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The Direct Stiffness Method Part I

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The DirectStiffness MethodPart IIntroduction to FEMIFEM Ch 2 – Slide 1The Direct Stiffness Method (DSM)A democratic method, works the same no matter what the element:Obvious decision: use the truss to teach the DSMImportance: DSM is used by all major commercial FEM codesBar (truss member) element, 2 nodes, 4 DOFsTricubic brick element, 64 nodes, 192 DOFsQuadratic thin-plate element, 6 nodes, 12 DOFsIntroduction to FEMIFEM Ch 2 – Slide 2Model Based Simulation(a simplification of diagrams of Chapter 1)Physical systemModeling + discretization + solution errorDiscretization + solution errorSolution errorDiscrete modelDiscretesolutionMathematical modelIDEALIZATIONDISCRETIZATIONSOLUTIONFEMVERIFICATION & VALIDATIONIntroduction to FEMIFEM Ch 2 – Slide 3Introduction to FEMIdealization Process(a) Physical System(b) Idealized Sytem: FEM-Discretized Mathematical ModelIDEALIZATION;;;;;;jointsupportmemberIFEM Ch 2 – Slide 4;;;;;;;;FEM model:Remove loads & supports:Disassemble:Localize: longeronsbattensdiagonalslongeronsGeneric element:Introduction to FEMDSM: Breakdown StepsIFEM Ch 2 – Slide 5;;;;;;;;Solve for jointdisplacements:Merge:Apply loadsand supports:Formelements: longeronsbattensdiagonalslongeronsGlobalize:Introduction to FEMDSM: Assembly & Solution StepsIFEM Ch 2 – Slide 6The Direct Stiffness Method (DSM) StepsDisconnectionLocalizationMember (Element) FormationGlobalizationMergeApplication of BCsSolutionRecovery of Derived QuantitiesBreakdown(Chapter 2)Assembly & Solution(Chapter 3)Introduction to FEMStarting with: IdealizationIFEM Ch 2 – Slide 7A Physical Plane TrussIntroduction to FEMjointsupportmemberToo complicated to do by hand. We will use a simpler one to illustrate DSM stepsTypical of those used for building roofs and short span bridges.IFEM Ch 2 – Slide 8The Example Truss: Physical andPin-Jointed IdealizationIntroduction to FEM;;;;123Idealization as Pin-Jointed Trussand FEM DiscretizationIFEM Ch 2 – Slide 9The Example Truss - FEM Model:Nodes, Elements and DOFsIntroduction to FEM45o45o1(0,0)2(10,0)3(10,10)xyx3x3f , uy2y2f , uy1y1f , ux2x2f , ux1x1f , uy3y3f , u(1)(2)(3)(3)(3)(3)(3)E = 100, A = 2 2, L = 10 2, ρ = 3/20 (1)(1)(1)(1)E = 50, A = 2, L = 10, ρ = 1/5 (2)(2)(2)(2)E = 50, A = 1, L = 10, ρ = 1/5 IFEM Ch 2 – Slide 10The Example Truss - FEM Model BCs:Applied Loads and Supports (Saved for Last)Introduction to FEM;;;;;;;;f = 2x3f = 1y3123xy(1)(2)(3)IFEM Ch 2 – Slide 11Master (Global) Stiffness Equationsf =fx1fy1fx2fy2fx3fy3u =ux1uy1ux2uy2ux3uy3fx1fy1fx2fy2fx3fy3=Kx1x1Kx1y1Kx1x2Kx1y2Kx1x3Kx1y3Ky1x1Ky1y1Ky1x2Ky1y2Ky1x3Ky1y3Kx2x1Kx2y1Kx2x2Kx2y2Kx2x3Kx2y3Ky2x1Ky2y1Ky2x2Ky2y2Ky2x3Ky2y3Kx3x1Kx3y1Kx3x2Kx3y2Kx3x3Kx3y3Ky3x1Ky3y1Ky3x2Ky3y2Ky3x3Ky3y3ux1uy1ux2uy2ux3uy3f = KuNodal forcesMaster stiffness matrix NodaldisplacementsLinear structure:orIntroduction to FEMIFEM Ch 2 – Slide 12Member (Element) Stiffness Equations¯f = K¯u¯fxi¯fyi¯fxj¯fyj=¯Kxixi¯Kxiyi¯Kxixj¯Kxiyj¯Kyixi¯Kyiyi¯Kyixj¯Kyiyj¯Kxjxi¯Kxjyi¯Kxjxj¯Kxjyj¯Kyjxi¯Kyjyi¯Kyjxj¯Kyjyj¯uxi¯uyi¯uxj¯uyjIntroduction to FEM_IFEM Ch 2 – Slide 13First Two Breakdown Steps:Disconnection and Localization Introduction to FEMThese steps are conceptual (not actually programmed as part of the DSM)123(3)(1)(2)y_(1)x_(1)y_(2)x_(2)y_(3)x_(3);;;;123Remove loads and supports,and disconnect pinsxyIFEM Ch 2 – Slide 14The 2-Node Truss (Bar) Element Introduction to FEMiijjdLxEquivalent spring stiffness−FFf , u xixi__f , u xjxj__f , u yjyj__f , u yiyi__y__sk = EA / LIFEM Ch 2 – Slide 15Truss (Bar) Element Formulation by Mechanics of Materials (MoM)K =EAL10−1000 00−10 1000 00F= ksd =EALdF =¯fxj=−¯fxi,,d =¯uxj−¯uxi¯fxi¯fyi¯fxj¯fyj=EAL10−1000 00−10 1000 00¯uxi¯uyi¯uxj¯uyjExercise 2.3Element stiffnessmatrix in local coordinatesElement stiffnessequations in local coordinatesIntroduction to FEMfrom whichIFEM Ch 2 – Slide 16Where We Are So Far in the DSMDisconnectionLocalizationMember (Element) FormationGlobalizationMergeApplication of BCsSolutionRecovery of Derived QuantitiesBreakdown(Chapter 2)Assembly & Solution(Chapter 3)Introduction to FEMwe are done with this ...we finish Chapter 2 withIFEM Ch 2 – Slide 17¯uxi= uxic + uyis, ¯uyi=−uxis + uyicγ¯uxj= uxjc + uyjs, ¯uyj=−uxjs + uyjcγNode displacements transform asixyc = cosϕs = sinϕin whichGlobalization: Displacement TransformationIntroduction to FEMxyjϕuxiuyiuxjuyjuyi___uxi_uxj_uyj_IFEM Ch 2 – Slide 18Displacement Transformation (cont'd)In matrix form oreeeNote:global on RHS,local on LHSIntroduction to FEMu = T uuuuu=c−s00sc 0000c−s00scuxixiuyiyiuxjxjuyjyj_____IFEM Ch 2 – Slide 19Globalization: Force TransformationNode forces transform asorxyijϕfxifyifxjfyjfxifyifxjfyj=c−s00sc0000c−s00scNote:global on LHS,local on RHSIntroduction to FEMfxifyifxjfyj____fyi_fxi_fxj_fyj_f = (T ) feeTe_IFEM Ch 2 – Slide 20ue==Teueuefe= TeTfeKe= TeT¯¯¯fe¯KeT¯KeeKe=EeAeLec2sc −c2−scsc s2−sc −s2−c2−sc c2sc−sc −s2sc s2Globalization: Congruential Transformationof Element Stiffness MatricesExercise 2.8Introduction to FEM(())IFEM Ch 2 – Slide 21The Example Truss - FEM Model(Recalled for Convenience)Insert the geometric &physical properties ofthis model intothe globalized memberstiffness equationsIntroduction to FEM45o45o1(0,0)2(10,0)3(10,10)xyx3x3f , uy2y2f , uy1y1f , ux2x2f , ux1x1f , uy3y3f , u(1)(2)(3)(3)(3)(3)(3)E = 100, A = 2 2, L = 10 2, ρ = 3/20 (1)(1)(1)(1)E = 50, A = 2, L = 10, ρ = 1/5 (2)(2)(2)(2)E = 50, A = 1, L = 10, ρ = 1/5 IFEM Ch 2 – Slide 22We Obtain the Globalized Element Stiffness Equations of the Example Trussfx1fy1fx2fy2= 1010−100000−10100000ux1uy1ux2uy2fx2fy2fx3fy3= 50000010−100000 −10 1ux2uy2ux3uy3fx1fy1fx3fy3=


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