elmnts agr upon this, traditionally howvr, most ducation in Finit Elmnts is givn in sparat courss. Why not try to tach Finit Elmnts in clos connction to whr th basic matrial is taught. That is, intgrat Finit Elmnts togthr with basic matrial in th v

8 sam cours! Of cours, Finit Elmnts can b taught as a wightd rsidual mthod for approximat solutions of sts of coupld partial diffrntial quations without discussing any physical application and just focusing on xistnc, uniqunss and rror bounds of th solution. This is of cours also important but for most studnts studying diffrnt nginring disciplins Finit Elmnts will b a tool for trying to undrstand and prdict th bhavior of rality. An important focus in studis of finit lmnt formulations of diffrnt nginring disciplins is to b awar what should b xpctd of th quality of th approximat numrical solution. This can only b larnt by knowing th most important dtails of th mathmatical background and thn solving numrical problms having an analytical solution to compar with. Anothr aspct important for nginrs working with th mthod as a daily tool, is how to us th mthod as fficintly as possibl both from a timconsuming and a computr rsourcs point of viw. Ovr a priod of at last 15 yars I hav workd with th graphical finit lmnt nvironmnt TRINITAS. This is a stand-alon tool for optimization, concptual dsign and ducation as wll as for gnral linar lasticity and hat transfr problm both as stady-stat, transint or as ignvalu problms. It is an Objct Orintd program basd on a graphical usr intrfac for manipulation of th databas of th program. This program contains procdurs for gomtry modling, domain proprty and boundary conditions dfinition, msh gnration, finit lmnt analysis and rsult valuation. Th program is usd in duction at diffrnt lvls. It is usd in basic courss in Finit Elmnts at an undrgraduat lvl and also in advancd cours whr th studnts add thir own routins for instanc; lmnt stiffnss matrix, strss calculations in lasticity problms or utilizing rady-to-us routins for crack propagation analysis. This finit lmnt nvironmnt has also bn usd for tsting of diffrnt rsarch idas and for solving of diffrnt industrial nginring applications. This program will b usd throughout th txt in this book as a tool for analysis of all xampls givn during th discussion of diffrnt finit lmnt applications. Th idas and th argumnts givn abov hav bn th main driving forc for doing this work. Hopfully, this book will prov usful as both an introduction of th mthod and also a standard tool or companion to b usd during daily finit lmnt work. Bo Torstnflt Novmbr, 27

9 Radr s Instruction Radrs that hav nvr studid Finit Elmnts ar rcommndd to first rad th bar chaptr (chap. 3) from th arly bginning to th vry nd vry carfully. It is th author s blif that this chaptr is dtaild nough to srv as a stand-alon bas for slf-studis whr th radr is rcommndd to, during rading th txt, prform a complt rwriting of th basic mathmatical chain. Evry chaptr to com is writtn in a mannr and with an aim to b mor or lss slf-containd for th radr with sufficint pr-qualifications. Typical rquird qualifications ar 2 yars studis at undrgraduat lvl of any of th most common nginring programms. Concrning th layout of th txt; thr ar important kywords which will appar as Margin txt. As a studnt rading th txt for th first tim on should, aftr rading a crtain chaptr or sction, go back and us ths margin txts as rmindrs for having rachd a sufficint lvl of undrstanding of diffrnt important concpts. On major challng whn trying to dscrib Finit Elmnts is to giv sufficint dtail without making it too lngthy. That is th rason why som in-dpth matrial is givn at th nd of th book rathr than whr it appars for th first tim. Margin txt vii

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11 Notation Notation principls usd in this book ar summarizd blow. If a lttr or symbol is usd twic with a diffrnt maning, th lttr or symbol will b givn twic in this list. As such, if a concpt dfind by a lttr or symbol has svral diffrnt usd nams dscribing th sam and quivalnt intrprtation, all will b givn blow. Gnral mathmatical symbols a a a i A C C 1 L S W G M n sd n l n n n n n f n p A scalar valu A column vctor writtn as a bold lowr-cas lttr A cofficint in a vctor A matrix writtn as a bold uppr-cas lttr Th st of all continuous functions Th st of all functions having a continuous first-ordr drivativ Local quations Strong formulation Wak formulation Galrkin formulation Matrix problm Numbr of spatial dimnsions Numbr of lmnts blonging to th msh Numbr of nods blonging to th msh Numbr of nods blonging to an lmnt Numbr of unknown frdoms in th msh Numbr of prscribd frdoms in th msh ix

12 Latin symbols A An ara or cross-sctional ara a Th global unknown vctor, th global displacmnt vctor, th global dgrs of frdom (d.o.f.) vctor a An lmnt-local unknown vctor, lmnt-local dgrs of frdom (d.o.f.) vctor b, b Load pr unit volum or lngth B B C Global kinmatic matrix Elmnt-local kinmatic matrix Boolan connctivity matrix c, c i An arbitrary vctor usd for th wight function c c p x i, y i, z i D D E E f f d f g f h f r G G Elmnt nodal coordinat vctor Spcific hat cofficint Nodal coordinat componnts Elasticity matrix in flxibility form Thrmal conductivity matrix Young s modulus of lasticity Elasticity matrix in stiffnss form Global load vctor Global load vctor from intrnal distributd forcs Global load vctor from ssntial boundary conditions Global load vctor from natural boundary conditions Global raction forc vctor Shar modulus Global oprator matrix, gradint matrix g, g Essntial boundary conditions (multi-dim or 1D) h, h Natural boundary conditions (multi-dim or 1D) I J K K L M N Ara momnt of inrtia Jacobian matrix Global stiffnss or conductivity matrix Elmnt stiffnss or conductivity matrix Lngth Bnding momnt Global shap function matrix

13 N i N Ni N Q A global shap function Elmnt-local shap function matrix A lmnt-local shap function Axial forc in bars or bams Hat gnration pr unit lngth q, q Hat flux pr unit surfac (multi-dim or 1D) q n q R r T T T t t t t i S S S S g S h s u Hat flux prpndicular to th surfac Load pr unit lngth Rsidual vctor Unbalancd forc vctor or discrt rsidual vctor Tmpratur Surrounding tmpratur Shar forc Tim Thicknss Traction vctor A tst function Surfac Statical momnt, th first momnt Strss tnsor Th part of th surfac whr ssntial boundary conditions ar known Th part of th surfac whr natural boundary conditions ar known Strss vctor Displacmnt vctor u, v, w Displacmnt vctor componnts w, w Wight function (vctor-valud or scalar-valud) V Volum x, y, z Global coordinats Grk symbols α α δ ij Thrmal xpansion cofficint Thrmal convction cofficint Kronckr dlta

14 φ i ε ε θ i λ ν ϱ σ ij τ ij A wight function Normal strain Strain componnt vctor Nodal rotation componnt Hat conductivity Poisson s ratio Dnsity Normal strss componnt Shar strss componnt ξ, η, ζ Local coordinats

15 Chaptr 1 Prlud Nowadays Finit Elmnts ar th standard tools for doing simulations in a larg varity of nginring disciplins. Finit Elmnts ar no mor a tool for just a limitd numbr of nthusiastic xprts; thy ar somthing all of us as nginrs hav to larn. On rason for why this mthod still, to som xtnt, is lookd upon as a tchniqu which you as an nginr can dcid not to larn is probably bcaus it is blivd to b too difficult and tim-consuming. It is now tim to chang this option onc for all. All of us can larn Finit Elmnts. Evry nginr must know at last som basic facts from Finit Elmnts applid to som of th most important filds of application. Whn trying to larn Finit lmnts it is important and usful to hav a solid knowldg of th physical problm, modls of it and thir analytical solutions. That is why Finit Elmnts should b studid in clos connction to ovrall basic studis of a crtain nginring disciplin. Finit Elmnts is just a approximat numrical tool for solving som basic local quations constituting a mathmatical modl of rality. A rason for why this tchniqu is still lookd upon as difficult to larn is probably that most txt books ar writtn by ddicatd rsarchr in diffrnt filds of finit lmnt applications. As an author on probably tnds to dscrib th mthod from a mathmatical point of viw as consistntly as possibl and with a notation prhaps nvr prviously sn by th studnts. Th tchniqu is now so wll-stablishd that from a mathmatical point of viw most faturs and dficincis ar known in a varity diffrnt mathmatical formulations of diffrnt important problms. In this txt w will larn why th mthod works, how th mthod works both analytically and numrically, how to us th mthod in typical daily nginring applications and, probably most important lsson what proprtis on should 1

16 BACKGROUND xpct form th numrical approximation of th unknown ntitis. It is th author s intntion to writ a txt covring dtails from th arly bginning of a discussion of th modl of rality, which w as nginrs would lik analyz, to th vry nd whr w hav th rsults from th finit lmnt analysis. Gnral Faturs of th txt: Evry finit lmnt application will start from th arly bginning of its application with a discussion concrning which ar th basic quations, why must thy hold and what ar th basic physical assumptions. Evry important concpt and xprssion will b dducd and th mathmatical chain will b unbrokn throughout th txt. Th mathmatical languag will b simpl and concis Th txt will not b wighd down by any rigorous mathmatical proof of important statmnts. Evry finit lmnt application will nd up in on or mor solvd xampls by th finit lmnt program TRINITAS. Th txt will also srv as a thortical dscription of what is implmntd in this program 1.1 Background Finit Elmnts hav bn dscribd ovr th last dcad in svral diffrnt ways. In th arly bginning it was dscribd as a Rayligh-Ritz mthod for lasticity problms and latr on as a gnral tool for solving of partial diffrntial quations of various kinds, always basd on a so calld wak formulation. From a mathmatical point of viw, th first dscription was basd on Calculus of variations and a modrn formulation is now basd on Functional analysis and th thory of linar vctor spacs. Important basic work was don by Courant during th first part of th 194 s and th word finit lmnt was coind in 196 by Clough. Intrst from nginrs working with diffrnt aronautical industrial applications was on of th main driving forcs during th dvlopmnt of th finit lmnt mthod. During th 197 s th first gnral-purpos commrcial finit lmnt packags wr availabl and othr nginring disciplins startd to us th mthod. Th dvlopmnt of diffrnt computr basd support activitis such as prprocssing of finit lmnt input and postprocssing of finit lmnt output, and th ovrall succss of th mthod has bcom possibl du th fast incrasing computr powr which has bn going on in paralll. Today Finit Elmnts ar on important cornrston in th ntir Computr-Aid Enginring (CAE) nvironmnt containing most nginring activitis ndd to b don in most nginring branchs.

17 CHAPTER 1. PRELUDE Th Big Pictur Nowadays Finit Elmnts ar usd in a larg varity of nginring disciplins. Typical filds ar lasticity and hat transfr problms in solid bodis and acoustics and fluid flow problms in fluids. A larg numbr of diffrnt linar or non-linar, stady stat or transint problm classs xist. All ths applications ar somtims calld Computational mchanics. If th scop is vn furthr xtndd, us of Finit Elmnts is also possibl and straightforward in magntic fild and diffusion problms tc.. This txt will concntrat on lasticity and hat transfr problms which ar th most important applications of Finit Elmnts among all diffrnt computational mchanics disciplins. A rathr limitd numbr of physical ntitis wll-known by most mchanical nginrs will b usd in ths formulations. In lasticity problms th displacmnt vctor u and in hat transfr problms th tmpratur T is of grat importanc. In fluid flow problms th vlocity vctor v, th prssur p and th dnsity ρ ar basic unknowns. In acoustic problms th prssur p onc again is of grat importanc. Plas obsrv that th vlocity v is just th tim drivativ of th displacmnt u. In svral transint (that is tim-dpndnt) problms w will also hav nd for furthr tim drivativs such th acclration vctor a. In lasticity problms th strss componnts σ ij and th strain componnts ε ij will b important ingrdints. In hat transfr problms w will also hav to put focus on th hat flux vctor q. To b vry dtaild th list can b mad longr but th gnral conclusion so far is that th total numbr of physical ntitis ndd to b familiar with is rathr limitd vn if w ar discussing th ntir fild of computational mchanics. Typical to modls of all thos disciplins is that thy consist of a limitd numbr of quations of diffrnt typs. Th first group of quations to b brought up in this discussion is th Balanc laws motivatd from basic bhaviour of natur. Thr is th Nwton s scond law, f = ma rquiring that all forcs acting on a body most b in quilibrium. This balanc law is th bas for lasticity problms. In hat transfr problms th govrning balanc law is th Consrvation of Enrgy, th first law in Thrmodynamics. This quation only mans that nrgy is undstroyabl. Thr is also a third important balanc law govrning fluid flow problms; this is Consrvation of Mass. Ths thr balanc laws govrn most computational mchanics applications. In mor complx, and probably non-linar applications, somtims svral or all of ths balanc laws hav to b utilizd. A scond group of quations is th Constitutiv rlations. Typical to ths quations ar that thy all ar mpirical quations stablishd through xprimntal studis. Common to ths quations ar also that thy try to dscrib th bhavior of a solid matrial or a fluid in trms of som usful masurs. In lasticity problms a first choic is th gnralizd Hook s law and in hat transfr th Fourir s law is qually common. In fluid flow calculations a Nw- Computational mchanics Balanc law Constitutiv rlation

18 THE BIG PICTURE Scalar-valud Vctor-valud Compatibility rlation Systm of linar algbraic quations Linar Eignvalu problms tonian fluid flow bhavior is th first and simplst choic for domain proprty charactrization. Anothr important classification of a typical finit lmnt formulation is whthr th problm nds up in a Scalar-valud or Vctor-valud problm. In th discussion to com w will find out that th displacmnt vctor u will b th basic unknown and in th hat transfr problm th tmpratur T will b th basic unknown. That is th lasticity problm is a vctor-valud problm and hat transfr problm is scalar-valud problm which b dscribd in dtail latr on. In cass whr w ar studying vctor-valud problms, thr is also a nd for a rlation coming from a third group of quations. Th group rfrrd to hr is th group of Compatibility rlations. Typical to this group of quations is that thy try to prdict how dformations in a mattr will tak plac. Such quations will always put up som rlations for how diffrnt componnts must rlatd to ach othr. In scalar-valud problms thr is nvr nd for any rlation blonging to this group. What has bn discussd so far is what is typical or is in common btwn diffrnt mathmatical formulations of diffrnt filds of application of th finit lmnt mthod. Also, from a numrical point of viw, svral ovrall important commnts can b mad for what is typical or shard btwn diffrnt finit lmnt applications. As a usr of a finit lmnt program it is probably qually important to b awar of what is going on in th computr during diffrnt typs of finit lmnt analysis. In lasticity and hat transfr stady-stat problms w will find out what th computr has to solv of Systm of linar algbraic quations. In cass of studying tim-dpndnt problms our mathmatical discussion will nd up in systms of algbraic coupld ordinary diffrntial quations in tim which hav to b solvd numrically by any of som tim intgration schm. An important aspct of such tim intgration schms is if th schm is implicit or xplicit. Ths schms hav diffrnt mrits and whr filds of application rarly ovrlaps. From a numrical point of viw w will also find anothr typical group of Linar Eignvalu problms. Th most important applications ar dynamic ignvalu problms and linar buckling problms. In non-linar problms on soonr or latr has to introduc a linarization of th quations and from a numrical point of viw an itrativ schm basd on Nwton s mthod has to b mployd. This sction only tris to giv th radr an ovrviw of th topic. Prhaps som of th kywords discussd hav bn touchd upon in som othr courss or contxts. Som of th algorithms and numrical tchniqus ndd hr probably hav bn studid in prvious mathmatical courss. If som of th matrial discussd hr is hard to undrstand it is vry natural bcaus this is an ovrviw and mor dtail will b givn latr on. This sction will probably srv qually wll as a summary and not only as an introduction of th topic.

