CHAP 4 FINITE ELEMENT ANALYSIS OF BEAMS AND FRAMES - PowerPoint Presentation

1. CHAP 4 FINITE ELEMENT ANALYSIS OF BEAMS AND FRAMESFINITE ELEMENT ANALYSIS AND DESIGNNam-Ho Kim

2. INTRODUCTIONWe learned Direct Stiffness Method in Chapter 2Limited to simple elements such as 1D barsIn Chapter 3, Galerkin Method and Principle of Minimum Potential Energy can be applied to more complex elementswe will learn Energy Method to build beam finite elementStructure is in equilibrium when the potential energy is minimumPotential energy: Sum of strain energy and potential of applied loadsInterpolation scheme:Beam deflectionInterpolationfunctionNodalDOFPotential of applied loadsStrain energy

3. BEAM THEORYEuler-Bernoulli Beam Theorycan carry the transverse loadslope can change along the span (x-axis)Cross-section is symmetric w.r.t. xy-planeThe y-axis passes through the centroidLoads are applied in xy-plane (plane of loading)LFxyFPlane of loadingyzNeutral axisA

4. BEAM THEORY cont.Euler-Bernoulli Beam Theory cont.Plane sections normal to the beam axis remain plane and normal to the axis after deformation (no shear stress)Transverse deflection (deflection curve) is function of x only: v(x)Displacement in x-dir is function of x and y: u(x, y)yy(dv/dx) q = dv/dxv(x) LFxyNeutral axis

5. BEAM THEORY cont.Euler-Bernoulli Beam Theory cont.Strain along the beam axis:Strain exx varies linearly w.r.t. y; Strain eyy = 0Curvature:Can assume plane stress in z-dir basically uniaxial statusAxial force resultant and bending momentMoment of inertia I(x)EA: axial rigidityEI: flexural rigidity

6. BEAM THEORY cont.Beam constitutive relationWe assume P = 0 (We will consider non-zero P in the frame element)Moment-curvature relation:Sign conventionPositive directions for applied loadsMoment and curvature is linearly dependent+P+P+M+M+Vy+Vyyxp(x)F1F2F3C1C2C3yx

7. GOVERNING EQUATIONSBeam equilibrium equationsCombining three equations together:Fourth-order differential equation

8. STRESS AND STRAINBending stressThis is only non-zero stress component for Euler-Bernoulli beamTransverse shear strainEuler beam predicts zero shear strain (approximation)Traditional beam theory says the transverse shear stress isHowever, this shear stress is in general small compared to the bending stressBending stress

9. POTENTIAL ENERGYPotential energyStrain energyStrain energy densityStrain energy per unit lengthStrain energyMoment of inertia

10. POTENTIAL ENERGY cont.Potential energy of applied loadsPotential energyPotential energy is a function of v(x) and slopeThe beam is in equilibrium when P has its minimum valuePvv*

11. RAYLEIGH-RITZ METHODAssume a deflection shape Unknown coefficients ci and known function fi(x)Deflection curve v(x) must satisfy displacement boundary conditionsObtain potential energy as function of coefficientsApply the principle of minimum potential energy to determine the coefficients

12. EXAMPLE – SIMPLY SUPPORTED BEAMAssumed deflection curveStrain energyPotential energy of applied loads (no reaction forces)Potential energyPMPE:p0E,I,L

13. EXAMPLE – SIMPLY SUPPORTED BEAM cont.Exact vs. approximate deflection at the centerApproximate bending moment and shear forceExact solutions

14. EXAMPLE – SIMPLY SUPPORTED BEAM cont. DeflectionBendingmomentShear forceError increases

15. EXAMPLE – CANTILEVERED BEAMAssumed deflectionNeed to satisfy BCStrain energyPotential of loadsFC–p0E,I,L

16. EXAMPLE – CANTILEVERED BEAM cont.Derivatives of U:PMPE:Solve for c1 and c2: Deflection curve: Exact solution:

17. EXAMPLE – CANTILEVERED BEAM cont. DeflectionBendingmomentShear forceError increases

18. FINITE ELEMENT INTERPOLATIONRayleigh-Ritz method approximate solution in the entire beamDifficult to find approx solution that satisfies displacement BCFinite element approximates solution in an elementMake it easy to satisfy displacement BC using interpolation techniqueBeam elementDivide the beam using a set of elementsElements are connected to other elements at nodesConcentrated forces and couples can only be applied at nodesConsider two-node bean elementPositive directions for forces and couplesConstant or linearlydistributed loadF1F2C2C1p(x)x

19. FINITE ELEMENT INTERPOLATION cont.Nodal DOF of beam elementEach node has deflection v and slope qPositive directions of DOFsVector of nodal DOFsScaling parameter sLength L of the beam is scaled to 1 using scaling parameter sWill write deflection curve v(s) in terms of sv1v2q2q1Lx1s = 0x2s = 1x

20. FINITE ELEMENT INTERPOLATION cont.Deflection interpolationInterpolate the deflection v(s) in terms of four nodal DOFsUse cubic function:Relation to the slope:Apply four conditions:Express four coefficients in terms of nodal DOFs

21. FINITE ELEMENT INTERPOLATION cont.Deflection interpolation cont.Shape functionsHermite polynomialsInterpolation propertyN1N3N2/LN4/L

22. FINITE ELEMENT INTERPOLATION cont.Properties of interpolationDeflection is a cubic polynomial (discuss accuracy and limitation)Interpolation is valid within an element, not outside of the elementAdjacent elements have continuous deflection and slopeApproximation of curvatureCurvature is second derivative and related to strain and stressB is linear function of s and, thus, the strain and stressAlternative expression:If the given problem is linearly varying curvature, the approximation is accurate; if higher-order variation of curvature, then it is approximateB: strain-displacement vector

23. FINITE ELEMENT INTERPOLATION cont.Approximation of bending moment and shear forceStress is proportional to M(s); M(s) is linear; stress is linear, tooMaximum stress always occurs at the nodeBending moment and shear force are not continuous between adjacent elementsLinearConstant

24. EXAMPLE – INTERPOLATIONCantilevered beamGiven nodal DOFsDeflection and slope at x = 0.5LParameter s = 0.5 at x = 0.5LShape functions:Deflection at s = 0.5:Slope at s = 0.5:Lv1v2q2q1

25. EXAMPLEA beam finite element with length LCalculate v(s)Bending momentBending moment cause by unit force at the tip

26. FINITE ELEMENT EQUATION FOR BEAMFinite element equation using PMPEA beam is divided by NEL elements with constant sectionsStrain energySum of each element’s strain energyStrain energy of element (e)p(x)F1F2F3C1C2C3yxF4F5C4C512345

27. FE EQUATION FOR BEAM cont.Strain energy cont.Approximate curvature in terms of nodal DOFsApproximate element strain energy in terms of nodal DOFsStiffness matrix of a beam element

28. FE EQUATION FOR BEAM cont.Stiffness matrix of a beam elementStrain energy cont.AssemblySymmetric, positive semi-definiteProportional to EIInversely proportional to L

29. EXAMPLE – ASSEMBLYTwo elementsGlobal DOFsF3F2yx1232EIEI2LL

30. FE EQUATION FOR BEAM cont.Potential energy of applied loadsConcentrated forces and couplesDistributed load (Work-equivalent nodal forces)

31. EXAMPLE – WORK-EQUIVALENT NODAL FORCESUniformly distributed loadpL/2pL/2pL2/12pL2/12pEquivalent

32. FE EQUATION FOR BEAM cont.Finite element equation for beamOne beam element has four variablesWhen there is no distributed load, p = 0Applying boundary conditions is identical to truss elementAt each DOF, either displacement (v or q) or force (F or C) must be known, not bothUse standard procedure for assembly, BC, and solution

33. PRINCIPLE OF MINIMUM POTENTIAL ENERGYPotential energy (quadratic form)PMPEPotential energy has its minimum whenApplying BCThe same procedure with truss elements (striking-the-rows and striking-he-columns)Solve for unknown nodal DOFs {Q}[Ks] is symmetric & PSD[K] is symmetric & PD

34. BENDING MOMENT & SHEAR FORCEBending momentLinearly varying along the beam spanShear forceConstantWhen true moment is not linear and true shear is not constant, many elements should be used to approximate itBending stressShear stress for rectangular section

35. EXAMPLE – CLAMPED-CLAMPED BEAMDetermine deflection & slope at x = 0.5, 1.0, 1.5 m Element stiffness matricesF2 = 240 Nyx121 m31 m

36. EXAMPLE – CLAMPED-CLAMPED BEAM cont.Applying BCAt x = 0.5 s = 0.5 and use element 1At x = 1.0 either s = 1 (element 1) or s = 0 (element 2)Will this solution be accurate or approximate?

37. EXAMPLE – CANTILEVERED BEAMOne beam elementNo assembly requiredElement stiffnessWork-equivalent nodal forcesC = –50 N-mp0 = 120 N/mEI = 1000 N-m2L = 1m

38. EXAMPLE – CANTILEVERED BEAM cont.FE matrix equationApplying BCDeflection curve:Exact solution:

39. EXAMPLE – CANTILEVERED BEAM cont.Support reaction (From assembled matrix equation)Bending momentShear force

40. EXAMPLE – CANTILEVERED BEAM cont.ComparisonsDeflectionSlopeBending momentShear force

41. PLANE FRAME ELEMENTBeamVertical deflection and slope. No axial deformationFrame structureCan carry axial force, transverse shear force, and bending moment (Beam + Truss)AssumptionAxial and bending effectsare uncoupledReasonable when deformation is small3 DOFs per nodeNeed coordinate transfor-mation like plane trussFp1234123

42. PLANE FRAME ELEMENT cont.Element-fixed local coordinatesLocal DOFs Local forcesTransformation between local and global coord.12fxyLocal coordinatesGlobal coordinates

43. PLANE FRAME ELEMENT cont.Axial deformation (in local coord.)Beam bendingBasically, it is equivalent to overlapping a beam with a barA frame element has 6 DOFs

44. PLANE FRAME ELEMENT cont.Element matrix equation (local coord.)Element matrix equation (global coord.)Same procedure for assembly and applying BC

45. PLANE FRAME ELEMENT cont.Calculation of element forcesElement forces can only be calculated in the local coordinateExtract element DOFs {q} from the global DOFs {Qs}Transform the element DOFs to the local coordinateThen, use 1D bar and beam formulas for element forcesAxial forceBending momentShear forceOther method: