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Curso de Analisis por elementos finitos

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  • 1. EMAE 415 Lectures Finite Element Analysis - Basic Concepts Based in part on slides posted by Dr. Ara Arabyan, University of Arizona; Dr. Kenneth Youssefi, San Jose State University
  • 2. Basics of Finite Element Analysis
    • FEA in Biomechanics
    • FEA models have proven to be powerful tools in analyzing biomechanical structures and to evaluate designs for implants, prostheses and musculoskeletal constructs.
    • The advantages are the ability to account for complex geometries and material behaviors.
    • Always remember, FEA is a numerical technique. It’s answers are only as good as the formulation of the problem.
  • 3. Basics of Finite Element Analysis
    • FEA is a mathematical representation of a physical system and the solution of that mathematical representation.
    • Always remember, FEA is a numerical technique. It’s answers are only as good as the formulation of the problem.
    • GIGO (Garbage in-garbage out)
  • 4. FEA Stress Analysis
    • After approximating the object by finite elements, each node is associated with the unknowns to be solved.
    • For a two dimensional solid (e.g. a beam) the displacements in x and y would be the unknowns.
    • This implies that every node has two degrees of freedom and the solution process has to solve 2n degrees of freedom.
    • Once the displacements have been computed, the strains are derived by partial derivatives of the displacement function and then the stresses are computed from the strains.
  • 5. Finite Element Analysis
    • Consider the cantilever beam
    • shown.
    Finite element analysis starts with an approximation of the region of interest into a number of meshes (triangular elements). Each mesh is connected to associated nodes (black dots) and thus becomes a finite element.
  • 6. Displacement Fields
    • The displacements of different points in the structure set up a displacement field
    • The displacement field expresses the displacement of any point in the structure as a function of its position measured in a reference frame
  • 7. Strains versus Displacement Fields (2-dimensional case) Knowing the displacement fields allows you to determine the strain fields.
  • 8. Stiffness of a Uniform Rod
    • Recall from elementary mechanics of solids that a uniform rod of length L , cross sectional area A , and elasticity modulus E can modeled as a linear spring of stiffness k eq
    • with
    A, E L P P P P
  • 9. Nodal Displacements, Forces
    • Consider a linear spring of stiffness k . Let the displacements of its two ends, called nodes , be denoted by u i and u j , known as nodal displacements . Let the forces acting at its two ends, called nodal forces , be denoted by f i and f j .
    i j f j f i Nodes y x u j u i Nodal force Nodal displacement Reference frame
  • 10. Force-Displacement Relations
    • The relationships between the nodal forces and displacements (as shown below) are given by:
    f j f i u j u i
  • 11. Element Stiffness Matrix
    • These relations can be written in matrix form as
    • or more briefly as
  • 12. Element Stiffness Matrix (cont’d)
    • In this relation
    • is known as the element stiffness matrix (always symmetric) ;
    • is known as the element nodal displacement vector; and
  • 13. Element Stiffness Matrix (cont’d)
    • is known as the element nodal force vector.
    • The element nodal displacements are also known as element nodal degrees of freedom (DOF)
  • 14. Singularity of Element Stiffness Matrix The equation ku = f cannot be solved for the nodal displacements for arbitrary f because the matrix k is singular . Physically this means that, in static equlibrium, the displacements of the endpoints of a spring cannot be determined uniquely for an arbitrary pair of forces acting at its two ends. One of the ends must be fixed or given a specified displacement; the displacement of the other end can then be determined uniquely.
  • 15. Solution for Single Rod Element
    • If node i is fixed (i.e. its displacement is set to 0) then ku = f reduces to
    • and the displacement of node j is easily determined as
    • which is the expected solution
    f j u j u i =0
  • 16. Matrix Reduction
    • Note that when a displacement or DOF is set to zero, rows and columns of k associated with that displacement are eliminated and only the remaining set is solved
    Row(s) associated with u i Column(s) associated with u i
  • 17. Multiple Elements
    • Now consider two springs of different stiffness linked to each other
    F 1 1 2 3 F 2 F 3 Globally numbered nodes Globally numbered elements
  • 18. Continuity Relations
    • When two elements are joined together the joined nodes become one and must have the same displacement
    • where the subscript denotes the global node number , the superscript denotes the global element number , and i and j denote local element numbers
  • 19. Force Balance Relations
    • The external nodal forces acting at each node must equal the sum of the element nodal forces at all nodes
    • where F 1 , F 2 , F 3 are external nodal forces numbered globally
  • 20. Assembly of Equations
    • When these continuity and force balance relations are imposed the resulting global equilibrium equations are
    • More briefly this can be written as
  • 21. Global Stiffness Matrix
    • In this relation
    • is known as the global stiffness matrix (always symmetric) ;
    • is known as the global nodal displacement vector; and
  • 22. Global Stiffness Matrix (cont’d)
    • is known as the global nodal force vector or the global load vector
    • The global nodal displacements are also known as global degrees of freedom (DOF)
  • 23. Assembly of Global Stiffness Matrix
    • Note that the global stiffness matrix is assembled from element matrices as follows
    Stiffness terms from two matrices add at coinciding DOF Stiffness matrix from element Stiffness matrix from element
  • 24. Singularity of Global Stiffness Matrix As in the case of individual element matrices the global stiffness matrix K is singular . (You can check this out for this small example by calculating the determinant of the matrix; the result will be zero.) Some nodes of the structure need to be constrained (i.e. fixed or given known displacements) to make it statically determinate or overconstrained. Then the remaining DOF can be determined. Constraining some nodes in the structure corresponds to applying boundary conditions.