19 Part I Linar Static Elasticity 5

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21 Chaptr 2 Introduction In most nginring activitis whr Solid Mchanics considrations hav to b takn into account, a good start is to assum a linar structural rspons and a load that is applid in a quasi-static mannr. This is on of th simplst modls to study and such an analysis can b classifid as a Linar Static Elasticity analysis. A hug majority of all nginring analysis work don, with th purpos of trying to invstigat Solid Mchanics proprtis of a structur, blongs to this class of analysis and in many cass such an analysis will srv as a propr final rsult from which most ovrall nginring dcisions can b takn. In this part of th txt linar strain-displacmnt rlations (small displacmnts) and linar lastic strss-strain displacmnts will b assumd. If also all boundary conditions ar constant and indpndnt of th applid load th structur will show a linar rspons. In this part of th txt th discussion will also b limitd to problms with quasi-static load application; no inrtia forcs will b includd. In Solid Mchanics thr xists a squnc of approximation lvls basd on diffrnt displacmnt assumptions giving a tru 3D dformabl body mor or lss frdom to dform. In th following, svral of th most important of ths basic displacmnt assumption idas will b discussd in trms of th basic local quations, strong and wak formulations, and finally appropriat finit lmnt formulations. In th txt blow th discussion will start with th bar assumption which is th assumption that givs a ral 3D body th last dforming possibilitis. In th following chaptrs thr ar also finit lmnt formulations givn for bams, 2D and 3D solids and finally, Mindlin-Rissnr shll lmnts. In all ths chaptrs, motivatd by diffrnt basic displacmnt assumptions, th ntir chain of quations will b givn. Th xprincd radr will quickly look through this and undrstand that vry much of th structur and ovrall Linar Static Elasticity 7

22 8 basic natur of th quations ar closly rlatd in-btwn ths diffrnt formulations for bars, bams and solids. That is, basic rlations could hav bn writtn mor gnrally onc and only rfrrd to in th nxt chaptrs. But th txt to com is, as alrady mntiond, writtn with a goal that a chaptr or an application should b slf-containd with minimum rquirmnt for jumping in on dirction or anothr in th txt. Anothr typical fatur for th txt is that vry discussion will start at th arly bginning of th application by a thorough discussion of th local quations constituting th modl of rality. Aftr ths lmnt-spcific topics th txt will continu with gnral discussions concrning how to assmbl and solv th systm of linar algbraic quations. Diffrnt dirct and itrativ algorithms and tchniqus for finding th solution will b givn. In th last chaptrs in this Linar static lasticity part of th txt ar discussions of som furthr important aspcts concrning how to trat and analyz linar static lasticity problms. Somtims thr is nd for transformations of diffrnt kinds. For xampl, on probably would lik to introduc a skw support not paralll to any of th global dirctions; that is thr will b nd for a transformation of on or svral lmnt stiffnss matrics. In a larg typical industrial finit lmnt analysis thr is likly to b a nd for combining diffrnt lmnt typs to ach othr. This can b don by th imposing of constraints on th systm of quations. A larg varity of diffrnt possibilitis xist. Th txt will also covr how to numrically solv systms of quations containing constraints. Th last sctions in th Linar static lasticity part of th txt will actually discuss a problm which is non-linar. That is frictionlss contact problms whr th basic problm is to find th xtnt of th contact surfac. Th contact surfac is th part of th boundary whr two contacting dformabl bodis only transmit comprssiv normal strsss. In a gnral cas th xtnt of this surfac is a rsult of th analysis and it has to b stablishd by itrations. A forc-displacmnt rlation is in th gnral cas non-linar bcaus of chang of contact surfac. An obvious typical ral situation is a ball or rollr baring.

23 Chaptr 3 Bars Considr a straight slndr body with a smoothly changing cross sction A(x) and with a lngth L. Lt us now assum that all loads applid to th body act in th dirction of th xtnsion of th body, which is th local horizontal x-dirction, s figur 3.1. Thr is a distributd load, h pr unit surfac [N/m 2 ] at th right nd and a distributd load, b(x) pr unit lngth [N/m], acting in th intrior of th body. That is, th body will not b xposd to any bnding loads and th body will only b strtchd in it s own dirction. If it is ncssary to includ bnding of such a slndr structur w hav to mov to th bam displacmnt assumption discussd in th nxt sction. In figur 3.1 th lft nd of th bar has bn givn a known prscribd displacmnt g, whr g << L, and E(x) is th Young s modulus of th matrial. A(x), E(x) g b(x) h x =. x = L x Figur 3.1: A typical bar structur 3.1 Th Bar Displacmnt Assumption Undr th circumstancs dscribd abov th Bar displacmnt assumption is 9 Bar displacmnt assumption

24 THE LOCAL EQUATIONS applicabl. That is, vry plan prpndicular to th x-axis is assumd to undrgo just a constant translation in th x-dirction and th initial plan will rmain flat in it s dformd configuration. By introducing this assumption th displacmnt u will b a function of x only as illustratd in figur 3.2. Undformd body u(x) Figur 3.2: A typical bar dformation Dformd body This mans that only on strss componnt σ(x) and on strain componnt ε(x) will b non-zro at vry cut x through th bar. That is, from a mathmatical point of viw this problm is locally on-dimnsional. This modl of rality will only involv thr diffrnt unknown functions, th displacmnt u(x), th strain ε(x) and th strss σ(x) in th intrior of th body which has to b calculatd undr considration of influnc from th boundary conditions g and h. 3.2 Th Local Equations To b abl to analyz this modl thr is nd for at last thr diffrnt quations. As mntiond in th introduction, all modls proposd for studying diffrnt physical phnomna always hav to fulfill at last on balanc law. In this cas a statical quilibrium rlation will srv as th balanc law. Equilibrium for a x N(x) b(x) x N(x+ x) Figur 3.3: Forcs acting on a slic x of a 1D bar modl short slic of lngth x of th bar rquirs N(x + x) N(x) + b(x) x = (3.1)

25 CHAPTER 3. BARS 11 whr Taylor s formula givs N(x + x) N(x) + dn(x) x (3.2) dx and th axial forc N(x) can b xprssd in th strss σ(x) and th crosssctional ara A(x) as follows N(x) = A(x)σ(x). (3.3) Ths thr quations (3.1) to (3.3) dfins a Balanc Law in trms of th strss σ(x) and aftr division by x w hav d (A(x)σ(x)) + b(x) =. (3.4) dx Typically, this quation always must hold indpndnt from what strss-strain or strain-displacmnt rlations will b assumd latr on. In this contxt, as w alrady hav indicatd, a linar lastic Constitutiv Rlation (th 1D Hook s law) σ(x) = E(x)ε(x) (3.5) Balanc Law Constitutiv Rlation will b usd and a linar compatibility rlation (small displacmnts) can b dducd by using th displacmnt u at th two positions x and x + x in figur 3.4. x u(x+ x) Undformd body Dformd body u(x) Figur 3.4: Typical bar dformation Th linar strain masur ε(x) is dfind as th chang in lngth ovr th initial lngth x as follows ε(x) = u(x + x) u(x) x u(x) + du(x) x u(x) = dx x = du(x) dx (3.6)

26 A STRONG FORMULATION Compatibility Rlation and this will srv as a Compatibility Rlation for a linar 1D bar structur. Ths thr basic local quations 3.4 to 3.6 can b summarizd in th box L as follows Box: L Local Equations in 1D Linar Static Elasticity d (A(x)σ(x)) + b(x) = dx σ(x) = E(x)ε(x) ε(x) = du(x) dx and ths quations hav to b fulfilld at any position insid th opn domain Ω =]; L[. On obvious rmark is of cours that thr is no influnc form th boundary conditions so far. 3.3 A Strong Formulation On of svral possibl ways to start th analytical work for solving this systm of quations is to liminat th strss σ(x) and th strain ε(x) by putting th constitutiv rlation and th compatibility rlation into th balanc law. Aftr introducing th boundary conditions from figur 3.1 th following wllposd boundary valu problm S can b stablishd. Box: S Strong form of 1D Linar Static Elasticity Givn b(x), h and g. Findu(x) such that ( d A(x)E(x) du(x) ) + b(x) = x Ω dx dx u() = g on S g E(L) du(l) dx = h on S h Rmarks: 1D scond-ordr mixd Boundary-Valu Problm This formulation S constituts a Strong formulation of a linar static 1D Bar problm and from a mathmatical point of viw this is a 1D scondordr mixd Boundary-Valu Problm

27 CHAPTER 3. BARS 13 u() = g is a non-homognous Essntial boundary condition. If g = th Essntial boundary condition is homognous. Th total surfac S consists in this 1D cas only of th two nd cross sctions S h and S g. Th diffrntial quation is an xampl of a scond ordr ordinary on. E(L)du(L)/dx = h is a Natural boundary condition. Natural Th boundary valu problm is mixd bcaus thr ar both ssntial and natural boundary conditions. Latr on w will b awar of that som ssntial boundary conditions always hav to xist to b abl to guarant th uniqunss of th solution of th matrix problm M. 3.4 A Wak Formulation A Strong formulation can always b transfrrd into an quivalnt Wak formulation by multiplication of an arbitrary Wight function w(x) and an intgration ovr th domain. L ( ) d du L w (AE dx dx ) + b dx = w d du L (AE dx dx ) dx + wb dx = (3.7) Wight function Aftr partial intgration of th first trm th following is obtaind. [ wae du ] L L dx dw du AE dx dx dx + L wb dx = (3.8) Th first trm in th quation abov can b rwrittn as th natural boundary condition h can b idntifid form box S as [ wae du ] L dx = w(l)a(l) E(L) du(l) dx } {{ } = h w() }{{} = A()E() du() dx (3.9) By putting on spcific rstriction on th wight function w(x) and no longr ltting th function w(x) b compltly arbitrary an infinit st V of functions can b dfind whr vry choic of wight function w(x) must b qual to zro on th part of th boundary whr ssntial boundary conditions (S g ) ar dfind. V = {w(x) w(x) = on S g } (3.1) An appropriat Wak formulation W of this 1D Bar problm can b summarizd as follows.

28 A GALERKIN FORMULATION Box: W Wak form of 1D Linar Static Elasticity Givn b(x), h and g. Findu(x) such that L dw(x) dx A(x)E(x)du(x) dx dx = u() = g on S g L w(x)b(x) dx + w(l)a(l)h for all choics of wight functions w(x) which blongs to th st V Rmarks: This wak formulation W srvs as an fficint platform for applying numrical tchniqus such as wightd rsidual mthods for solving this bar problm approximatly. Th partial intgration stp is prformd bcaus it opns th possibility to nd up in a symmtric systm of linar algbraic quations that is mor fficintly solvd in th computr compard to a non-symmtric systm. Th natural boundary condition is now implicitly containd in th intgral quation. It is possibl to show that th Strong and Wak formulations ar quivalnt. 3.5 A Galrkin Formulation Wightd Rsidual Mthods Th basic rason for first turning th local quations into a Strong formulation and aftr that transfr th problm into an quivalnt Wak formulation is that th wak form can b utilizd as a bas for a varity of diffrnt Wightd Rsidual Mthods that all ar capabl of solving our basic bar problm, at last approximatly. Gnral to ths mthods ar that both th unknown function u(x) and th wight function w(x) ar built up from finit sums of n functions. n u(x) u h (x) = t 1 (x)a 1 +t 2 (x)a t n (x)a n +t (x) = t i a i +t (x) (3.11) i=1 and n w(x) = φ 1 (x)c 1 + φ 2 (x)c φ n (x)c n = φ i c i (3.12) i=1

