2y ago

119 Views

22 Downloads

9.93 MB

63 Pages

Transcription

The Finite Element Method for the Analysis ofNon-Linear and Dynamic SystemsProf. Dr. Eleni ChatziLecture 1 - 16 September, 2014Institute of Structural EngineeringMethod of Finite Elements II1

Course InformationInstructorProf. Dr. Eleni Chatzi, email: chatzi@ibk.baug.ethz.chOffice Hours: HIL E14.3, Wednesday 10:00-12:00 or by emailAssistantAdrian Egger, HIL E13.1, email: egger@ibk.baug.ethz.chCourse WebsiteLecture Notes and Homeworks will be posted at:http://www.ibk.ethz.ch/ch/educationSuggested ReadingNonlinear Finite Elements for Continua and Structures by T.Belytschko, W. K. Liu, and B. Moran, John Wiley and Sons, 2000The Finite Element Method: Linear Static and Dynamic FiniteElement Analysis by T. J. R. Hughes, Dover Publications, 2000The Finite Element Method Vol. 2 Solid Mechanics by O.C.Zienkiewicz and R.L. Taylor, Oxford : Butterworth Heinemann, 2000Institute of Structural EngineeringMethod of Finite Elements II2

Course OutlineReview of the Finite Element method - Introduction toNon-Linear AnalysisNon-Linear Finite Elements in solids and Structural Mechanics-Overview of Solution MethodsContinuum Mechanics & Finite DeformationsLagrangian FormulationStructural ElementsDynamic Finite Element Calculations- Integration Methods- Mode SuperpositionEigenvalue ProblemsSpecial Topics- Boundary Element & Extended Finite Element methodsInstitute of Structural EngineeringMethod of Finite Elements II3

Grading PolicyPerformance Evaluation - Homeworks (100%)HomeworkHomeworks are due in class 2-3 weeks after assignmentComputer Assignments may be done using any coding language(MATLAB, Fortran, C, MAPLE) - example code will beprovided in MATLABCommercial software such as CUBUS, ABAQUS and SAP willalso be used for certain AssignmentsHomework Sessions will be pre-announced and it is advised to bringa laptop along for those sessionsInstitute of Structural EngineeringMethod of Finite Elements II4

Lecture #1: StructureLecture #1: StructureReview of the Finite Element MethodStrong vs. Weak FormulationThe Finite Element (FE) formulationThe Iso-Parametric MappingExamplesThe Bar ElementThe Beam ElementInstitute of Structural EngineeringMethod of Finite Elements II5

Review of the Finite Element Method (FEM)Classification of Engineering SystemsDiscreteContinuousq y dyq xdydxq x dxq yh1h2Flowof waterLPermeable SoilImpermeable RockF KXDirect Stiffness Methodk 2φ 2x 2φ 2y 0Laplace EquationFEM: Numerical Technique for approximating the solution of continuoussystems. We will use a displacement based formulation and a stiffnessbased solution (direct stiffness method).Institute of Structural EngineeringMethod of Finite Elements II6

Review of the Finite Element Method (FEM)How is the Physical Problem formulated?The formulation of the equations governing the response of a system underspecific loads and constraints at its boundaries is usually provided in theform of a differential equation. The differential equation also known as thestrong form of the problem is typically extracted using the following setsof equations:123Equilibrium EquationsaL axex. f (x) R (L x)2Constitutive RequirementsEquationsex. σ E Kinematics Relationshipsduex. dxInstitute of Structural EngineeringAxial bar Exampleq(x) axRxaLaxRf(x)L-xMethod of Finite Elements II7

Review of the Finite Element Method (FEM)How is the Physical Problem formulated?Differential Formulation (Strong Form) in 2 DimensionsQuite commonly, in engineering systems, the governing equations are of a2second order (derivatives up to u 00 or 2 ux ) and they are formulated in termsof variable u, i.e. displacement:Governing Differential Equation ex: general 2nd order PDE222 u u uA(x, y ) 2 ux 2B(x, y ) x y C (x, y ) 2 yu φ(x, y , u, y, y )Problem ClassificationBoundary Condition ClassificationB 2 AC 0 ellipticB 2 AC 0 parabolicB 2 AC 0 hyperbolicInstitute of Structural EngineeringEssential (Dirichlet): u(x0 , y0 ) u0order m 1 at most for C m 1 uNatural (Neumann): y(x0 , y0 ) u̇0order m to 2m 1 for C m 1Method of Finite Elements II8

