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Basic FEA Procedures Structural Mechanics Displacement-based Formulations

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Computational Procedure Element Matrices: – Generate characteristic matrices that describe element behavior Assembly: – Generate the structure matrix by connecting elements together Boundary Conditions: – Impose support conditions, nodes with known displacements – Impose loading conditions, nodes with known forces Solution: – Solve system of equations to determine unknown nodal displacements Gradients: – Determine strains and stresses from the nodal displacements

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Element Physics Truss (bar) element, axial tension or compression, not bending or torsion Arbitrarily oriented in the (x,y) plane Uniform cross-sectional area and material properties Two nodes per element, linear displacement variation along element y x u2 v2 u1 v1 N1 N2 E y x F2 x F2y F1y F1x N1 N2 E

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Element Matrices [k] An element is fundamentally a matrix that relates nodal displacements with nodal forces – It reflects equilibrium between concentrated forces applied at nodes and the spring-like response when the bar changes length – It embodies constitutive behavior in the forma of a modulus value – It maintains compatibility between displacements and strains

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Element Matrices Complexity arises quickly: here is a 3D Timoshenko beam element These are “simple” elements that can be formulated directly They embody exact variations of the field quantities (displ’s, rotations)

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Assembly Establish geometry and connectivity, number nodes and elements Calculate characteristic matrix for each element independently Load into a structure matrix [sdof x sdof] in size (sdof = num nodes x dof per node) E1 E2 E3 N1 N2 N3 E1 0.00 u2 0.0020000000.000.00-20000000.00v2 0.00 u3 0.00-20000000.000.0020000000.00v3 n2xn2yn3xn3y E2 5760027.44-7680008.00-5760027.447680008.00u3 -7680008.0010239972.567680008.00-10239972.56v3 -5760027.447680008.005760027.44-7680008.00u1 7680008.00-10239972.56-7680008.0010239972.56v1 n3xn3yn1xn1y E3 26666666.670.00-26666666.670.00u1 0.00 v1 -26666666.670.0026666666.670.00u2 0.00 v2 n1xn1yn2xn2y [K] structure 32426694.11-7680008.00-26666666.670.00-5760027.447680008.00u1 -7680008.0010239972.560.00 7680008.00-10239972.56v1 -26666666.670.0026666666.670.00 u2 0.00 20000000.000.00-20000000.00v2 -5760027.447680008.000.00 5760027.44-7680008.00u3 7680008.00-10239972.560.00-20000000.00-7680008.0030239972.56v3 n1xn1yn2xn2yn3xn3y

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Structure Matrix [K] Coefficients on the diagonal are positive Symmetric if loads are linearly related to displacements (superimposable) Sparse (most coefficients = 0) for problems with more than a few elements Represents a set of equations for simultaneous solution, order can be changed Can be solved more efficiently if non-zero entries cluster along the diagonal Is singular (non-invertible, non-solvable) with inadequate support (free to move) Once it is solvable, it is “just” a (generally) very big linear algebra problem 32426694.11-7680008.00-26666666.670.00-5760027.447680008.00 -7680008.0010239972.560.00 7680008.00-10239972.56 -26666666.670.0026666666.670.00 20000000.000.00-20000000.00 -5760027.447680008.000.00 5760027.44-7680008.00 7680008.00-10239972.560.00-20000000.00-7680008.0030239972.56

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Boundary Conditions Specification of known values associated with nodal degrees of freedom Two types, two different approaches to implementation Displacements – This is the quantity we are solving for – They are the “essential” boundary conditions in that a set must be specified to create adequate support conditions for the structure – Non-support conditions can also be specified, e.g. a known nodal displacement Forces – These are “right hand side” values for our set of equations – They are the “natural” boundary conditions in that they “load” the structure They are mutually exclusive – You cannot specify displacement and force for the same degree of freedom – If one is specified (e.g. displacement = 0), the other develops (e.g. reaction force = ?)

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Example B.C’s Displacements are handled moving the reaction influences to the right hand side and creation of equations that directly reflect the condition Forces are simply added into the right hand side E1 E2 E3 N1 N2 N3 1000 32426694.11-7680008.00-26666666.670.00-5760027.447680008.00u1 = F1x -7680008.0010239972.560.00 7680008.00-10239972.56v1F1y -26666666.670.0026666666.670.00 u2F2x 0.00 20000000.000.00-20000000.00v2F2y -5760027.447680008.000.00 5760027.44-7680008.00u3F3x 7680008.00-10239972.560.00-20000000.00-7680008.0030239972.56v3F3y 32426694.11-7680008.000.00 7680008.00u1 = 0.00 -7680008.0010239972.560.00 -10239972.56v1-1000 0.00 1.000.00 u20.00 1.000.00 v20.00 1.000.00u30.00 7680008.00-10239972.560.00 30239972.56v30.00 This is it! Solve for the nodal displacements … 32426694.11-7680008.007680008.00u1 = 0 -7680008.0010239972.56-10239972.56v1-1000 7680008.00-10239972.5630239972.56v30 - or -

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Solving 1 Rarely is {D} = [K] -1 {R} actually carried out [K] -1 is “full” even if [K] is “sparse”, therefore storage is an issue for large models The structure stiffness inverse is not needed to find {D} Solution by Gauss elimination, LU decomposition, Cholesky decomp, etc. Solution can start as [K] is assembled (wavefront methods) Element numbering does make a difference in storage and efficiency – Most codes offer renumbering schemes to minimize storage and/or maximize solution speed

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Solving 2 To iterate, or not to iterate … Direct solvers are most useful for modest-size problems with many load cases Iterative solvers are often used even if direct solution is possible A convergence criteria is required to judge when a “solution” is found The convergence rate is related to the condition number of [K] Condition number is related to the ratio of highest to lowest Eigenvalues Many factors produce poor [K] conditioning (bad ele shape, mat prop extremes)

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Strain and Stress Calculation For bar/truss elements with just nodal boundary conditions: – Find axial elongation L from differences in node displacements – Find axial strain from the normal strain definition – Find axial stress from the stress-strain relationship

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Element Loads This is a surprisingly complicated issue Distributed axial loads, self-weight, pressure, etc. It is a place where FEA can go wrong and give you bad results Element loads must be converted to nodal loads “Lost” loads and inaccurate stresses must be dealt with – What if one of the nodes below is constrained? – Is the element stress state the same for the two representations? These loading states are “consistent”, but require special handling

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CAD and Finite Element Analysis Most ME CAD applications require a FEA in one or more areas: –Stress Analysis –Thermal Analysis –Structural Dynamics –Computational.

CAD and Finite Element Analysis Most ME CAD applications require a FEA in one or more areas: –Stress Analysis –Thermal Analysis –Structural Dynamics –Computational.

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