29 CHAPTER 3. BARS 15 Rmarks: Th functions t i (x) ar calld Tst functions. Latr on furthr dtails Tst functions and ruls will b givn concrning how to slct ths functions and what proprtis thy must fulfill. Th function t (x) must b thr to scur that th non-homognous ssntial boundary condition u() = g will b fulfilld. Furthr dtails will b givn blow. All a i ar unknown scalar constants. In th cas whn all tst functions has bn stablishd th only unknowns ar all ths a i. Th arbitrarinss of th wight function slction w(x) is by this tchniqu furthr limitd to th choic of th n functions φ i (x) and th valu of ach of th scalars constants c i. By this introduction of finit sris consisting of n functions our problm turns ovr from a Continuous on with infinit numbr unknowns to a Discrt on with a limitd numbr of unknowns On of th most popular wightd rsidual mthods is th Galrkin mthod. On rason for this is that this mthod always will gnrat symmtric systms of linar algbraic quation which is mor fficintly solvd in th computr compard to non-symmtrical ons. Hr th basic ida is Continuous Discrt Galrkin mthod t i (x) = φ i (x) = N i (x) i = 1, 2,,, n (3.13) that if a slction is don of th tst functions t i vry function φ i also is dfind and vic vrsa. Plas obsrv, that from hr ths functions (th tst and th wight functions) most oftn will b calld Shap functions and th notation N i (x) will b usd. By moving ovr from sums to a matrix notation th approximation u h (x) and th wight function w(x) can b rwrittn as follow Shap functions u h (x) = N(x)a + t (3.14) w(x) = N(x)c w(x) = c T N T (x) (3.15) whr a 1 c 1 N(x) = [ N 1 (x) N 2 (x)... N n (x) ] a 2 c 2.., a =., c =. (3.16). Concrning quation (3.15) th two altrnativs ar qual and actually th latr will b mostly usd. a n c n

30 A GALERKIN FORMULATION Bfor som gnral and mathmatically mor prcis ruls will b givn concrning what proprtis a crtain choic of shap functions N i hav to fulfill, on possibl choic among many othrs, will b givn and discussd from an intuitiv point of viw. In this particular bar problm w hav now accptd an ida whr an approximation is introducd for th displacmnt u(x) in th bar. Latr on w will find out that this will of cours also gnrat approximat solutions for th strss and strain in th bar. Th simplst possibl assumption is to think of th displacmnt approximation u h (x) as a pic-wis linar polygon chain. Such a function is continuous in its slf but th first drivativ is discontinuous. Th qustion is now how to u(x) Exact solution g Pic-wis linar approximation x =. x = L x Figur 3.5: A 1D bar displacmnt approximation assumption Nods Finit lmnt xprss such a pic-wis linar function as convnint and fficint as possibl with n linar indpndnt paramtrs typically stord in th column vctor a. Lt a numbr of n + 1 so calld Nods x i b dfind insid and at th nds of th domain Ω from to L. Th intrval btwn two nods is calld a Finit lmnt. As a first choic of functions N i a st of pic-wis linar functions 1. N 1 (x) N 1 (x) N i-1 (x) N i (x) N i+1 (x) N n (x) x =. x 1 x 2 x i-2 x i-1 x i x i+1 x i+2 x n-1 x n Figur 3.6: On possibl choic of shap functions N i for a 1D bar problm will srv as a bas for furthr discussions and so far thy ar only dfind from intuitiv rason and from figur 3.6 as follows

31 CHAPTER 3. BARS 17 N 1 (x) = { (x1 x)/(x 1 x ). x x 1. x 1 x x n (3.17). x x x i 1 (x x N i (x) = i 1 )/(x i x i 1 ) x i 1 x x i (3.18) (x i+1 x)/(x i+1 x i ) x i x x i+1. x i+1 x x n { N n (x) =. x x x n 1 (x x n 1 )/(x n x n 1 ) x n 1 x x n (3.19) Rmarks: In a typical intrval th unknown function will b approximatd by a linar function as follows u(x) u h (x) = N i 1 (x)a i 1 + N i (x)a i x i 1 x x i (3.2) whr only two shap functions at th tim will b non-zro and influnc th approximation at an arbitrary point insid th intrval. All ths functions N i hav a unit valu at on nod and ar zro at all othr nods. That is, th following holds N i (x j ) = δ ij = { 1. i = j. i j (3.21) which mans that th shap functions ar linar indpndnt at th nods. That is, th vctor a will rprsnt th displacmnt in th nods. It is also possibl to show that ths shap functions N i (x) ar linarly indpndnt at an arbitrary position insid th intrvals. From a gnral mathmatical point it is possibl to show that such a linarly indpndnt choic of shap functions N i (x) will span a n-dimnsional subspac from which th approximation will b rcivd. Anothr important proprty that has to b fulfilld by a crtain choic of a shap function N i (x) is that th function must blong to th st C which consists of all continuous function N i (x) which fulfills ( ) 2 dni (x) dx <. (3.22) Ω dx

32 A GALERKIN FORMULATION Th function t (x) can now b constructd from th N 1 (x) function as follows t (x) = N 1 (x)g t (x x 1 ) =. (3.23) By putting th quations (3.14), (3.15) and (3.23) into th wak formulation W th following discrt Galrkin formulation will b achivd. L d dx (ct N T )AE d dx (Na + N 1 g) dx = Th vctor c T can b brought out as follows c T { L d dx N T AE d dx (Na + N 1 g) dx L L c T N T b dx + c T N T (L)A(L)h N T b dx N T (L)A(L)h } = and a matrix B(x) can b dfind as B(x) = dn(x) [ dn1 (x) dn = 2 (x) dx dx dx dn n (x) dx ] (3.24) which thn can b insrtd into th quation abov and th following is obtaind Global stiffnss matrix K Global load vctor f { L c T B T AEBdx a ( L N T b dx + N T (L)A(L)h L )} B T AE d dx N 1 g dx =. Th Global stiffnss matrix K and th Global load vctor f can b idntifid from this quation as K = L B T (x)a(x)e(x)b(x)dx (3.25) f = L N T (x)b(x) dx + N T (L)A(L)h L B T (x)a(x)e(x) d dx N 1 (x)g dx. (3.26) whr th matrix K is a symmtric matrix with n rows and columns and th vctor f is a column vctor containing on load cas. A discrt Galrkin formulation for this 1D problm now rads

33 CHAPTER 3. BARS 19 Box: G Galrkin form of 1D Linar Static Elasticity Find a such that c T (Ka f) = c T r = for all choics of th vctor c (th wight function) Not ncssary hr, but convnint in th discussions to com is to introduc th following gnral split of th global load vctor f into thr diffrnt load vctor contributions. f = f d + f h f g (3.27) Th first part f d coms from intrnal distributd forcs and in this 1D cas it is qual to f d = L N T (x)b(x) dx (3.28) and two othr parts ar from ssntial and natural boundary conditions on S h and S g. f h = N T (L)A(L)h (3.29) f g = L B T (x)a(x)e(x) d dx N 1 (x)g dx. (3.3) Th vctor f h can always b valuatd, indpndnt from th xplicit choic of shap functions, as follows. f h = N T (L)A(L)h = A(L)h. (3.31) 1 and vctor f g is shap function dpndnt. In th cas with linar shap functions, as discussd so far, and a constant cross sction A and Young s modulus E, w hav f g = L B T AE d dx N 1 g dx = AE L g (3.32) whr L 1 is th lngth of th first lmnt. From ths xprssions it is asy to conclud that th product A(L)h and th product AEg/L 1 ar both forcs.

34 A MATRIX FORMULATION 3.6 A Matrix Formulation Unbalancd rsidual forcs It is obvious from abov that th Galrkin formulation mans a scalar product btwn th column vctor c and anothr column vctor r and it is still on singl quation. Th vctor r is calld th rsidual and it can b intrprtd as Unbalancd rsidual forcs. From th basic ida of involving an arbitrary wight function w(x) in th wak formulation now only rmains a vctor c. This vctor still must b possibl to slct compltly arbitrary. From this rquirmnt it is obvious that th vctor r must b qual to a zro vctor r = (3.33) which mans that th structur is in quilibrium. Plas obsrv that this fulfillmnt of quilibrium is hr said to b in a wak sns which mans that w hav quilibrium masurd in nodal forcs! A matrix problm consisting of n linar algbraic quations can now b idntifid. Box: M Matrix form of 1D Linar Static Elasticity Find a such that Ka = f whr K and f ar known quantitis By solving this systm of quations th vctor a will rprsnt th displacmnts at th nods at quilibrium. Th vry last stp in th analysis is to calculat th, in th strong formulation liminatd strsss and strains, by making us of th compatibility and constitutiv rlations from th local quations (S box L). ε(x) = d dx (N(x)a + N 1 (x)g) = B(x)a + d N 1 (x) dx g (3.34) Finit Elmnt Program σ(x) = E(x) d dx (N(x)a + N 1 (x)g) = E(x)(B(x)a + d N 1 (x) g) (3.35) dx This is always don in an lmnt-by-lmnt fashion. What now lacks is a numrical procdur for stablishing of th matrics K and f and solving of th matrix problm M for th vctor a. Such a numrical procdur is typically implmntd as a computr program which can b calld a Finit Elmnt Program.

35 CHAPTER 3. BARS 21 Rmarks: In th bginning of this discussion thr ar thr unknown functions of x. Ths ar th strss σ(x), th strain ε(x) and th displacmnt u(x) which all now can b calculatd at last in an approximativ mannr. Most of th mathmatical work don so far is of an analytical natur and nds only to b don onc (whn trying to larn and undrstand why th finit lmnt mthod works bfor it coms to us of a computr program) Both th matrix K and vctor f ar compltly dfind by givn data in figur 3.1, th numbr of shap functions N i (lmnts) and th bhavior of ths shap functions (th lmnt typ) From a mathmatical point of viw this discussion can b summarizd as L S W G M and sourcs for rrors in this mathmatical modl of rality ar dviations from rality in th constitutiv and th compatibility rlations, dviations in th slctd boundary conditions and numrical rrors du to us of a limitd numbr of Finit Elmnts with a spcific bhavior in ach lmnt. On can show that th solution to th matrix problm M always xists and has a uniqu solution if th global stiffnss matrix K is non-singular. If thr xist at last on ssntial boundary condition which prvnts rigid body motion th stiffnss matrix will b non-singular. That is th global stiffnss matrix K is positiv dfinit and th following holds a T Ka > a dt(k) > In this 1D cas th matrix K is symmtric with a thr-diagonal population. 3.7 A 2-Nod Elmnt Stiffnss Matrix This 1D finit lmnt analysis discussion is now approaching th nd of th analytical part of th analysis and w ar clos to a position whr w hav to put in numbrs and start th numrical part of th analysis. This is normally prformd by a computr program basd on this analytical discussion. What still has to b discussd is how to valuat th intgrals in box M. Aftr that th numbr and th bhavior of shap functions N i is dcidd, ths intgrals only contain known givn quantitis and th basic qustion is how to valuatd ths as fficint as possibl! Plas obsrv, whn it coms to practical us of a finit lmnt program on always has to slct a crtain numbr of finit lmnts of a crtain lmnt typ which mans xactly th sam as slcting th numbr and th bhavior of th shap functions N i.

36 A 2-NODE ELEMENT STIFFNESS MATRIX On of th cornrstons in a finit lmnt formulation is that th ntir domain is split into a finit numbr of sub-domains, so calld finit lmnts. Du to th natur of th shap functions as linar indpndnt and only nonzro ovr vry limitd parts of th ntir domain it is convnint to prform th intgral ovr on lmnt (on sub-domain) at a tim and w hav n l K = i=1 xi+1 x i B T (x)a(x)e(x)b(x)dx (3.36) whr this is a sum of n l matrics whr n l is th numbr of finit lmnts. Each of ths sub-matrics will only contain 4 non-zro cofficints symmtrically positiond around th main-diagonal of th sub-matrix. This is bcaus th vctor B valuatd for x-valus insid th intrval x i and x i+1 will only contain 2 non-zro positions and th non-zro part of th product B T B is a symmtric 2 row and 2 column matrix. Lt us now study such a sub-intrval in mor dtail. W will now mov ovr to an lmnt-local notation, s figur 3.7. Th two linar parts of th global 1. N i (x) N 1 N 2 N i+1 (x) x i-1 x i x i+1 x i+2 x Figur 3.7: Th rlation btwn lmnt-local and global shap functions shap function N i and N i+1 ovr th intrval from x i to x i+1 has bn givn th closly rlatd notations N1 and N2 which is an lmnt-local numbring from 1 to 2 ovr th numbr of nods associatd with this lmnt. Th following nw lmnt-local vctors can now b dfind { } N (x) = [ N1 (x) N2 (x) ] and a a = 1 a (3.37) 2 and b usd for an lmnt-local xprssion of th displacmnt approximation u (x) as follows u (x) = N 1 (x)a 1 + N 2 (x)a 2 = N (x)a. (3.38) Th lmnt-local strain approximation ε (x) can now b writtn as [ ] ε (x) = du (x) dn = 1 (x) dn2 (x) a = B (x)a (3.39) dx dx dx } {{ } =B (x)

37 CHAPTER 3. BARS 23 and th global matrix B will hr appar in an lmnt-local vrsion B. A rlation btwn th global unknown vctor a and th lmnt-local unknown vctor a is asily stablishd as a 1. { } [ ]. a a = 1 1 a a = i 1 = C a (3.4) 2 1 a i } {{ } =C. a n whr th matrix C is a Boolan Matrix populatd by only unity or zro valus. lmnt-local Boolan Matrix Rmarks: Th lmnt-local vctor a is always a subst of th global vctor a. Thr is always on unity valu in ach row of th matrix C as long as non of th nods in th lmnt blongs to th boundary S g, whr w hav known valus of th displacmnts. In cass whr on or svral nods ar associatd to th boundary S g w can so far think of a matrix C prsrving its numbr of rows and whr a zro row without any unit valu is introducd corrsponding to th givn valu g. A mor thorough and dpnd discussion of this topic can b found in chaptr 5 undr sction 5.8. It can now b shown that th global stiffnss matrix K can b built from a sum of small 2x2 matrics which ar xpandd by a pr- and post-multiplication of th boolan matrix C. n l K = i=1 C T i xi+1 B T (x)a(x)e(x)b (x) dx x } i {{ } =K i C i (3.41) Such a small matrix is an important and oftn discussd topic calld th Elmnt Stiffnss Matrix K. Th subscript i will only b usd whn a spcific lmnt i is discussd. In this analytical discussion it is now tim to prform th vry last analytical stps. Lt us xprss th two local shap functions N1 and N2 in an lmnt-local coordinat axis ξ and in accordanc to th figur 3.8 and w typically hav Elmnt Stiffnss Matrix N1 (ξ) = 1 2 (1 ξ); N 2 (ξ) = 1 (1 + ξ). (3.42) 2