Review of the Finite Element Method (FEM)Differential Formulation (Strong Form) in 2 DimensionsThe previous classification corresponds to certain characteristics for eachclass of methods. More specifically,Elliptic equations are most commonly associated with a diffusive ordispersive process in which the state variable u is in an equilibriumcondition.Parabolic equations most often arise in transient flow problems wherethe flow is down gradient of some state variable u. Often met in theheat flow context.Hyperbolic equations refer to a wide range of areas includingelasticity, acoustics, atmospheric science and hydraulics.Institute of Structural EngineeringMethod of Finite Elements II9

Strong Form - 1D FEMReference ProblemConsider the following 1 Dimensional (1D) strong form (parabolic) duc(x) f(x) 0dxd c(0) u(0) C1dxu(L) 0ddx(Neumann BC)(Dirichlet BC)Physical Problem (1D)Diff. EquationOne dimensional Heatflow𝑑𝑑𝑇 𝐴𝑘 𝑄 0𝑑𝑥𝑑𝑥𝑑𝑑𝑢 𝐴𝐸 𝑏 0𝑑𝑥𝑑𝑥Axially Loaded BarInstitute of Structural EngineeringQuantitiesT temperatureA areak thermalconductivityQ heat supplyu displacementA areaE Young’smodulusB axial loadingMethod of Finite Elements IIConstitutiveLawFourier𝑞 𝑘 𝑑𝑇/𝑑𝑥𝑞 heat fluxHooke𝜎 𝐸𝑑𝑢/𝑑𝑥𝜎 stress10

Weak Form - 1D FEMApproximating the Strong FormThe strong form requires strong continuity on the dependent fieldvariables (usually displacements). Whatever functions define thesevariables have to be differentiable up to the order of the PDE thatexist in the strong form of the system equations. Obtaining theexact solution for a strong form of the system equation is a quitedifficult task for practical engineering problems.The finite difference method can be used to solve the systemequations of the strong form and obtain an approximate solution.However, this method usually works well for problems with simpleand regular geometry and boundary conditions.Alternatively we can use the finite element method on a weakform of the system. This weak form is usually obtained throughenergy principles which is why it is also known as variational form.Institute of Structural EngineeringMethod of Finite Elements II11

Weak Form - 1D FEMFrom Strong Form to Weak formThree are the approaches commonly used to go from strong to weakform:Principle of Virtual WorkPrinciple of Minimum Potential EnergyMethods of weighted residuals (Galerkin, Collocation, LeastSquares methods, etc)*We will mainly focus on the third approach.Institute of Structural EngineeringMethod of Finite Elements II12

Weak Form - 1D FEMFrom Strong Form to Weak form - Approach #1Principle of Virtual WorkFor any set of compatible small virtual displacements imposed on the bodyin its state of equilibrium, the total internal virtual work is equal to thetotal external virtual work.ZWint T τ dΩ Wext ΩZūT bdΩ ΩZūST TS dΓ ΓXūiT RC iiwhereTS : surface traction (along boundary Γ)b: body force per unit areaRC : nodal loadsū: virtual displacement : virtual strainτ : stressesInstitute of Structural EngineeringMethod of Finite Elements II13

Weak Form - 1D FEMFrom Strong Form to Weak form - Approach #2Principle of Minimum Potential EnergyApplies to elastic problems where the elasticity matrix is positive definite,hence the energy functional Π has a minimum (stable equilibrium).Approach #1 applies in general.The potential energy Π is defined as the strain energy U minus the work ofthe external loads WΠ U WZ1 T C dΩU 2 ΩZZW ūT bdΩ ΩΓTūST Ts dΓT XiūTi RCi(b Ts , RC as defined previously)Institute of Structural EngineeringMethod of Finite Elements II14

Weak Form - 1D FEMFrom Strong Form to Weak form - Approach #3Galerkin’s MethodGiven an arbitrary weight function w, whereS {u u C 0 , u(l) 0}, S 0 {w w C 0 , w (l) 0}C 0 is the collection of all continuous functions.Multiplying by w and integrating over ΩZlw (x)[(c(x)u 0 (x))0 f (x)]dx 00[w (0)(c(0)u 0 (0) C1 ] 0Institute of Structural EngineeringMethod of Finite Elements II15

Weak Form - 1D FEMUsing the divergence theorem (integration by parts) we reduce theorder of the differential:Zl0wg dx [wg ]l00Z lgw 0 dx0The weak form is then reduced to the following problem. Also, inwhat follows we assume constant properties c(x) c const.Find u(x) S such that:Zl00Zlw cu dx 0wfdx w (0)C10S {u u C 0 , u(l) 0}S 0 {w w C 0 , w (l) 0}Institute of Structural EngineeringMethod of Finite Elements II16