  • 25. Solution for Global Structure
    • If node 1 is fixed (i.e. its displacement is set to 0) then the equilibrium equations reduce to
    1 2 3 F 2 F 3
  • 26. Solution for Global Structure (cont’d)
    • The displacements of nodes 2 and 3 can now be found from
    • It can be shown that constrained global stiffness matrix is not singular
  • 27. Matrix Reduction
    • Note that when a DOF is set to zero rows and columns of K associated with that DOF are eliminated and only the remaining set is solved
    Row(s) associated with u 1 Column(s) associated with u 1
  • 28. Rod Elements
    • The spring models introduced thus far constitute one class of finite elements and are known as rod, spar (ANSYS), or truss (ALGOR) elements
    • In the form shown these elements can be used to model only unidimensional (one dimensional) problems
    • The more general form of these elements can be used to model two or three dimensional problems
  • 29. Example
    • An aluminum ( E = 10.4  10 6 psi) rod of variable cross section is subjected to a point load of 1000 lb at its narrower end. Determine the deflection of the rod at its loaded end.
    12 in 1000 lb A l = 0.250 in 2 A r = 0.125 in 2
  • 30. Example, contd.
    • The variable cross section rod can be approximated as a number of rods of constant cross section. For this example let us choose three rods of equal length that span the length of the original rod.
    1000 lb 4 in 4 in 4 in A 1 A 2 A 3
  • 31. Example, contd.
    • The cross-sectional area of each rod can be assumed to be the average of the segment each rod spans:
  • 32. Example, contd.
    • The equivalent stiffness of each rod can now be computed from
    • which results in
  • 33. Example, contd.
    • The spring (finite element) model of the problem is now
    1000 lb 1 2 3 4
  • 34. Example, contd.
    • The equilibrium equations can now be written as:
    Row and column associated with u 1
  • 35. Example, contd.
    • Solving the remaining equations we obtain
    • Since the deflection of the right end is required the answer to the problem is
  • 36. Comparison with Exact Results
    • Using “exact” analytical methods the displacement of any point ( u(x) ) on the rod is given by
    x 1000 lb
  • 37.
    • Using this expression the displacement of the three nodes in the finite element model can be computed as
    • Evidently the results from the finite element method are very close to those produced by “exact” methods
    Comparison with Exact Results (cont’d)
  • 38. Comparison with Exact Results (cont’d)
    • The plot below charts the variation of displacements across the bar for the two solutions
  • 39. Formulation of the Finite Element Method f B – Body forces (forces distributed over the volume of the body; gravitational, inertia, or magnetic forces) f S – surface forces (pressure of one body on another, or hydrostatic pressure) f i – Concentrated external forces
  • 40. Formulation of the Finite Element Method Denote the displacements of any point (X, Y, Z) of the object from the unloaded configuration as U T The displacement U causes the strains and the corresponding stresses The goal is to calculate displacement, strains, and stresses from the given external forces.
  • 41. Example – a plate under load
    • Derive and solve the system equations of a plate loaded as
    • shown.Plate thickness is 1 cm and the applied load P y is constant.
    using two triangular elements,
  • 42. Example, cont’d
    • Displacement within the triangular element with three nodes can be assumed to be linear.
  • 43. Example, cont’d
    • Displacement for each node,
  • 44. Example, cont’d
    • Solve the equations simultaneously for α and β ,
  • 45. Example, cont’d
    • Substitute x 1 = 0, y 1 = 0, x 2 =10, y 2 = 0, x 3 = 0, y 3 =4 to obtain displacements u and v for element 1,
    Element 1
  • 46. Example, cont’d
    • Rewriting the equations in the matrix form,
  • 47. Example, cont’d
    • Similarly the displacements within element 2 can be expresses as
  • 48. Example, cont’d
    • The next step is to determine the strains using 2D strain-displacement relations,
  • 49. Example, cont’d Differentiate the displacement equation to obtain the strain,
  • 50. Example, cont’d Element 2
  • 51. Example, cont’d
    • Using the stress-strain relations for homogeneous, isotropic
    • plane-stress, we have
  • 52. Formulation of the Finite Element Method Equilibrium condition and principle of virtual displacements The left side represents the internal virtual work done and the right side represents the external work done by the actual forces as they go through the virtual displacement.   The above equation is used to generate finite element equations. And by approximating the object as an assemblage of discrete finite elements, these elements are interconnected at nodal points.