38 A 2-NODE ELEMENT STIFFNESS MATRIX 1. N 1 (ξ) N 2 (ξ) x i x i ξ x Figur 3.8: Th lmnt-local coordinat systm ξ Th mapping btwn th two coordinat systms can b writtn as x(ξ) = 1 2 (x i+1 x i )ξ (x i+1 + x i ). (3.43) whr th lngth of th lmnt L = (x i+1 x i ). Diffrntiation and th chain rul thn givs dx = 1 2 L dξ and dn i (ξ(x)) dx = N i (ξ) dξ dξ dx = 2 Ni (ξ) L dξ (3.44) and it is asy to valuat th B matrix as follows [ ] B dn = 1 dn2 = 2 [ dn 1 dn ] 2 dx dx L = 1 [ ] 1 1 dξ dξ L (3.45) whr th B matrix in this simpl cas is indpndnt from th local coordinat systm. Th lmnt stiffnss matrix K is thn K = 1 [ 1 L 2 1 ] [ ] A(x)E(x) L dξ (3.46) 1 2 and if th th cross sction A(x) and th Young s modulus E(x) ar constants with rspct to x and ξ w hav K = EA L [ ] 1 1 dξ 2 1 } {{ } =2 which finally can b summarizd in th box blow. Box: 1D 2-nod bar lmnt stiffnss matrix K = EA [ ] 1 1 L 1 1 (3.47)

39 CHAPTER 3. BARS 25 This is xactly what has to b implmntd and valuatd numrically in th computr program and th nd of th analytical discussion is rachd at last for this lmnt typ. Evn if th cross sction A(x) changs ovr th domain on typically us th valu of th cross sction at th mid-point of th lmnt. That is, th th cross sction is modld as a stp-wis constant function. 3.8 A 3-Nod Elmnt Stiffnss Matrix Th slction of shap functions discussd so far is actually th simplst possibl with its pic-wis linar natur with a discontinuous first-ordr drivativ. Lt us now introduc a scond choic of shap functions, still with a discontinuous first-ordr drivativ, rquiring a nod at th mid-point of ach lmnt. By doing so our approximation of th displacmnt u(x) will b nhancd by a scond-ordr trm and th approximation will b a pic-wis parabolic polynomial chain. 1. N 1 (ξ) N 2 (ξ) N 3 (ξ) N i (x) N i+2 (x) x i x i+1 x i ξ x Figur 3.9: Elmnt local shap functions for 3-nod lmnt In a typical lmnt of lngth L = x i+2 x i w hav now dfind thr lmnt-local shap functions in accordanc to figur 3.9 and lmnt-local displacmnt approximation can b writtn as u(ξ) = N 1 (ξ)a 1 + N 2 (ξ)a 2 + N 3 (ξ)a 3 = N a. (3.48) whr N = [ N1 (ξ) N2 (ξ) [ ξ N3 (ξ) ] = 2 (ξ 1) 1 ξ2 ξ (ξ + 1) 2 ]. (3.49) It is now possibl to valuat th B matrix as follows [ ] B dn = 1 dn2 dn3 = 2 [ dn 1 dn2 dx dx dx L dξ dξ dn 3 dξ ] (3.5) whr th B matrix in this parabolic cas will b dpndnt on th local coordinat systm. Aftr introducing drivativs of th shap functions with rspct

40 AN ELEMENT LOAD VECTOR to ξ w hav B = 2 L [ ξ 1/2 2ξ ξ + 1/2 ]. (3.51) Th lmnt stiffnss matrix K will in this cas b a 3x3 matrix and in a cas with constant cross sction and Young s modulus w hav K = 4AE 1 ξ 1/2 2ξ [ ξ 1/2 2ξ ξ + 1/2 ] L dξ (3.52) L ξ + 1/2 K = 2AE L 1 1 (ξ 1/2)2 2ξ(ξ 1/2) (ξ 1/2)(ξ + 1/2) 4ξ 2 2ξ(ξ + 1/2) sym. (ξ + 1/2) 2 dξ (3.53) By solving six diffrnt intgral ovr polynomials in ξ w nd up with an lmnt stiffnss matrix for a 1D 3-nod lmnt for scond-ordr problms as dfind in th box blow. Box: 1D 3-nod bar lmnt stiffnss matrix K = EA L An Elmnt Load Vctor Elmnt Load Vctor According to th global load vctor f, it is also possibl to idntify such a typical lmnt contribution calld th Elmnt Load Vctor f which can b valuatd ovr on lmnt intrval at a tim. n xi+1 f = N T b(x) dx +f h f g (3.54) i=1 C T i x i } {{ } =f i In th gnral cas, w will find latr on that lmnt nodal loads gnratd from diffrnt distributd load contributions can b intgratd ovr th lmnt domain. Th global load vctor contributions f h and f g will b discussd furthr in th nxt sction. Concrning th lmnt load vctor calculations from diffrnt typs of distributd loads on typically rstrict th variation insid th lmnt to th variation usd for th displacmnt approximation. That is, in our 1D cas and

41 CHAPTER 3. BARS 27 focusing on th 2-nod lmnt with th distributd load/unit lngth b(x), w hav b(x) = N b = [ ] { } N1 N2 b 1 b = (1 ξ)b (1 + ξ)b 2 (3.55) whr th rquird input to th lmnt ar th intnsity of th distributd load at th two nods, b 1 and b 2. This tratmnt lads to what is calld a Consistnt Load Vctor. This mans that th load must not chang mor rapidly than th displacmnt. Also th lmnt load vctor f is most convnintly valuatd in a local coordinat systm ξ. Thus f = 1 1 [ N 1 (ξ) N 2 (ξ) ] [ N 1 (ξ) N2 (ξ) ] { L b 2 dξ 1 b 2 } (3.56) Consistnt Load Vctor which mans intgration of ach cofficint in a symmtric 2x2 matrix as follows f = L N 2 1 (ξ)dξ sym. 1 N1 (ξ)n2 (ξ)dξ { } b 1. (3.57) N 2 2 (ξ)dξ Ths thr intgrals ar simpl polynomials to intgrat ovr a symmtric intrval and w rciv th following numrical valus b N1 2 (ξ)dξ = N2 2 (ξ)dξ = and 1 1 N 1 (ξ)n 2 (ξ)dξ = 1 3. (3.58) By putting ths valus into th lmnt load vctor f xprssion, w obtain f = L 6 [ ] { b 1 b 2 }. (3.59) Lt us simplify this xprssion to a situation with a constant load intnsity b 1 = b 2 = b, w gt f = b L 2 { 1 1 } (3.6) which has th obvious physical intrprtation that th total forc gnratd by th load intnsity b tims th lmnt lngth L is split into two qual concntratd forcs acting at th nds of th lmnt.

42 THE ASSEMBLY OPERATION 3.1 Th Assmbly Opration Now all analytical dtails ar discussd and w know what is ndd for doing an implmntation of such a finit lmnt algorithm for analysis of 1D linar static lasticity problms in som programming languag. Th ovrall matrix problm in box M with th global stiffnss quations Ka = f can now b rwrittn in trms of lmnt stiffnss matrics K i and lmnt load vctor contributions f i as follows n l i=1 n l C T i K i C i a = i=1 C T i f i + f h f g (3.61) Assmbly Opration and boundary condition trms f h and f g. Aftr calculation of ach lmnt contribution an xpansion and adding of ths lmnt contributions to th global stiffnss matrix K and th global load vctor f is prformd. This numrical procss is calld th Assmbly Opration of th global stiffnss quations. Hr w typically only calculat and stor non-zro cofficints in th uppr right part of th global stiffnss matrix in th computr mmory. This systm of quations can b solvd by svral diffrnt solution tchniqus such as both dirct or itrativ algorithms. Aftr such a solution procdur w hav numrical valus in th global frdom vctor a and th vry last stp in th analysis is to calculat th strains and th strsss Strss and Strain Calculations Th strain and th strss calculation is, as alrady mntiond, prformd on th lmnt lvl. By making us of quations 3.34, 3.35 and 3.4 w hav ε = B a = B C a and σ = Eε. (3.62) In th cass with th 2-nod linar lmnt th strain and th strss approximation will b rough. Bcaus th B matrix is constant and indpndnt of spac th strain and th strss will rciv a constant valu in ach lmnt. That is, th strain and strss approximation will b pic-wis constant ovr th domain and th strain and strss fild ar discontinus ovr th lmnt bordrs. This is gnral also in multi-dimnsional linar lasticity problms. From a mor gnral point of viw, on important and obvious qustion in this contxt is which points should b usd to comput th strsss and th strains? Th answr is that associatd to a typical lmnt (or shap function slction) thr ar always a limit numbr of wll-dfind points insid th lmnt which givs th most accurat strain and strss approximation. Such points ar calld suprconvrgnt points which for th 2-nod lmnt is at th cntr of th lmnt and for th 3-nod lmnt thr ar two suprconvrgnt points at ξ = ±1/ 3.

43 CHAPTER 3. BARS 29 Anothr important aspct possibl to discuss alrady in this 1D contxt is that ths jumps in th strain and th strss approximation can b usd as a masur of th rror in th numrical approximation. Such an rror masur can b utilizd in a so calld Adaptiv finit lmnt analysis whr an itrativ procdur is usd and th domain is rmshd with th rror in th rgion form th prvious calculation as a masur for what siz th lmnts should hav in that rgion of th domain. Adaptiv finit lmnt analysis 3.12 Multi-dimnsional Truss Fram Works This finit lmnt formulation of 1D lasticity problms discussd so far is probably not that important as a tool in th daily nginring work but it is vry important as a first application of th finit lmnt mthod for gtting usd to and familiar with why and how th mthod works. This 1D formulation can asily b xtndd to multi-dimnsional cass of fram works of bars oftn calld Truss lmnts. Such a truss lmnt dos not transmit momnt at th nd points and th ovrall raction forc in such multidimnsional lmnt always acts in th dirction of th lmnt. A fram work modl with its momnt-fr connctions is a simpl, fast and powrful tool for analysis of ral fram works in a varity of diffrnt nginring structurs such as buildings, bridgs, crans and mast structurs. As an analyst on has to b carful not to introduc non-physical mchanisms bcaus of th momnt-fr connctions btwn th mmbrs in th modl of th fram work which sldom is prsnt in th ral structur. Truss lmnts (x 2 ; y 2 ) v 2 y (x 1 ; y 1 ) v 1 θ u 2 u 1 x Figur 3.1: 1D bar lmnt usd in a 2D situation Lt us now think of our 1D coordinat systm as a local dirction in a global 2D or 3D coordinat systm. Th displacmnt of a 1D bar lmnt xprssd in a 2D global coordinat systm can b writtn as follows a = { a 1 a 2 } = [ x ȳ x ȳ ] u 1 v 1 u 2 v 2 = T 2D a 2D (3.63)

44 MULTI-DIMENSIONAL TRUSS FRAME WORKS or in a 3D cas w rciv whr a = [ x ȳ z x ȳ z ] u 1 v 1 w 1 u 2 v 2 w 2 x = (x 2 x 1 )/L ȳ = (y 2 y 1 )/L z = (z 2 z 1 )/L L = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 + (z 2 z 1 ) 2. = T 3D a 3D (3.64) In th txt to com, w will omit th subscripts 2D and 3D and rly on a contxt dpndnt notation whr it is sufficint to drop ths subscripts. Plas obsrv, th assumd displacmnt in figur 3.1 sms to b not vry gnral but it is. It is th longation of th bar that is important and as long as th rotation is small thr ar an infinit numbr of dformations nding up in th sam longation. Th rotation dosn t influnc th lngth of th bar in this linar small displacmnt analysis. In figur 3.11 an arbitrary dformd and rotatd lmnt is sktchd with a scald-up dformation. A dformd lmnt v 2 y v 1 u 2 u 1 A constant longation x Figur 3.11: An arbitrary dformd lmnt with a crtain longation Lt us pick up th quilibrium quation on th lmnt lvl and for just on typical 1D lmnt w hav K a = f EA [ } } 1 1 L (3.65) 1 1 ] { a 1 a 2 = { f 1 f 2 which always is two idntical quations with opposit signs. Th 1D lmntlocal displacmnt vctor a can b liminatd and in th 2D cas w multiply with T T 2D from th lft and obtain T T 2DK T 2D d 2D = T T } {{ } 2Df (3.66) =K 2D

45 CHAPTER 3. BARS 31 from which w can idntify th lmnt stiffnss matrix for a 2D bar (or truss) lmnt as follows x K 2D = EA [ ] [ ] ȳ 1 1 x ȳ L x. (3.67) 1 1 x ȳ ȳ Aftr matrix multiplication w obtain K 2D = EA L x 2 x ȳ x 2 x ȳ ȳ 2 x ȳ ȳ 2 x 2 x ȳ sym ȳ 2 Th 3D cas is vry similar as follows K 3D = EA L. (3.68) x 2 x ȳ x z x 2 x ȳ x z ȳ 2 ȳ z x ȳ ȳ 2 ȳ z z 2 x z ȳ z z 2 x 2 x ȳ x z sym ȳ 2 ȳ z z 2. (3.69) Ths two lmnt stiffnss matrics ar implmntd as thy ar dscribd abov in most finit lmnt packags such as in th TRINITAS program. Whn it coms to hand calculations for purpos of undrstanding th thory and th numrical work flow in a typical finit lmnt analysis it is possibl in 2D cass to idntify an angl θ (S figur 3.1) whr x = cos θ = c ȳ = sin θ = s and th lmnt stiffnss matrix K 2D can b rwrittn in a mor asy to us form for hand calculations. c 2 cs c 2 cs K 2D = EA s 2 cs s 2 L c 2 cs (3.7) sym s 2