Weak FormNotes:1Natural (Neumann) boundary conditions, are imposed on thesecondary variables like forces and tractions. uFor example, y(x0 , y0 ) u̇0 .2Essential (Dirichlet) or geometric boundary conditions, are imposedon the primary variables like displacements.For example, u(x0 , y0 ) u0 .3A solution to the strong form will also satisfy the weak form, but notvice versa.Since the weak form uses a lower order of derivatives it canbe satisfied by a larger set of functions.4For the derivation of the weak form we can choose any weightingfunction w , since it is arbitrary, so we usually choose one that satisfieshomogeneous boundary conditions wherever the actual solutionsatisfies essential boundary conditions. Note that this does not holdfor natural boundary conditions!Institute of Structural EngineeringMethod of Finite Elements II17

FE formulation: DiscretizationHow to derive a solution to the weak form?Step #1:Follow the FE approach:Divide the body into finite elements, e, connected to each otherthrough nodes.𝑥1𝑒𝑒𝑥2𝑒Then break the overall integral into a summation over the finiteelements:"#Z xeX Z x2e2w 0 cu 0 dx wfdx w (0)C1 0ex1ex1eInstitute of Structural EngineeringMethod of Finite Elements II18

1D FE formulation: Galerkin’s MethodStep #2: Approximate the continuous displacement using a discreteequivalent:Galerkin’s method assumes that the approximate (or trial) solution, u, canbe expressed as a linear combination of the nodal point displacements ui ,where i refers to the corresponding node number.Xu(x) u h (x) Ni (x)ui N(x)uiwhere bold notation signifies a vector and Ni (x) are the shape functions.In fact, the shape function can be any mathematical formula that helps usinterpolate what happens at points that lie within the nodes of the mesh.In the 1-D case that we are using as a reference, Ni (x) are defined as 1stdegree polynomials indicating a linear interpolation.As will be shown in the application presented in the end of this lecture, for thecase of a truss element the linear polynomials also satisfy the homogeneousequation related to the bar problem.Institute of Structural EngineeringMethod of Finite Elements II19

1D FE formulation: Galerkin’s MethodShape function Properties:Bounded and ContinuousOne for each nodeNie (xje ) δij , where 1 if i jδij 0 if i 6 jThe shape functions can be written as piecewise functions of the xcoordinate:This is not a convenient notation. x xInsteadof using the global coordinatei 1 ,xi 1 x xi x, things become simplified when xi xi 1xi 1 xNi (x) using coordinate ξ referring to the, xi x xi 1 x xlocalsystem of the element (see page i 1i 0,otherwise25).Institute of Structural EngineeringMethod of Finite Elements II20

1D FE formulation: Galerkin’s MethodStep #2: Approximate w (x) using a discrete equivalent:The weighting function, w is usually (although not necessarily) chosen tobe of the same form as uXw (x) w h (x) Ni (x)wi N(x)wii.e. for 2 nodes:N [N1 N2 ]u [u1u2 ]Tw [w1w2 ]TAlternatively we could have a Petrov-Galerkin formulation, where w (x) isobtained through the following relationships:w (x) Xhe dNi(Ni δ)wiσ dxiδ coth(Pe e2) e2PeInstitute of Structural Engineeringcoth e x e xe x e xMethod of Finite Elements II21

1D FE formulation: Galerkin’s MethodNote: Matrix vs. Einstein’s notation:In the derivations that follow it is convenient to introduce theequivalence between the Matrix and Einstein’s notation. So far wehave approximated:Xu(x) Ni (x)ui (Einstein0 s notation) N(x)u (Matrix notation)iw (x) XNi (x)wi N(x)w(similarly )iAs an example, if we consider an element of 3 nodes:u(x) 3XNi (x)ui N1 u1 N2 u2 N3 u3 i u1u(x) [N1 N2 N3 ] u2 N(x)uu3Institute of Structural EngineeringMethod of Finite Elements II22

1D FE formulation: Galerkin’s MethodStep #3: Substituting into the weak formulation and rearrangingterms we obtain the following in matrix notation:Z lZ l00w cu dx wfdx w (0)C1 0 0Z0l(wT NT )0 c(Nu)0 dx 0ZlwT NT fdx wT N(0)T C1 00Since w, u are vectors, each one containing a set of discrete valuescorresponding at the nodes i, it follows that the above set of equations canbe rewritten in the following form, i.e. as a summation over the wi , uicomponents (Einstein notation):ZlX0iZ lf0dNi (x)uidxX! c wj Nj (x)dx jInstitute of Structural EngineeringXj dNj (x) wjdxdxXjwj Nj (x)C1 0x 0Method of Finite Elements II23

