1 / 44

The Finite Element Method Introduction

The Finite Element Method Introduction. Linear Structural Analyse -        Truss Structure -        Beam -        Shell -        3-D Solid Material nonlinear -        Plasticity (Structure with stresses above yield stress) -        Hyperelasticity (ν = 0.5, i.e. Rubber)

didier
Télécharger la présentation

The Finite Element Method Introduction

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. The Finite Element Method Introduction

  2. Linear Structural Analyse -        Truss Structure -        Beam -        Shell -        3-D Solid Material nonlinear -        Plasticity (Structure with stresses above yield stress) -        Hyperelasticity (ν = 0.5, i.e. Rubber) -        Creep, Swelling

  3. Geometric nonlinear -       Large Deflection -       Stress Stiffening Dynamics -       Natural Frequency -       Forced Vibration -       Random Vibration

  4. Stability -       Buckling Field Analysis -       Heat Transfer -       Magnetics -       Fluid Flow -       Acoustics

  5. Evolution of the Finite Element Method 1941 HRENIKOFF, MC HENRY, NEWMARK Approximation of a continuum Problem through a Framework 1946    SOUTHWELL Relaxation Methods in theoretical Physics 1954    ARGYRIS, TURNER Energy Theorems and Structural Analysis (general Structural Analysis for Aircraft structures) 1960    CLOUGH FEM in Plane Stress Analysis

  6. FE = Finite Element i, j, k = Nodal points (Nodes) of an Element -       Dividing a solid in Finite Elements -       Compatibility between the Elements through a displacement function -       Equilibrium condition through the principal of virtual work

  7. The stiffness relation: [K] {d} = {F} or K d = F K = Total stiffness matrix d = Matrix of nodal displacements F = Matrix of nodal forces

  8. K d = F dT = [u1 v1 w1 . . . un vn wn] FT = [Fx1 F y1 . . . F xn F yn F zn] K is a n x n matrix K is a sparse matrix and symmetric

  9. K d = F Solving the stiffness relation by: -       CHOLESKY – Method -       WAVE – FRONT – Method

  10. k F1 F2 1 2 u1 u2 Spring Element F1 = k (u1 – u2) F2 = k (u2 – u1) 1, 2 = Nodes F1, F2 = Nodal forces k = Spring rate u1, u2 = Nodal displacements

  11. Element stiffness matrix

  12. k1 k2 1 2 3 F1 F3 F2 u1 u2 u3 Spring System Element stiffness matrices

  13. the stiffness relation by using superposition Total stiffness matrix

  14. y u2 F2  2 u1 A  F1 x 1 Truss Element  = length A = cross-sectional area E = Young´s modulus Spring rate of a truss element Element stiffness matrix c = cosα s = sinα

  15. y Fy3 3 Fx3 A E A E   2 = 1350 1 = 450 2 1 x Element :Element : Node 1  1 Node 1  2 Node 2  3 Node 2  3 Stiffness relation

  16. y y 1 2 2 1 v1 v2 M1 M2 EJ x x 1 2 Q1 Q2  Beam Element Forces Displacements A = Cross – sectional area E = Young’s modulus I = Moment of inertia  = Length

  17. the stiffness relation

  18. Example for practical FEM application Engineering system Possible finite element model

  19. v3 y 3 u3 v2 v1 u2 2 1 u1 x Plane stress Triangular Element Equilibrium condition: Principal of virtual work Compatibility condition: linear displacement function

  20. General displacements (Displacement function) u(x,y) = α1 + α2x + α2y v(x,y) = α4 + α5x + α6y Nodal displacements u1= α1 + α2x1+ α3y1 v1= α4 + α5x1+ α6y1 similar for node 2 and node 3.

  21. u = N d General displacements to nodal displacements ε = B d Strains to nodal displacements σ = D ε Stresses to strains σ = D B d Stresses to nodal displacements

  22. 3 5 2 4 1 Other displacement functions Triangular element with 6 nodes 6 quadratic displacement function u(x,y) = α1 + α2x + α3y+ α4x2 + α5y2+α6xy v(x,y) = α7 + α8x + α9y+ α10x2 + α11y2+α12xy

  23. 7 6 10 2 5 4 1 3 8 Triangular element with 10 nodes 9 cubic displacement function -       stress field can be better approximated -       more computing time -       less numerical accuracy -       geometry cannot be good approximated

  24. Principal of Virtual Work δU + δW = 0 δU = virtual work done by the applied force δW = stored strain energy σ = stress matrix p = force matrix ε = strain matrix u = displacement matrix

  25. Element stiffness matrix D = Elasticity matrix

  26. b1 = y2 – y3 c1 = x3 – x2 b2 = y3 – y1 c2 = x1 – x3 AΔ = Area of element b3 = y1 – y2 c3 = x2 – x1 linear displacement function yields : -       linear displacement field -       constant strain field -       constant stress field

  27. m1 m2 m0 k1 k2 c1 c2 u0 u1 u2 F0 F1 F2 Dynamics Equation of motion

  28. or M = Mass matrix C = Damping matrix K = Stiffness matrix d = Nodal displacement matrix = Nodal velocity matrix = Nodal acceleration matrix

  29. for a continuum u = N d ε = B d

  30. the element matrices ρ = Mass density μ = Viscosity matrix

  31. General Equation of Motion Types of dynamic solution o    Modal analysis o    Harmonic response analysis - Full harmonic - Reduced harmonic o    Transient dynamic analysis - Linear dynamic - Nonlinear dynamic

  32. Modal Analysis Purpose: To determine the natural frequencies and mode shapes for the structure Assumptions: Linear structure (M, K, = constant) No Damping (c = 0 ) Free Vibrations (F = 0)

  33. for harmonic motion: d =d0 cos (ωt) (-ω2M + K) d0 = 0 Eigenvalue extraction procedures Transformation methods Iteration methods JACOBI INVERSE POWER GIVENS INVERSE POWER WITH SHIFTS HOUSEHOLDER SUB – SPACE ITERATION Q – R METHOD

  34. Harmonic Response Analysis Purpose: To determine the response of a linear system Assumptions: Linear Structure (M, C, K = constant) Harmonic forcing function at known frequency

  35. Forcing funktion F = F0e-iωt Response will be harmonic at input frequency d = d0 e-iωt (-ω2M – iωC + K) d = F0 is a complex matrix d will be complex (amplitude and phase angle)

  36. Limiting cases: ω = 0 : K d = F0 Static solution C = 0 : (-ω2M + K) d = F0 Response in phase C = 0, ω = ωn : (-ωn2M + K) d = F0 infinite amplitudes C = 0, ω = ωn : (-ωn2M - iωnC + K) d = F0 finite amplitudes, large phase shifts

  37. Transient Dynamic Analysis F(t) = arbitrary time history forcing function periodic forcing function

  38. impulsive forcing function Earthquake in El Centro, California18.05.1940

  39. Two major types of integration: -  Modal superposition -  Direct numerical integration

  40. 1 2 T0 T1 T2 Q0 Q1 Q2 1, A1 2, A2  0 1 2 , A   0ne-dimensional heat flow principle ,  = conductivity elements  = convection element 0, 1, 2 = temperature elements A = Cross-sectional area = Length λ = Conductivity Aα = Convection surface T = Temperature Q = Heat flow C = Specific heat α = Coefficient of thermal expansion

  41. Heath flow through a conduction element: Heat stored in a temperature element: cp = specific heat capacity C = specific heat Heat transition for a convection element: Q = A(T – T2)

  42. Heat balance or

  43. C = specific heat matrix K = conductivity matrix Q = heat flow matrix T = temperature matrix = time derivation of T For the stationary state with = 0 KT = Q

More Related