46 NUMERICAL EXAMPLES 3.13 Numrical Exampls A numbr of numrical xampls will b studid blow. Rsults from hand calculations and from computr calculations ar prsntd. A 1D bar problm Considr an linar lastic bar of lngth L with a varying cross sction A(x) ( A(x) = A 1 x ) 2 2L and a givn Young s modulus E. Th bar is rigidly supportd at th lft nd and at th right a concntratd forc F is applid. Th cross sction variation is dfind as follows A(x), E F x =. x = L x Figur 3.12: Th givn bar gomtry Th xact solution for th displacmnt u(x) can b achivd by solving of th strong formulation dfind in box S whr th givn displacmnt g and th distributd forc/unit lngth b(x) ar qual to zro. Putting in th cross sction xprssion into th diffrntial quation givs d dx (A ( 1 x 2L ) 2 E du(x) dx ) = and on intgration of both sids ( EA 1 x ) 2 du(x) 2L dx = C 1 whr C 1 is a unknown constant. In th right nd w hav th boundary condition EA(L) du(l) dx = A(L)h = F C 1 = F. A scond intgration thn givs u(x) = 2F L 1 ( EA 1 x ) + C 2 2L

47 CHAPTER 3. BARS 33 whr th boundary condition u() = C 2 = 2F L EA and th analytical solution for th displacmnt u(x) is u(x) = x 2L ( 1 x 2L ) 2F L EA and for th strss σ(x) w hav σ(x) = E du(x) dx = 1 F ( 1 x ) 2. A 2L This xprssion will b usd latr on as a comparison. Lt us now solv this problm numrically by making us of th finit lmnt mthod. Us th following finit lmnt msh consisting of thr 2-nod 1D a a 1 a 3 a 2 F x =. x = L/3 x = 2L/3 x = L x Figur 3.13: Th givn bar gomtry lmnts of qual lngth L i = L/3 whr i = 1, 2, 3. Concrning th cross sction of th lmnts w hr will us lmnts with a constant cross sction calculatd at th cntr of ach lmnt from th givn xprssion. That is, A 1 = A(x = L/6) = 121A /144, A 2 = A(x = L/2) = 81A /144 and A 3 = A(x = 5L/6) = 49A /144 which givs th following lmnt stiffnss matrics K 1 = 121EA [ ] 1 1 K 48L = 81EA [ ] L 1 1 K 3 = 49EA [ ] L 1 1 Bfor assmbling th global stiffnss quations Ka = f w slct a global nod numbring in th unknown vctor a as follows a 1 a = a 2 a 3

48 NUMERICAL EXAMPLES and th Boolan matrics C i can b idntifid as follows [ ] [ ] [ C 1 = C = C = 1 Rwriting of th global stiffnss quation into a sum of xpandd lmnt stiffnss matrics givs and 3 i=1 EA 48L C T i K i C i a = f which nds up in th following systm to solv EA a a 2 48L = a 3 whr th solution is a 1 a = a 2 = a 3 F L EA F. Finally, w can calculat strains and strsss in th lmnts from σi = Eε i = EB a i = E [ ] { } a L a i 2 which givs σ 1 = 3E L σ 2 = 3E L σ 3 = 3E L { [ ] F L EA 6354 { [ ] F L EA [ 1 1 ] F L 1683EA { i ]. a 1 a 2 a 3 } = 6354F/53361A } = 94864F/53361A } = F/53361A. = F Plas not that both th strain and th strss approximations insid this lmnt typ ar constant. If w mov ovr to 3-nod lmnt a pic-wis linar variation would b obtaind.

49 CHAPTER 3. BARS 35 Lt us now put in th numrical valus L = 1. m, F = 1. N, E = Pa and A =.1 m 2 which gnrats th following numrical rsults a m a m a m σ P a σ P a σ P a and th analytical rsults ar u(l/3) = m u(2l/3) = m u(l) = m σ(l/6) = P a σ(l/2) = P a σ(5l/6) = P a Th sam analysis has bn don by th TRINITAS program and rsults ar shown in figur 3.14 blow. All frdom prpndicular to th bar hav bn fixd bcaus thr is no 1D bar lmnt implmntd in th program. Displacmnt Lngth = x-comp. = y-comp. =.9844E E - 5. Axial Strss =.2939E+7 Figur 3.14: A TRINITAS analysis Rmarks: Th approximation of th displacmnt always givs th bst agrmnt at th nods. In this cas w hav a rlativ rror about on prcnt. A comparison of th strsss at th cntr of th lmnt show an xact agrmnt. It can b shown that this is not just luck in this cas and that this always holds for xactly this typ problm. In th gnral cas on should conclud that th strsss ar lss accuratly approximatd compard to th displacmnts. Th strss fild is in th gnral cas always discontinuous at th lmnt bordrs and no quilibrium quation in strss is fulfilld at lmnt boundaris.

50 NUMERICAL EXAMPLES A 2D truss problm In this xampl w will focus on a simpl truss fram work consisting of thr bar mmbrs with momnt-fr connctions. This problm will b numrically analyzd by hand calculations in accordanc to th finit lmnt mthod. Th P E, A, L E, A, L y x E, A, L Figur 3.15: A 2D truss problm fram work is modld by thr 2D bar lmnts and th global numbring of frdoms and lmnts ar don in accordanc to figur Th contnts of v 2 u 2 P 1 2 v 1 v 3 u 1 u3 3 Figur 3.16: A 2D finit lmnt msh th global frdom vctor a can b stablishd from this finit lmnt msh numbring. W hav a = u 2 v 2 u 3

51 CHAPTER 3. BARS 37 whr th squnc in-btwn th unknowns is slctd by th analyst and th slction will influnc th apparanc but not rsult of th ntir analysis. Th lmnt stiffnss matrics K i can b stablishd from quation K 1 = EA 4L K 3 = EA L and th Boolan matrics ar C 1 = 1 1 C 2 = K 2 = EA 4L C 3 = W ar now rady to stablish th global stiffnss quation as a sum of xpandd lmnt stiffnss matrics and w obtain as an intrmdiat rsult 3 C T i K i C i a = f EA 4L i= and th final systm to solv thn rads 2 1 EA u L v 2 = v 3 which has th following solution. u 2 a = v 2 = F L 4EA u 3 9 1/ F 4 Th axial strss in ach of th bar lmnts can b calculatd from σ i = EB a i = EB T i d i = E L i u 2 v 2 u 3 [ 1 1 ] [ x ȳ x ȳ. = ] i u 1 v 1 u 2 v 2 F i

52 NUMERICAL EXAMPLES σ i = E L i [ x ȳ x ȳ ] which givs for th thr diffrnt lmnts usd in this analysis. σ1 = F [ 1 8A 3 1 ] 3 9 1/ = F A 3 σ2 = F [ 1 8A ] 3 9 1/ = F A 3 σ3 = F [ 1 4A 1 ] = F 2 2A A numrical analysis of this problm has also bn don by th finit lmnt program TRINITAS. Usd numrical valus ar F = 1. N, L = 1. m, E = P a, A =.1 m 2 and th rsults ar shown in figur 3.17 i u 1 v 1 u 2 v 2 i Axial Strss =.1E+8 Displacmnt Lngth = x-comp. = y-comp. = E-5 Axial Strss = -.1E+8 Figur 3.17: A TRINITAS 2D truss analysis

53 CHAPTER 3. BARS 39 A 3D truss problm Considr th following simpl 3D truss fram work analyzd by th TRINITAS programs. Th msh consists of 4 nods in 6 plans which mans 24 nods and 5(12 + 6) = 9 2-nod 3D truss lmnts. All lmnts hav a cross sctional ara A =.5 m 2 and a Young s modulus E = P a. Th applid load is a concntratd forc vctor P = { } T N. Th gomtrical positions of th nods and som rsults ar shown in figur Th gomtry is dfind by th distancs a x = a y = a z =.4 m. All 4 nods at th bottom ar fully fixd in all dirctions. a x a y P Displacmnt Lngth = x-comp. = y-comp. = z-comp. = a z z Axial Strss =.158E+9 y x Figur 3.18: A TRINITAS 3D truss analysis

54 COMMON PITFALLS AND MISTAKES 3.14 Common Pitfalls and Mistaks As alrady mntiond rigid body motion always hav to b prvntd. That is at last on fixd or prscribd frdom in 1D problms, 3 fixd or prscribd frdoms in 2D cass or 6 fixd or prscribd frdoms in 3D cass hav to b prsnt. Plas b vry carful whn it coms to supprssing of th rigid body rotations spcially in 3D cass bcaus it can actually b rathr tricky. Bsids ths ncssitis to prvnt rigid body motion du to th uniqunss of th solution a coupl of common pitfalls will b discussd blow. Lt us start th discussion by considring a mistak oftn mad whn applying symmtric boundary conditions. Figur 3.19 shows a symmtric 2D truss. Complt modl Symmtry plan Singular modl Corrct symmtric modl Figur 3.19: A 2D symmtric truss problm Th conclusion from this is that a singl lmnt can nvr b lft without support in both nds. Th lowr-lft lmnt in th singular modl abov is fr to rotat bcaus th momnt-fr connction at th right nd and th boundary condition at th lft nd which maks it possibl to mov frly in th vrtical dirction without gnrating any strain th lmnt. From an intuitiv point of viw most nginrs hav a fling that th lmnt will stiffn bcaus it will b longr but this ffct is a non-linar ffct approximatd away and not takn into account in our linar compatibility rlation. Anothr closly rlatd problm is mchanisms du to too fw diagonal mm-

55 CHAPTER 3. BARS 41 brs in th structur. A simpl 2D bridg-lik structur will srv as an xampl. A singular bridg modl A nonsingular bridg modl Figur 3.2: A 2D mchanism problm Th conclusion is that it is not ncssary to put in diagonal mmbrs vrywhr in th modl. It is nough to prvnt all ovrall possibl mchanisms in th structur. In 3D cass both thos two typs of problms discussd in th 2D problms abov will b vr mor frqunt. As a vry last xampl w discuss th th 3D cas studid in th prvious xampl. Think of a cas whr w also put in a nod in th crossing point of ach pair of diagonal mmbrs in th original structur. In this cas ach of ths diagonal mmbrs will b split into two truss lmnt and th modl will b singular. Th rason is that all ths nw xtra nods will not hav any stiffnss in a dirction prpndicular to th plan spannd by th nighboring lmnts. This can asily b curd by just putting in a fixd boundary condition in th appropriat dirction. Gnrally on can conclud that us of truss fram work modls, spcially in 3D, rquirs a thorough undrstanding of what such a modl with its momntfr connctions mans and how this proprty, which in most cass dviats from rality, ffcts th analysis. In many 3D cass it is oftn asir to crat a bam modl of th fram work but this is mor tim-consuming to solv compard to a bar modl and it is oftn possibl to draw th sam ovrall nginring conclusions from th fastr bar modl.

56 SUMMARY 3.15 Summary In vry plan in a 1D bar w hav thr unknowns and thr local quations Box: L Local Equations in 1D Linar Static Elasticity d (A(x)σ(x)) + b(x) = dx σ(x) = E(x)ε(x) ε(x) = du(x) dx to fulfill. Ths quations can b turnd ovr to a wll-posd strong formulation by limination of th strain and th strss and imposing of boundary conditions. Box: S Strong form of 1D Linar Static Elasticity Givn b(x), h and g. Findu(x) such that ( d A(x)E(x) du(x) ) + b(x) = x Ω dx dx u() = g on S g E(L) du(l) dx = h on S h An quivalnt wak form can b achivd by introduction of a wight function and aftr on partial intgration w hav Box: W Wak form of 1D Linar Static Elasticity Givn b(x), h and g. Findu(x) such that L dw(x) dx A(x)E(x)du(x) dx dx = L w(x)b(x) dx + w(l)a(l)h u() = g on S g for all choics of wight functions w(x) which blongs to th st V whr th natural boundary conditions ar implicitly containd. By introducing

57 CHAPTER 3. BARS 43 that th tst functions and th wight functions ar slctd qually w nd up in th following discrt Galrkin formulation. Box: G Galrkin form of 1D Linar Static Elasticity Find a such that c T (Ka f) = c T r = for all choics of th vctor c(th wight function) whr K = f = L L B T (x)a(x)e(x)b(x)dx N T (x)b(x) dx + N T (L)A(L)h L B T (x)ae d N 1 (x) dx g dx. This is still only on singl quation whr now th ssntial boundary conditions ar implicitly containd. Bcaus of th arbitrarinss of th vctor c which implis that th rsidual r must b qual to zro th following systm of linar algbraic quations is obtaind. Box: M Matrix form of 1D Linar Static Elasticity Find a such that Ka = f whr K and f ar known quantitis n l K = Ci T K i C i K i = i=1 n l f = i=1 C T i f i + f h f g f i = xi+1 x i xi+1 x i B T (x)a(x)e(x)b (x) dx N T (x)b(x) dx Th numrical procdur starts from hr and th following work flow can b idntifid. Numrical Work Flow: Split th ntir domain into a numbr of Finit Elmnts of a crtain typ Dfin domain proprtis such as th Young s modulus E and th cross sction A in th lmnts Dfin ssntial boundary conditions such as fixd or prscribd nod displacmnts

58 SUMMARY Dfin natural boundary condition as givn concntratd or distributd loads Mak a global numbring squnc of all involvd unknown frdoms (Normally don automatically by th program) Calculat all lmnt stiffnss matrics K i and xpand and add thos stiffnss cofficints into th appropriat positions in th global stiffnss matrix K Calculat all lmnt load vctors f i and xpand and add thos load contributions into th appropriat positions in th global load vctor f Solv th systm of linar algbraic quation Ka = f by som Gauss limination or LR-factorization look-a-lik procdur. (Furthr discussions concrning how to calculat th unknown vctor a will b givn latr on in th chaptrs to com) Pick up th lmnt frdom vctor a i from th global on a for on lmnt at th tim and calculat th strains and th strsss Invstigat th rsults, hopfully in trms of a nic color pictur showing th dformd structur with th strss lvls in color, and try to xamin th rlvanc of th achivd approximation. In most finit lmnt analyss, aftr th rsults hav bn accptd from an ovrall nginring point of viw, on also hav to accpt th analysis from a numrical point of viw which normally mans a rfinmnt of th msh in som critical part of th domain trying to find out if th accuracy of th numrical rsults is sufficint. At this vry nd, on should not forgt that th rason for doing th finit lmnt analysis was som ovrall nginring qustion concrning how to dsign a crtain pic of quipmnt and th finit lmnt analysis only givs som hints concrning th siz of displacmnts, strains and strsss in a modl of rality.

59 Chaptr 4 Bams Lt us now onc again considr a straight slndr body with a minor changing cross sction A(x) and ara momnt of inrtia I(x). Obsrv, that th gomtry of this structur may b th sam as th on w studid in th prvious bar problm. But in this cas, w no mor rstrict all loads to only act in th local x-dirction. Hr w will instad considr a distributd load q(x) pr unit lngth [N/m] acting prpndicular to th local dirction x and an important unknown is th displacmnt v(x) in th local y-dirction. y, v(x) q(x) I(x), E(x) x Figur 4.1: A typical bam structur Indpndnt from th boundary conditions w will introduc latr it will hav to carry th load q(x) by bnding. That is, th bar displacmnt assumption is no mor applicabl and in this cas w hav to turn ovr to a displacmnt assumption which givs th structur som xtndd possibility to dform. 45

60 THE BEAM DISPLACEMENT ASSUMPTION 4.1 Th Bam Displacmnt Assumption Bam displacmnt assumption Lt us focus on a plan prpndicular to th x-axis and assum that this plan will rmain flat. W now also giv this plan th possibility to undrgo both rotation and translation in th xy-plan as shown in figur 4.2. This is th so calld Bam displacmnt assumption. This discussion will b limitd to plan bnding. That is th load q(x) and displacmnt v(x) will rmain in th xyplan. Not that this also rquirs a cross sction which is symmtric around th z-axis. To includ a gnral 3D dformation of a bam is straight forward as A θ(x) y Undformd body y Dformd body x y A u(x,) v(x) Man surfac Figur 4.2: A typical bam dformation Man surfac long as th shar cntr and th cntr of gravity of th cross sction coincid. Th Man surfac is th plan which is xposd to nithr tnsil strsss nor to comprssiv strsss, if th bam carris a bnding momnt only. This also mans that th fibrs in this surfac will kp its initial lngth and th strain is qual to zro. This man surfac plan always coincid with th xz-plan in th coordinat systm. 4.2 Th Local Equations An xprssion for th horizontal displacmnt u(x, y) in a gnral position A in a bam subjctd to gnral loadings can b dfind by th valu of th y-axis and th slop of th dformd man surfac of th bam. u(x, y) = u(x, ) y dv(x) dx (4.1) This approximation is valid as long as θ is small (say lss than 1 dgr) and w hav θ dv(x)/dx. Calculation of th normal strain in th local x-dirction ε x from this displacmnt assumption is straight forward. ε x (x, y) = u(x, y) x = du(x, ) dx y d 2 v(x) dx 2 (4.2)

61 CHAPTER 4. BEAMS 47 This is th compatibility rlation in this Eulr-Brnoulli bam thory displacmnt assumption. Th physical intrprtation of th first trm is th strain gnratd by an axial forc N carrid by th bam. As a constitutiv rlation w still us a linar lastic matrial and Hook s 1D law applis. On can conclud from this that th strss σ x and strain ε x both varis as linar functions with rspct to th y-dirction. σ x (x, y) = E(x)ε x (x, y) (4.3) For a balanc law w onc again us static quilibrium. First, th rlationship btwn th bnding momnt M(x) and th strss σ x (x, y), and th normal forc N(x) and th strss σ x (x, y) is stablishd. From figur 4.3 w hav for th y da=dydz σ x (x,y) z x Cross sction A(x) Figur 4.3: A singl symmtric bam cross sction bnding momnt M(x) M(x) = yσ x da = A(x) A(x) yσ x dydz (4.4) and for th axial normal forc N(x) N(x) = σ x da = σ x dydz. (4.5) A(x) A(x) Elimination of th strss and th strain by putting quations 4.3 and 4.2 into th quations 4.4 and 4.5 givs th following two quations du(x, ) M(x) = E(x) ydydz E(x) d 2 v(x) dx dx 2 y 2 dydz (4.6) and A(x) } {{ } =S(x)= A(x) } {{ } =I(x)

62 THE LOCAL EQUATIONS =N(x) { }} { du(x, ) N(x) = E(x) dydz dx A(x) } {{ } =A(x) d 2 v(x) dx 2 A(x) ydydz } {{ } =S(x)= (4.7) From this w can draw svral important conclusions. First, w introduc th following closly rlatd gomtrical dfinitions for th cross sction of th bam. A(x) = dydz; S(x) = ydydz; I(x) = y 2 dydz (4.8) A(x) A(x) A(x) Ths ar th cross sction A(x), th first momnt (statical momnt) S(x) and th ara momnt of inrtia I(x). Furthr on, w find that S(x) must b qual to zro whn this intgral is takn ovr th ntir cross sction. Th rason is that th man surfac of th bam always coincid with th cntr of gravity for th cross sction. From quation 4.9 w thn obtain M(x) = E(x)I(x) d 2 v(x) dx 2 (4.9) which mans that th bnding momnt is proportional to th curvatur κ d 2 v(x)/dx 2 of th bam for small displacmnts. Concrning th bnding strss σ x (x, y) in th bam an usful quation is rcivd by putting quation 4.2 and 4.9 into quation 4.3 σ x (x, y) = N(x) A(x) + M(x)y I(x) (4.1) Lt us now study quilibrium for a short slic x of a gnrally loadd bam whr vrtical forc and momnt quilibrium quations giv us q(x) y T(x) T(x+Δx) x N(x) M(x) M(x+Δx) N(x+Δx) Δx Figur 4.4: Forcs and momnts acting on a slic x of th bam

63 CHAPTER 4. BEAMS 49 T (x + x) T (x) + q(x) x = (4.11) M(x + x) M(x) + T x + q(x) x x =. (4.12) 2 Ths quations can b simplifid by us of Taylor s formula, thus T (x + x) T (x) + dt (x) x (4.13) dx M(x + x) M(x) + dm(x) x. (4.14) dx Putting th quations 4.13 and 4.14 into quations 4.11 and 4.12 and ltting x go to zro givs th following gnral quilibrium quations. dt (x) dx = q(x) (4.15) dm(x) = T (x) (4.16) dx By putting quations 4.9 and 4.15 into a drivativ with rspct to x of quation 4.16 w hav liminatd both th bnding momnt M(x) and th shar forc T (x) and w will nd up with th wll known d 2 ( dx 2 E(x)I(x) d 2 ) v(x) dx 2 = q(x) (4.17) Elastic curv diffrntial quation. This is a fourth-ordr diffrntial quation in th transvrsal displacmnt v(x) which is prpndicular to th man surfac of th bam. In ordr to b abl to solv this quation w nd to apply boundary conditions. Four diffrnt boundary conditions, two at ach nd of th bam ar rquird. Elastic curv 4.3 A Strong Formulation Aftr having a known st of boundary conditions it is possibl to stablish a wll-posd strong formulation of this bam dflction problm. A larg varity of th diffrnt combinations of boundary conditions ar possibl at last if w also includ statically indtrminat bams into th discussion. Whn it coms to statically dtrminat systms thr ar mainly two diffrnt combinations. On can hav a momnt-fr support at both nds or a rigid support at on nd and th othr nd fr. Hr w will us this vry last cas as an xampl for th discussion. That is, th boundary conditions ar a known displacmnt v() = g 1 and a known rotation dv()/dx = g 2 at th lft nd and a givn momnt M(L) = M = h 1

64 A WEAK FORMULATION and a givn vrtical concntratd forc T (L) = T = h 2 acting in th right nd. Th lft nd of th intrval will b givn th notation S g and th right nd will b calld S h. Ths four boundary conditions and th Elastic curv diffrntial quation constituts a wll-posd fourth-ordr Boundary-Valu problm S. Box: S Strong form for an Eulr-Brnoulli Bam Givn q(x), h 1, h 2, g 1 and g 2. Findv(x) such that d 2 ( dx 2 E(x)I(x) d 2 ) v(x) dx 2 q(x) = x Ω = ], L[ v() = g 1 on S g E(L)I(L) d 2 v(l) dx 2 = h 1 on S h dv() dx = g 2 on S g d ( E(L)I(L) d 2 ) v(l) dx dx 2 = h 2 on S h Rmarks: Boundary Valu Problm Essntial Natural This formulation S is a Strong formulation of a linar static 1D Bam dflction problm and from a mathmatical point of viw this is a 1D fourth-ordr mixd Boundary Valu Problm. v() = g 1 and dv()/dx = g 2 ar Essntial boundary conditions. If g 1 or g 2 th boundary condition is a non-homognous on. Th total surfac S consists in this 1D cas only of th two nd cross sctions S h and S g. h 1 = M and h 2 = T ar Natural boundary conditions. Th boundary valu problm is mixd bcaus thr ar both ssntial and natural boundary conditions. Latr on w will b awar of that som ssntial boundary conditions always hav to xist to b abl to guarant th uniqunss of th solution of th matrix problm M. 4.4 A Wak Formulation Wight function This Strong formulation can always b transfrrd into an quivalnt Wak formulation by multiplication of an arbitrary Wight function w(x) and intgration ovr th domain Ω. L ( ( d 2 w(x) dx 2 E(x)I(x) d 2 ) ) v(x) dx 2 q(x) dx = (4.18)

65 CHAPTER 4. BEAMS 51 Aftr partial intgration of th first trm w obtain [ w d ( EI d 2 )] L v L ( dw d dx dx 2 EI d 2 ) v dx dx dx 2 dx and aftr a scond partial intgration of this first trm [ w d ( EI d 2 )] L [ v dw dx dx 2 dx EI d 2 v dx 2 ] L L L d 2 w + dx 2 EI d 2 v dx 2 dx wqdx = (4.19) L wqdx =. (4.2) Lt us now introduction som spcific rquirmnts on th choic of wight functions w(x). Mak th choic from an infinit st V of functions whr all mmbrs w i (x) must xplor th following proprtis V = {w i ( ) w i ( ) =, w i( ) = on S g } (4.21) whr w i (x) = dw(x)/dx. By this rstriction on th wight function two of six trms in quation 4.2 abov will vanish and th following Wak formulation is summarizd in box W blow. Box: W Wak form for an Eulr-Brnoulli Bam Givn q(x), T, M, g 1 and g 2. Findv(x) such that L d 2 w(x) dx 2 E(x)I(x) d 2 v(x) dx 2 dx = v() = g 1 dv() dx = g 2 L w(x)q(x) dx + w(l)t dw(l) dx M Rmarks: This wak formulation W will srv a bas for applying a wightd rsidual mthod such as th Galrkin mthod. Two tims partial intgration is prformd bcaus that opns a possibility to latr on nd up in a symmtric systm linar algbraic quations that is mor fficintly solvd in th computr compard to a non-symmtric on. It always rducs th rquirmnts on rgularity of th approximation of th unknown function v(x). Th natural boundary condition is implicitly containd in th intgral quation. It can b shown that th Strong and th Wak formulations ar quivalnt.

66 A GALERKIN FORMULATION 4.5 A Galrkin Formulation Shap functions In all Wightd Rsidual Mthods th approximation of th unknown function, in this cas v(x), is built up from a sum of tst t i functions and a known function t taking car of non-homognous ssntial boundary conditions. n v(x) v h (x) = t 1 (x)a t n (x)a n + t (x) = t i a i + t (x) (4.22) Th wight function w(x) can also b built up as sum of functions φ i as follows n w(x) = φ 1 (x)c 1 + φ 2 (x)c φ n (x)c n = φ i c i (4.23) What is typical to a Galrkin formulation is, as w alrady hav sn, that both th approximation and th wight function ar composd from th sam st of functions, so calld Shap functions N i whr all N i s blong to th st V. whr t i (x) = φ i (x) = N i (x) (4.24) V = {N i ( ) N i ( ) =, N i( ) = on S g } (4.25) A discrt Galrkin finit lmnt formulation for bams can now b basd on th following basic xprssions v(x) = N(x)a + N 1 (x)g 1 + N 2 (x)g 2 (4.26) w(x) = N(x)c or quivalntly w(x) = c T N T (x) (4.27) whr a 1 c 1 N(x) = [ N 1 (x) N 2 (x)... N n (x) ] a 2, a =.., c = c 2... (4.28) i=1 i=1 a n c n Th functions N 1 and N 2 ar not shap functions taking part in th ovrall approximation of th solution. Ths functions ar only ndd for taking car of non-homognous ssntial boundary conditions. That is, g 1 and g 2 and in cass whr both ths known valus ar qual to zro w can drop th two last trms. Latr on w will s that th functions N 1 and N 2 will b closly rlatd to th choic of shap functions N i. But w rquir that ths functions fulfill th following basic proprtis. N 1 () = 1, d N 1 () dx = and N2 () =, d N 2 () dx = 1 (4.29) By putting this rstrictions to ths functions thy ar indpndnt to ach othr and N 2 dosn t influnc th approximation and N 1 dosn t influnc th first drivativ of th approximation v h (x).