1D FE formulation: Galerkin’s MethodThis is rewritten as,X"Zwj0jlXidNi (x) dNj (x)cuidxdx!# fNj (x)dx (Nj (x)C1 ) x 0 0The above equation has to hold wj since the weighting function w (x) isan arbitrary one. Therefore the following system of equations has to hold:!Z l XdNi (x) dNj (x)cui fNj (x)dx (Nj (x)C1 ) x 0 0 j 1, ., ndxdx0iAfter reorganizing and moving the summation outside the integral, thisbecomes:#"Z lX Z l dNi (x) dNj (x)cui fNj (x)dx (Nj (x)C1 ) x 0 0 j 1, ., ndxdx00iInstitute of Structural EngineeringMethod of Finite Elements II24

1D FE formulation: Galerkin’s MethodWe finally obtain the following discrete system in matrix notation:Ku fwhere writing the integral from 0 to l as a summation over thesubelements we obtain:eeK Ae K K Zx2eNT,x cN,x dxx1ef Ae f e f e Zx2ex1eZx2e BT cBdxx1eNT fdx NT h x 0where A is not a sum but an assembly (see page and, x denotesdifferentiation with respect to x.dN(x)is known as the strain-displacementIn addition, B N,x dxmatrix.Institute of Structural EngineeringMethod of Finite Elements II25

1D FE formulation: Iso-Parametric FormulationIso-Parametric MappingThis is a way to move from the use of global coordinates (i.e.in(x, y )) into normalized coordinates (usually (ξ, η)) so that the finallyderived stiffness expressions are uniform for elements of the sametype.𝑥1𝑒𝑥2𝑒𝑥 11𝜉Shape Functions in Natural Coordinatesx(ξ) XNi (ξ)xie N1 (ξ)x1e N2 (ξ)x2ei 1,21N1 (ξ) (1 ξ),2Institute of Structural Engineering1N2 (ξ) (1 ξ)2Method of Finite Elements II26

1D FE formulation: Iso-Parametric FormulationMap the integrals to the natural domain element stiffness matrix.Using the chain rule of differentiation for N(ξ(x)) we obtain:Z 1Z x2eNT,x cN,x dx (N,ξ ξ,x )T c(N,ξ ξ,x )x,ξ dξKe x1ewhere N,ξ and x,ξ 1d dξ1(121(12 ξ) ξ) 1212 x e x1ehdx 2 J (Jacobian) and h is the element lengthdξ22ξ,x dξ J 1 2/hdxFrom all the above,Ke cx2e x1e 1 1 11 Similary, we obtain the element load vector:Z x2eZ 1NT (ξ)fx,ξ dξ NT (x)h x 0fe NT fdx NT h x 0 x1e 1Note: the iso-parametric mapping is only done for the integral.Institute of Structural EngineeringMethod of Finite Elements II27

1D FE formulation: Galerkin’s MethodSo what is meant by assembly? (Ae )It implies adding the components of the stiffness matrix that correspond tothe same degrees of freedom (dof).In the case of a simple bar, it is trivial as the degrees of freedom (axialdisplacement) are as many as the nodes:2:K21:K112Red indicates the node eachcomponent corresponds to31221 1 Element Stiffness Matrices (2x2): 𝐾 1 ℎ 1 1𝑐121 1𝑐Total Stiffness Matrix (4x4): 𝐾 ℎ 1 1 10 1Institute of Structural Engineering12,30 1 131 1𝐾2 ℎ 1 1𝑐23123Method of Finite Elements II28

1D FE formulation: Galerkin’s MethodIn the case of a frame with beam elements, the stiffness matrix of the elements istypically of 4x4 size, corresponding to 2 dofs on each end (a displacement and arotation):φ2φ1u1*The process will be shown explicitlyduring the HW sessionsu2yφ211:K122:Ku2xGreen indicates the dof eachcomponent corresponds to23u3φ3u1𝐾1 111 𝐾121Element Stiffness Matrices (4x4): 𝐾 1𝐾13 1 ��231𝐾331𝐾34φ2φ212𝐾44 𝐾221Total Stiffness Matrix (2x2): 𝐾 𝐾340Institute of Structural Engineeringu2x1𝐾14𝐾2 u1 1112𝐾24 φ1 𝐾1221 u2y, 𝐾 2𝐾34𝐾13 1 φ22𝐾44 222𝐾232𝐾242𝐾132𝐾232𝐾332𝐾340 φ22 u2y 𝐾242u2x𝐾11Method of Finite Elements IIφ32𝐾14 2𝐾24 2 𝐾34 2𝐾44 u2xφ2u3φ3Fixed dofs arenot included inthe totalstiffness matrix29