  • 53. Formulation of the Finite Element Method The equilibrium equation can be expressed using matrix notations for m elements. where B (m) represents the rows of the strain displacement matrix C (m) is the elasticity matrix of element m H (m) is the displacement interpolation matrix U is a vector of the three global displacement components at all nodes F is a vector of the external concentrated forces applied to the nodes
  • 54. Formulation of the Finite Element Method The previous equation can be rewritten as follows, The above equation describes the static equilibrium problem. K is the stiffness matrix.
  • 55. Stiffness matrix for element 1
  • 56. Example, cont’d Calculating the stiffness matrix for element 2.
  • 57. Example, cont’d The stiffness of the structure as a whole is obtained by combing the two matrices.
  • 58. Example, cont’d The load vector R, equals R c because only concentrated loads act on the nodes where P y is the known external force and F 1x , F 1y , F 3x , and F 3y are the unknown reaction forces at the supports.
  • 59. Example, cont’d The following matrix equation can be solved for nodal point displacements
  • 60. Example, cont’d The solution can be obtained by applying the boundary conditions
  • 61. Example, cont’d The equation can be divided into two parts, The first equation can be solved for the unknown nodal displacements, U 3 , U 4 , U 7 , and U 8 . And substituting these values into the second equation to obtain unknown reaction forces, F 1x , F 1y , F 3x , and F 3y . Once the nodal displacements have been obtained, the strains and stresses can be calculated.
  • 62. Finite Element Analysis
    • FEA involves three major steps
      • Pre-Processing
      • Solving Matrix (solver)
      • Post-Processing
  • 63. Summary of Pre-Processing
    • Build the geometry
    • Make the finite-element mesh
    • Add boundary conditions; loads and constraints
    • Provide properties of material
    • Specify analysis type (static or dynamic, linear or non-linear, plane stress, etc.)
  • 64. FEA Pre-Processing Mesh Development The FEA mesh is your way of communicating geometry to the solver, the accuracy of the solution is primarily dependent on the quality of the mesh. The better the mesh looks, the better it is. A good-looking mesh should have well-shaped elements, and the transition between densities should be smooth and gradual without skinny, distorted elements.
  • 65. FEA Pre-Processing - Example Coarse mesh Refined mesh - better?…probably… depends on the loading and boundary conditions
  • 66. FEA Pre-Processing Finite elements supported by most finite-element codes:
  • 67. FEA Pre-Processing Material Properties Material properties can be specified for element regions, elements or even within elements in most large-scale FEA codes. The material properties required for an isotropic, linear static FEA are: Young’s modulus (E), Poisson’s ratio (v), and shear modulus (G). G = E / 2(1+v) Only two of the three properties can be specified independently.
  • 68. FEA Pre-Processing Nonlinear Material Properties A multi-linear model requires the input of stress-strain data pairs to essentially communicate the stress-strain curve from testing to the FE model Highly deformable, low stiffness, incompressible materials, such as rubber and other synthetic elastomers require distortional and volumetric constants or a more complete set of tensile, compressive, and shear force versus stretch curve. A creep analysis requires time and temperature dependent creep properties. Plastic parts are extremely sensitive to this phenomenon
  • 69. FEA Pre-Processing Boundary Conditions In FEA, the name of the game is “ boundary condition ”, that is calculating the load and constraints that each component experiences in its working environment. The results of FEA should include a complete discussion of the boundary conditions
  • 70. Boundary Conditions Loads Loads are used to represent inputs to the system. They can be in the forms of forces, moments, pressures, temperature, or accelerations. Constraints Constraints are used as reactions to the applied loads. Constraints can resist translational or rotational deformation induced by applied loads.
  • 71. Boundary Conditions Degrees of Freedom Spatial DOFs refer to the three translational and three rotational modes of displacement that are possible for any part in 3D space. A constraint scheme must remove all six DOFs for the analysis to run. Elemental DOFs refer to the ability of each element to transmit or react to a load. The boundary condition cannot load or constrain a DOF that is not supported by the element to which it is applied.
  • 72. Post-Processing View Animated Displacements View Displacement Fringe Plot View Stress Fringe Plot View Results Specific To the Analysis Review Boundary Conditions Review Load Magnitudes and Units Review Mesh Density and Quality of Elements Does the shape of deformations make sense? Are magnitudes in line with your expectations? Is the quality and mag. Of stresses acceptable? Yes Yes No No No Yes