67 CHAPTER 4. BEAMS 53 Rmarks: From th Wak formulation, aftr two tims partial intgration, w can conclud that both th tst function and th wight function must fulfill th basic rquirmnt that L d 2 w(x) d 2 v(x) dx 2 dx 2 dx < N i C 1 (4.3) whr th st C 1 is all functions with at last a continuous first-ordr drivativ. This mans that a usful approximation of th bam dflction must hav a continuous first-ordr drivativ ovr th lmnt bordrs. All Hrmitian polynomial xprssions hav this proprty. Hrmitian Bfor w procd by putting quations 4.26 and 4.27 into box W w pick up th opportunity to dfin th B matrix as follows d 2 [ v(x) d dx 2 = 2 N 1 (x) ] d 2 N n (x) dx 2 dx } {{ 2 } =B a + d2 N1 (x) dx 2 g 1 + d2 N2 (x) dx 2 g 2 = Ba + d2 N1 (x) dx 2 g 1 + d2 N2 (x) dx 2 g 2 (4.31) d 2 w(x) dx 2 = c T B T (x) (4.32) and w obtain th following discrt Galrkin formulation for bam problms. Box: G Galrkin form of an Eulr-Brnoulli Bam Find a such that c T (Ka f) = c T r = for all choics of th vctor c (th wight function) whr th global stiffnss matrix K and th global load vctor f ar idntifid as follows K = L B T (x)e(x)i(x)b(x) dx (4.33)

68 A MATRIX FORMULATION f = =f d { }} { L L =f h { }} { N T (x)q(x) dx + N T (L)T dn T (L) M dx ( d B T 2 N1 (x) (x)e(x)i(x) dx 2 ) g 1 + d2 N2 (x) dx 2 g 2 dx = f d + f h f g } {{ } =f g 4.6 A Matrix Formulation (4.34) Onc again th conclusion from th Galrkin formulation is that for all choics of th vctor c th scalar product with th unbalancd rsidual forc vctor r must b qual to zro. c T r = r = Ka f = (4.35) Th only solution to this is that th forc vctor r is qual to a zro vctor which mans that quilibrium is achivd, masurd at th nodal forcs. A matrix problm consisting of n linar algbraic quations can now b stablishd. Box: M Matrix form of an Eulr-Brnoulli Bam Find a such that Ka = f whr K and f ar known quantitis Rmarks: nodal rotations Concrning th global frdom vctor a in this bam formulation it will, as w will undrstand from th nxt sction, not only contain nodal displacmnts. It also includs th nodal rotations. Aftr th numbr of lmnts and th natur of shap functions N i has bn dcidd it is straight forward to calculat both th matrix K and vctor f From a mathmatical point of viw, and xactly as in th bar formulation, this discussion can b summarizd as L S W G M

69 CHAPTER 4. BEAMS 55. Sourcs for rrors in this mathmatical modl of rality ar dviations from rality in th constitutiv and th compatibility rlations, dviations in th slctd boundary conditions and numrical rrors du to us of a limitd numbr of Finit Elmnts with a spcific bhavior in ach lmnt. On can show that th solution to th matrix problm M always xists and has a uniqu solution if th global stiffnss matrix K is non-singular. In this bam formulation w hav to prvnt rigid body motions such as translation in th y-dirction and rotation around th y-axis. By doing so th global stiffnss matrix K is positiv dfinit and th following holds a T Ka > a dt(k) > Latr on w will add axial stiffnss from our 1D bar to this bam. Such an lmnt will hav 6 dgrs of frdoms and this lmnt is a complt 2D fram lmnt. This mans that w hav to prvnt rigid body motion in th global x-dirction. In this 1D bam cas th matrix K is symmtric with a clos to diagonal population with a bandwidth of A 2D 2-nod Bam Elmnt It is now tim to try to find out what is th natur and th bhavior of shap functions fulfilling what is rquird and postulatd so far in this discussion. On th lmnt lvl w hav to b abl to idntify shap functions Ni and frdoms a i associatd to th nods dscribing an approximation for th transvrsal displacmnts v = N1 a Nma m = [ ] a 1 N1... Nm.. = N a (4.36) whr m is th total numbr of frdoms associatd to on lmnt. In th xampl to com w will vry soon raliz that m = 4. As w alrady hav discussd, a propr shap function choic must blong to th st of functions C 1, which consists of all functions with continuous first-ordr drivativ. From this th conclusion is that for ach lmnt w hav to considr th translation in th y-dirction and th slop at both nds of th lmnt. In figur 4.5 an lmnt with four frdoms is sktchd. Th lmnt lngth L can b xprssd in th lmnt coordinats L = x i+1 x i. Lt us now mov focus ovr to a cubic polynomial xprssion such as v(x) = α 1 + α 2 x + α 3 x 2 + α 4 x 3 (4.37) a m

70 A 2D 2-NODE BEAM ELEMENT y, v(x) θ 1 v 1 θ 2 v 2 x x i x i+1 Elmnt intrval Figur 4.5: A 2-nod bam lmnt frdom st which has four unknown constants α 1 to α 4. Ths constants can always b liminatd and xprssd in th lmnt frdom vctor v1 a θ1 = v2. (4.38) θ2 By prforming th following stps it will b possibl to stablish all th lmntlocal shap functions N1 to N4 by idntifications. Lt us start with just rwriting quation 4.37 in a matrix notation, as follows v (x) = [ 1 x x 2 x 3 ] α 1 α 2 α 3 α 4 = F (x)α (4.39) and th slop dv (x)/dx of th bam is θ(x) dv (x) dx = [ 1 2x 3x 2 ] α 1 α 2 α 3 α 4 (4.4) which also is th angl of rotation θ(x) for small angls. Lt us now du to simplicity think of an lmnt with its lft nd at x i = and th right nd at x i+1 = L. By us of quations 4.39 and 4.4 for both nds th following four quations ar obtaind v 1 1 α 1 a θ = 1 = 1 α 2 v 2 1 L L 2 L 3 = Aα (4.41) α 3 θ 2 1 2L 3L 2 α 4 and w now hav a rlationship btwn th lmnt frdom vctor a and th unknown constants in th vctor α. If th invrs A 1 xists it is straight

71 CHAPTER 4. BEAMS 57 forward to liminat vctor α in-btwn quation 4.39 and quation 4.41 writtn as α = A 1 a which givs v (x) = F (x)α = F (x)a 1 a (4.42) and by comparing with quation 4.36 th lmnt-local shap functions N can b idntifid as N (x) = F (x)a 1. (4.43) Aftr stablishing th invrs A 1 th following N (x) = [ 1 x x 2 x ] /L 2 2/L 3/L 2 1/L 2/L 3 1/L 2 2/L 3 1/L 2 multiplication will nd up in th four lmnt-local shap functions whr (4.44) N (x) = [ N 1 (x) N 2 (x) N 3 (x) N 4 (x) ] (4.45) N 1 (x) = 1 3x 2 /L 2 + 2x 3 /L 3 N 2 (x) = x 2x 2 /L + x 3 /L 2 N 3 (x) = 3x 2 /L 2 2x 3 /L 3 N 4 (x) = x 3 /L 2 x 2 /L (4.46) which all ar cubic xprssions in x. In figur 4.6 ths four shap functions ar drawn. By putting quation 4.45 and 4.38 into 4.36 w obtain N i (x) 1. N 1 (x) N 3 (x) N 2 (x) x 1 = N 4 (x) x 2 = L x Figur 4.6: Th lmnt-local bam shap functions N 1 (x) to N 4 (x) v (x) = N (x)a = N 1 (x)v 1 + N 2 (x)θ 1 + N 3 (x)v 2 + N 4 (x)θ 2 (4.47)

72 A 2D 2-NODE BEAM ELEMENT from which w can conclud that th displacmnt v at th nods ar unaffctd by th valu of th rotation. Concrning th slop or th rotation at th nods w hav in a similar fashion that th valu of th displacmnt dosn t influnc th rotation. This is xactly th proprty on should xpct from a Hrmitian polynomial approximation. This can b summarizd as N1 (x 1 ) = N3 (x 2 ) = dn 2 (x 1 ) dx dn1 (x 1 ) = dn 3 (x 2 ) dx dx = dn 4 (x 2 ) dx = 1 = N 2 (x 1 ) = N 4 (x 2 ) = (4.48) and th gnral conclusion is that th approximation of th displacmnt and th rotation at th nods ar uncoupld. From quation 4.31 w can calculat th lmnt-local B matrix whr [ B d (x) = 2 N1 (x) dx 2 d 2 N1 (x) dx 2 d 2 N2 (x) dx 2 d 2 N3 (x) dx 2 d 2 N4 (x) dx 2 d 2 N 2 (x) dx 2 d 2 N 3 (x) dx 2 = B1(x) = 6/L x/L 3 = B2(x) = 4/L + 6x/L 2 = B3(x) = 6/L 2 12x/L 3 = B1 = B4(x) = 6x/L 2 2/L d 2 N 4 (x) dx 2 ] (4.49) (4.5) which, as xpctd, all ar linar functions in x. Now it is tim to calculat th lmnt stiffnss matrix K and for a cas with constant cross sction proprtis w hav K = L B T (x)eib (x)dx = L EI B 1 B 2 B 3 B 4 [ B 1 B 2 B 3 B 4 ] dx. (4.51) Expanding th intgral into ach position of th matrix product B T B only six diffrnt uniqu intgrals ar idntifid. This is bcaus of th symmtry and us of th rlation B 1 = B 3. Each of ths intgrals mans intgration of a

73 CHAPTER 4. BEAMS 59 scond-ordr polynomial xprssion in x. K = EI L B 2 1 dx L B 1B2dx L sym L B 2 1 dx B 2 2 dx L B 1B2dx L Lt us look into dtails concrning only th first on which is L L ( K11 = EI B1 2 dx = EI 6 L 2 [ 48x 3 EI L 6 L B 1B 4dx L B 2B 4dx B 2 1 dx L B 1B4dx + 12x ) 2 dx = L x L 4 72x2 L 5 ] L L B 2 4 dx (4.52) = 12 EI L 3 (4.53) and all th othrs ar achivd similarly. Th final rsult is th following 2D 2-nod 4-frdom bam lmnt. This lmnt is uslss as an lmnt to b usd in bam fram works bcaus it do not rsist any axial forcs. Box: 2D bam lmnt for transvrsal displacmnts only 12 6L 12 6L K = EI 4L 2 6L 2L 2 L L sym 4L 2 This lmnt is sldom implmntd in commrcial programs bcaus it is only of acadmic intrst to b usd during hand calculations. 4.8 An Elmnt Load Vctor Th global load vctor f has bn split into 3 diffrnt parts f = f d + f h f g (4.54) whr th part f d consists of contributions from th distributd load q(x) which has to b valuatd at th lmnt lvl and is givn as follows f d = n i=1 C T i xi+1 x i N T } {{ } =f i (x)q(x) dx. (4.55)

74 AN ELEMENT LOAD VECTOR whr C T i is a Boolan matrix. This matrix dfins to which global frdom ach of th lmnt-local frdoms blongs. (For furthr dtails s th Bar chaptr) Th lmnt load vctor f i can b analyzd furthr if w assum that th givn load q(x) at most varis as a linar function ovr ach lmnt intrval. That is th following linar xprssion q(x) = (1 x L )q 1 + x L q 2 (4.56) is usful whr q 1 and q 2 ar th intnsity of load q(x) at th nods. This ida mans actually an approximation of givn data of th problm which at th first glanc sm to b a bit stupid. What w gain from this ida is that w can procd on furthr stp analytically and th input to th numrical procdur will b th load intnsity at th nods only. From an nginring point of viw on can also argu that if th givn load q(x) varis rapidly with rspct to x w nd to us shortr lmnts anyway, if w would lik to try to catch dtails causd by th rapid load chang. Th lmnt load vctor f i can b rwrittn as f i = L 1 3x 2 /L 2 + 2x 3 /L 3 x 2x 2 /L + x 3 /L 2 3x 2 /L 2 2x 3 /L 3 x 3 /L 2 x 2 /L ( (1 x L )q 1 + x ) L q 2 dx (4.57) whr w hav four diffrnt intgrals to solv. Th polynomial xprssions ar all at most of a fourth ordr dgr. Th rsult from ths four intgrations givs f i = L (7q 1 + 3q 2)/2 L 2 (q 1/2 + q 2/3) L (3q 1 + 7q 2)/2 L 2 (q 1/3 + q 2/2) f i = q L /2 q L 2 /12 q L /2 q L 2 /12) (4.58) Consistnt lmnt load vctor whr th scond xprssion is th spcial cas whn th load q(x) is a constant, that is q 1 = q 2 = q. From this w can conclud that a lin load q(x) also gnrat momnts as wll as nodal loads. In this contxt on should also mntion that som popl argu for an ad hoc rducd lmnt load vctor without nodal momnts which somtims also is implmntd in commrcial programs. Th tratmnt shown hr is calld a Consistnt lmnt load vctor and this is what coms out from this mathmatical formulation.

75 CHAPTER 4. BEAMS 61 Concrning th boundary condition trms f h and f g w hav f h =. T M f g = 12EIg 1 /L 3 + 6EIg 2 /L 2 6EIg 1 /L 2 + 4EIg 2 /L 12EIg 1 /L 3 6EIg 2 /L 2 6EIg 1 /L 2 + 4EIg 2 /L.. (4.59) whr ths ar two global load vctor contributions with th sam numbr of rows as in th global unknown frdom vctor a. Plas obsrv that th dimnsion of th trms altrs btwn forc and momnt and in this discussion th ssntial boundary conditions occur at th lft nd of th bam and th natural bc. at th right nd. Svral othr combinations ar also of cours possibl. 4.9 A 2D Bam Elmnt with Axial Stiffnss In most ral spac fram structurs it is also ncssary to tak axial dformations into account. This is, as alrady mntiond, not includd in th prvious lmnt. Fortunatly it is asy to cur this dficincy of th lmnt. In cass with small dformations it is a good approximation saying that dformations in th axial and th transvrsal dirctions volvs indpndnt of ach othr. y, v(x) θ 1 v 1 u 1 x i x i+1 θ 2 v 2 u 2 x Elmnt lngth Figur 4.7: A 2D 2-nod fram lmnt frdom st That is th bar problm and th bam problm can b solvd indpndnt from ach othr using th sam finit lmnt msh. Such a bam lmnt rsisting axial stiffnss also is oftn calld a Fram lmnt Fram lmnt

76 A 3D SPACE FRAME ELEMENT Box: 2D 2-nod fram lmnt K = 1 L EA EA 12EI/L 2 6EI/L 12EI/L 2 6EI/L 4EI 6EI/L 2EI sym EA 12EI/L 2 6EI/L Plas obsrv that th squnc btwn th six frdoms in th lmnt is not critical. This slction is only on of svral othrs which will work qually wll. Most oftn txtbooks starts, as has bn don hr, with th axial frdom u 1 and continu with th two othr frdoms at th first nod v 1 and θ 1 and nd up with th sam frdoms at th scond nod of th lmnt. Still this lmnt lacks on important fatur to b a candidat for implmntation in a gnral purpos finit lmnt program. This 2D lmnt stiffnss matrix is stablishd along a local dirction and in a ral spac fram application on most likly will find mmbrs running in arbitrary dirctions in 2D or 3D spac. This dfct will b curd in th nxt sction whr w will discuss a 3D bam lmnt in an arbitrary dirction in 3D spac. 4EI 4.1 A 3D Spac Fram Elmnt To stablish a gnral 3D bam lmnt for any spac fram analysis is rathr straightforward. To add bnding out of th plan it is just a mattr of suprimposing of a scond load systm bnding th bam in th xy-plan. For an applid z y θ x1 w 1 v 1 θ x2 w 2 v 2 x θ y1 θ z1 u 1 A, E, I y, I z, K, L θ y2 θ z2 u 2 Figur 4.8: A 3D 12-frdom bam lmnt dfind in a local systm torqu trying to twist th bam th obvious choic is to us St. Vnant torsional thory. An lmnt stiffnss matrix K can now asily b stablish from what

77 CHAPTER 4. BEAMS 63 w alrady know, as follows a 1 a 1 b 1 b 2 b 1 b 2 c 1 c 2 c 1 c 2 d 1 d 1 c 3 c 2 c 4 K = b 3 b 2 b 4 a 1 b 1 b 2 c 1 c 2 d 1 s y m. c 3 b 3 (4.6) whr a 1 = EA/L d 1 = GK/L b 1 = 12EI z /L 3 b 2 = 6EI z /L 2 b 3 = 4EI z /L b 4 = 2EI z /L c 1 = 12EI y /L 3 c 2 = 6EI y /L 2 c 3 = 4EI y /L c 4 = 2EI y /L and G is th torsional modulus of th matrial and K is th torsional proportional constant for th cross sction. Rmarks: This lmnt works suprior with rspct to classical bam thory as long as th cross sction has two axs of symmtry. In cass with opn thin-walld cross sctions warping of th cross sction can b significant and on should b carful. This typically happns in cross sctions whr th cntroid and th shar cntr of th cross sction do not coincid. Th slctd squnc btwn th lmnt frdoms a is in this cas a T = { u 1 v 1 w 1 θ x1 θ y1 θ z1 u 2 v 2 w 2 θ x2 θ y2 θ z2 } On usful tchniqu for ovrcoming this problm is to us shll lmnt for modling of such cross sction. W now nd an lmnt which can b usd in an arbitrary dirction in 3D spac. This can b takn car of by th ruls for transformation of a vctor in 3D spac. Plas not that th prvious coordinat systm in figur 4.8 now has bcom a prim coordinat systm or local coordinat systm and th lmnt frdoms prsnt in figur 4.9 ar not th sam as th ons in figur 4.8.

78 STRESS AND STRAIN CALCULATIONS w 2 z z y y x x θ x1 θ y1 w 1 θ z1 v 1 u 1 A, E, I y, I z, K, L θ x2 θ y2 θ z2 v 2 u 2 Figur 4.9: A 3D global 12-frdom bam lmnt Lt us focus on a vctor u dfind by componnts in th global coordinat systm and th sam vctor now calld u dfind by componnts in th local coordinat systm. Th rlation btwn th componnts of this vctors is u = Λu (4.61) whr th matrix Λ is 3x3 matrix containing dirction cosins btwn th axs of th two systms. Th transformation of th lmnt frdom vctor a in th local systm to lmnt frdoms a dfind in th global systm is givn by a = Λ Λ Λ Λ = T a (4.62) and this can can b usd for transformation of th lmnt-local quilibrium quation K a = f. (4.63) whr th matrix K is quivalnt to what is calld K in quation 4.6. By putting quation 4.62 into quation 4.63 and multiplying with T T from th lft w can idntify a global 3D lmnt stiffnss matrix K and a global lmnt load vctor f for a bam or fram lmnt rsisting axial forcs and twisting momnt and thrfor usful in gnral 3D spac fram analysis. T f T } T K {{ T } a = T } {{ } (4.64) =K =f 4.11 Strss and Strain Calculations As for bar lmnts and all othr lmnts basd on basically th sam finit lmnt formulation, th strain and th strss is calculatd lmnt by lmnt

79 CHAPTER 4. BEAMS 65 at th nd of th numrical analysis. W hav by gnralizing quation 4.2 and us of quation 4.3 ( du(x, ) σ x (x, y, z) = Eε x (x, y, z) = E y d 2 v(x) dx dx 2 z d 2 ) w(x) dx 2 (4.65) whr th displacmnt componnt in th local y-dirction is w(x) and v(x) is th displacmnt in th y-dirction. Th first trm is th axial strss du to a forc acting along th local dirction of th bam. Th axial and th bnding strsss togthr can b xprssd in th lmnt frdom vctor as follows ( { } 1 u σ x (x, y, z) = E L [ 1 1 ] 1 u 2 y [ v1 ] B1 B2 B3 B4 θz1 v2 θz2 z [ ] B1 B2 B3 B4 w 1 θ y1 w 2 θ y2 ) (4.66) This xprssion is usd in th TRINITAS program for plotting of th strss lvl on th surfac of vry bam lmnt.

80 NUMERICAL EXAMPLES 4.12 Numrical Exampls A 2D consol bam Lt us considr a simpl standard cas which also has an analytical solution. Th lft nd of th bam has a compltly rigid support and th right nd is fr. A constant load q(x) = q pr unit lngth acts along th ntir lngth of th bam. At th right nd acts a concntratd forc P. That is, from a comparison with box S w hav th following valus of th boundary conditions g 1 = g 2 = h 1 = and h 2 = P. y, v(x) P q x I, E, L Figur 4.1: A 2D consol bam xampl Th xact solution to this problm is rcivd from th Elastic curv quation 4.17 which in this cas can b simplifid as follows d 4 v(x) dx 4 = q /EI. (4.67) bcaus both th cross sction gomtry and th matrial do not chang along th bam. Aftr intgrating four tims without boundaris w hav d 3 v(x) dx 3 = q x/ei + C 1 d 2 v(x) dx 2 = q x 2 /2EI + C 1 x + C 2 dv(x) dx = q x 3 /6EI + C 1 x 2 /2 + C 2 x + C 3 v(x) = q x 4 /24EI + C 1 x 3 /6 + C 2 x 2 /2 + C 3 x + C 4. Four unknown constants C 1 to C 4 hav to b dtrmind from th givn boundary conditions. Th lft nd conditions g 1 = g 2 = imply immdiatly that C 3 =

81 CHAPTER 4. BEAMS 67 C 4 =. Th conditions at th right nd givs h 1 = d 2 v(l) dx 2 = q L 2 /2EI + C 1 L + C 2 = h 2 = P d 3 v(l) dx 3 = q L/EI + C 1 = P/EI C 1 and C 2 qual to C 1 = P/EI q L/EI C 2 = P L/EI + q L/2 2 EI and all constants ar known and th solution v(x) is v(x) = q L 4 (6( x 24EI L )2 4( x L )3 + ( x L )4) + P ( L3 3( x 6EI L )2 ( x L )3). Lt us now solv this problm numrically by Finit Elmnts and hand calculations. Us only on lmnt with just a vrtical displacmnt v and a rotation θ in th right nd. This on lmnt problm with its lft nd fixd will nd up in th following two global stiffnss quations dscribing th bhavior of th consol bam. EI L 3 [ 12 6L 6L 4L 2 ] { v θ } = { q L/2 + P q L 2 /12 Th two global frdoms v and θ can b calculatd as follows { } v = 1 [ ] { } 4L 2 6L L 3 q L/2 + P θ 12L 2 6L 12 EI q L 2 /12 { } { } v q L = 4 /8EI + P L 3 /3EI θ q L 3 /6EI + P L 2. /2EI From this rsult w can conclud that both th displacmnt v(x = L) and th rotation θ = dv(x = L)/dx at th right nd ar xactly th sam as from th analytical solution abov. Lt us also compar th displacmnts along th bam at an arbitrary x-valu. By us of th valus of th frdoms v and θ abov puttd into th quations 4.36 and 4.46 w obtain v (x) = N 3 (x)v + N 3 (x)θ = which aftr simplification givs } (3( x L )2 2( x ( L )3) q L 4 8EI + P L3 3EI ( ( x L )3 ( x L )2) L v(x) = q L 4 (5( x 24EI L )2 2( x L )3) + P ( L3 3( x 6EI L )2 ( x L )3). ) + ( q L 3 6EI + P L2 2EI )

82 NUMERICAL EXAMPLES A scond important conclusion which can b drawn from this rsults is that th finit lmnt approximation in th intrior of th lmnt is xact for th part of th displacmnt gnratd by th concntratd load P but not for th part manating from th distributd load q. This is obvious and this conclusion could b drawn as a cubic polynomial was usd for th bam shap function which is not sufficint for catching th fourth ordr trm originating from th particular solution to th lastic curv quation for this xampl. That is, in linar static lastic bam modls it is nough using just on lmnt for intrvals without any distributd load. If w would lik to do a numrical comparison with a finit lmnt program w nd numrical valus. Us L = 1. m, P = 1. N, q = 1. N/m, E = Pa. Assum that th cross sction has a hight h and a width b whr b = 4h =.1 m which givs th ara momnt of inrtia I = m 4. At th tip of th bam w hav th bst agrmnt btwn this analytical solution and a finit lmnt solution and from th two load systms w rciv v(x = L) = m from both th analytical and a finit lmnt solution with only on lmnt. Displacmnt Lngth = x-comp. = y-comp. = z-comp. = Max Axial+Bnding Strss =.531E+9 Figur 4.11: A on lmnt TRINITAS analysis

83 CHAPTER 4. BEAMS Summary Aftr assuming that vry cut prpndicular to th man surfac of th bam will rmain plan th following local quations can b statd. Box: L Local Eulr-Brnoulli Bam Equations d 2 M(x) dx 2 + q(x) =, M(x) = σ x (x, y) = E(x)ε x (x, y) ε x (x, y) = du(x, ) dx y d 2 v(x) dx 2 A(x) yσ x (x, y)dydz By limination of th momnt M(x), th strain ε x (x, z) and th strss σ x (x, z) th following 1D fourth-ordr Boundary-Valu problm is rcivd. Box: S Strong form for an Eulr-Brnoulli Bam Givn q(x), h 1, h 2, g 1 and g 2. Findv(x) such that d 2 ( dx 2 E(x)I(x) d 2 ) v(x) dx 2 q(x) = x Ω = ], L[ v() = g 1 on S g E(L)I(L) d 2 v(l) dx 2 = h 1 on S h dv() dx = g 2 on S g d ( E(L)I(L) d 2 ) v(l) dx dx 2 = h 2 on S h Th following quivalnt wak formulation is stablishd aftr partial intgrations and rstricting th wight function to b qual to zro whr ssntial boundary conditions ar prsnt. Box: W Wak form for an Eulr-Brnoulli Bam Givn q(x), T, M, g 1 and g 2. Findv(x) such that L d 2 w(x) dx 2 E(x)I(x) d 2 v(x) dx 2 dx = v() = g 1 dv() dx = g 2 L w(x)q(x) dx + w(l)t dw(l) dx M

84 SUMMARY By rquiring that both th approximation and th wight function ar built up from th sam st of functions a discrt Galrkin formulation is achivd. Box: G Galrkin form of an Eulr-Brnoulli Bam Find a such that c T (Ka f) = c T r = for vry choic of th vctor c (th wight function) Th singl quation in th Galrkin formulation abov, which shall b intrprtd as a scalar product btwn a row vctor c T and a column vctor r, will now turn ovr into a systm of linar algbraic quations (M) bcaus th rsidual vctor r must b qual to a zro vctor. Box: M Matrix form of an Eulr-Brnoulli Bam Find a such that Ka = f whr K and f ar known quantitis In this bam cas discussion th solution vctor a to this matrix problm will contain both displacmnts and rotations at th nods. From hr th numrical procdur starts and th following work flow can b idntifid. Numrical Work Flow: Split th ntir domain into a numbr of finit lmnts Dfin domain proprtis such as th Young s modulus E, th cross sction A and th ara momnt of inrtia I in th lmnts Dfin ssntial boundary conditions such as fixd or prscribd nod displacmnts and/or rotations Dfin natural boundary condition as givn concntratd or distributd forcs or momnt loads Mak a global numbring squnc of all involvd unknown frdoms (Normally don automatically by th program)

85 CHAPTER 4. BEAMS 71 Calculat all lmnt stiffnss matrics K i and xpand and add ths stiffnss cofficints into th appropriat positions in th global stiffnss matrix K Calculat all lmnt load vctors f i and xpand and add ths load contributions into th appropriat positions in th global load vctor f Solv th systm of linar algbraic quation Ka = f by som Gauss limination or LR-factorization look-a-lik procdur. (Furthr discussions concrning how to calculat th unknown vctor a will b givn latr on in th chaptrs to com) Pick up th lmnt frdom vctor a i from th global on a for on lmnt at th tim and calculat th strains and th strsss Invstigat th rsults, hopfully in trms of a nic color pictur showing th dformd structur with th strss lvls in color, and try to xamin th rlvanc of th achivd approximation. In most finit lmnt analyss, aftr th rsults hav bn accptd from an ovrall nginring point of viw, on also hav to accpt th analysis from a numrical point of viw which normally mans a rfinmnt of th msh in som critical part of th domain trying to find out if th accuracy of th numrical rsults is sufficint. Finally, on should not forgt that th rason for doing th finit lmnt analysis was som ovrall nginring qustion concrning how to dsign a crtain pic of quipmnt and th finit lmnt analysis only givs som hints concrning th siz of displacmnts, strains and strsss in a modl of rality.