Axially Loaded Bar ExampleA. Constant End LoadGiven: Length L, Section Area A, Young’s modulus EFind: stresses and deformations.Assumptions:The cross-section of the bar does not change after loading.The material is linear elastic, isotropic, and homogeneous.The load is centric.End-effects are not of interest to us.Institute of Structural EngineeringMethod of Finite Elements II30

Axially Loaded Bar ExampleA. Constant End LoadStrength of Materials ApproachFrom the equilibrium equation, the axial force at a random point xalong the bar is:RAFrom the constitutive equation (Hooke’s Law):f(x) R( const) σ(x) σ(x)R EAEHence, the deformation δ(x) is obtained from kinematics as: (x) δ(x)Rx δ(x) xAENote: The stress & strain is independent of x for this case ofloading. Institute of Structural EngineeringMethod of Finite Elements II31

Axially Loaded Bar ExampleB. Linearly Distributed Axial Constant End LoadFrom the equilibrium equation, the axial force at random point xalong the bar is:f(x) R aL axa(L2 x 2 )(L x) R ( depends on x)22In order to now find stresses & deformations (which depend on x)we have to repeat the process for every point in the bar. This iscomputationally inefficient.Institute of Structural EngineeringMethod of Finite Elements II32

Axially Loaded Bar ExampleFrom the equilibrium equation, for an infinitesimal element: σdσAσ q(x) x A(σ σ) A {z}lim q(x) 0 A q(x) 0 xdx x 0Also, dud 2u, σ E , q(x) ax AE 2 ax 0dxdxStrong Formd 2u ax 0dx 2u(0) 0 essential BCAEf(L) R AEdudx R natural BCx LAnalytical Solutionu(x) uhom up u(x) C1 x C2 ax 36AEC1 , C2 are determined from the BCInstitute of Structural EngineeringMethod of Finite Elements II33

Axially Loaded Bar ExampleAn analytical solution cannot always be foundApproximate Solution - The Galerkin Approach (#3): Multiply by the weight functionw and integrate over the domainLZAE0d 2uwdx dx 2LZaxwdx 00Apply integration by parts Z Ld 2udu ldu dwAEwdx w AEdx 2dxdxdx dx000 Z LZ Ld 2udu dwduduAE 2 wdx AE(L)w(L) AE(0)w(0) AEdxdxdxdxdx dx00ZLAEBut from BC we have u(0) 0, AE du(L)w(L) Rw(L), therefore the approximatedxweak form can be written asLZAE0du dwdx Rw(L) dx dxInstitute of Structural EngineeringLZaxwdx0Method of Finite Elements II34

Axially Loaded Bar ExampleVariational Approach (#1)Let us signify displacement by u and a small (variation of the) displacement by δu. Thenthe various works on this structure are listed below:LZδWint Aσδεdx0δWext Rδu x LLZqδudxδWbody 0In addition, σ E dudxThen, from equilibrium: δWint δWext δWbodyLZ AE0du d(δu)dx dx dxZLqδudx Rδu x L0This is the same form as earlier via another path.Institute of Structural EngineeringMethod of Finite Elements II35

Axially Loaded Bar ExampleIn Galerkin’s method we assume that the approximate solution, u can be expressed asu(x) nXuj Nj (x)j 1w is chosen to be of the same form as the approximate solution (but with arbitrarycoefficients wi ),w(x) nXwi Ni (x)i 1Plug u(x),w(x) into the approximate weak form:LZAE0nXj 1ujZ L XnnnXdNj (x) X dNi (x)axwi Ni (x)dxwidx Rwi Ni (L) dxdx0i 1i 1i 1wi is arbitrary, so the above has to hold wi :n ZXj 10L Z LdNj (x)dNi (x)AEdx uj RNi (L) axNi (x)dxdxdx0i 1.nwhich is a system of n equations that can be solved for the u

The Finite Element Method: Linear Static and Dynamic Finite Element Analysis by T. J. R. Hughes, Dover Publications, 2000 The Finite Element Method Vol. 2 Solid Mechanics by O.C. Zienkiewicz and R.L. Taylor, Oxford : Butterworth Heinemann, 2000 Institute of Structural Engineering Method of Finite Elements II 2

Related Documents: