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Section 2: Finite Element Analysis Theory. Method of Weighted Residuals Calculus of Variations Two distinct ways to develop the underlying equations of FEA!. Section 2: FEA Theory. Some definitions:. V = volume of object A = surface area = A u + A s
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Section 2: Finite Element Analysis Theory • Method of Weighted Residuals • Calculus of Variations Two distinct ways to develop the underlying equations of FEA!
Section 2: FEA Theory • Some definitions: • V = volume of object • A = surface area = Au + As • Au = surface of known displacements • As = surface of known stresses • b = body force • t = surface stresses (tractions)
Section 2.1: Weighted Residual Methods • A group of methods that take governing equations in the strong form and turn them into (related) statements in the weak form. • Applicable to a wide class of problems (elasticity, heat conduction, mass flow, …). • A “purely mathematical” concept.
2.1: Weighted residual methods (cont.) • Need to write the equilibrium equations and boundary conditions in an abstract form as follows: Solve these for u(x)!
2.1: Weighted residual methods (cont.) • Let be the exact solution to the problem(differential equation and boundary conditions) • Then, for any choice of vectors W and W’:
2.1: Weighted residual methods (cont.) • Integrate these “results” over the entire volume and surface: • Previous expression is still true if W and W’ are functions of x (called weighting functions):
2.1: Weighted residual methods (cont.) • Now, consider an approximate solution to the same problem: • Matrix/vector form of this: Known functions Unknown constants
2.1: Weighted residual methods (cont.) • Plugging this approximate solution into the differential equation and boundary conditions results in some errors, called the residuals. • Repeating the previous process now gives us an integral close to but not exactly equal to zero!
2.1: Weighted residual methods (cont.) • Goal: Find the value of a that makes this integral as close as possible to zero – “best approximation”. • Idea: for ndifferent choices of the weighting functions, derive an equation for a by requiringthat the above integral equal zero: • Solve these equations for a!
2.1: Weighted residual methods (cont.) • Notes on weighted residual methods: • It is typical (but not required) to assume that the known functions satisfy the displacement boundary conditions exactly on Au. (Essential conditions) • In some methods, one must integrate the volume integral by parts to get “appropriate” equations. • Different methods result from different ideas about how to choose the weighting functions.
2.1: Weighted residual methods (cont.) • Collocation Method: • Assume only one PDE and one BC to solve! • Idea: pick n points in object (at least one in V and one on A) and require residual to be zero at each point!
2.1: Weighted residual methods (cont.) • Subdomain Method: • Assume only one PDE and one BC to solve! • Divide object up into n distinct regions (at least one in V and one on A). • Require integral overeach region to be zero.
2.1: Weighted residual methods (cont.) • Notes on collocation and subdomain methods: • Weighting functions for collocation method are the Dirac delta functions: • Weighting functions for subdomain method are the indicator functions: • Advantage: Simple to formulate. • Disadvantage: Used mostly for problems with only one governing equation (axial bar, beam, heat,…).
2.1: Weighted residual methods (cont.) • Least Squares Method: • Considers magnitude of residual over the object. • Finds minimum by setting derivatives to zero
2.1: Weighted residual methods (cont.) • Galerkin’s Method: • Idea: Project residual of differential equation onto original approximating functions! • To get W’, must integrate any derivatives in volume integral by parts!
2.1: Weighted residual methods (cont.) • Must use integrated-by-parts version of !
2.1: Weighted residual methods (cont.) • Notes on least square and Galerkin methods: • More widely used than collocation and subdomain, since they are truly global methods. • For least squares method: • Equations to solve for a are always symmetric but tend to be ill-conditioned. • Approximate solution needs to be very smooth. • For Galerkin’s method: • Equations to solve for a are usually symmetric but much more “robust”. • Integrating by parts produces “less smooth” version of approximate solution; more useful for FEA!
2.1: Weighted residual methods (cont.) • Example: 1D Axial Rod “dynamics” • Given: Axial rod has constant density ρ, area A, length L, and spins at constant rate ω. It is pinned at x = 0 and has applied force -F at x = L. The governing equation and boundary conditions for the steady-state rotation of the rod are: • Required: Using each of the four weighted residual methods and the approximate solution , estimate the displacement of the rod.
2.1: Weighted residual methods (cont.) • Some preliminaries: • Problem has an exact solution given by • Approximate solution satisfies essential boundary condition u(x = 0) = 0. • Two unknown constants → n = 2. • Notation:
2.1: Weighted residual methods (cont.) • Solution: • Collocation Method -- • Since n=2, have two collocation points. One must be at x = L (must have one on As). Assume other at x = L/3. • Equation #1: evaluate residual of E(u) at x = L/3: • Equation #2: evaluate residual of B(u)at x = L:
2.1: Weighted residual methods (cont.) • Solution: • Collocation Method -- • Solve simultaneous equations: • Plot results:
2.1: Weighted residual methods (cont.) • Solution: • Subdomain Method -- • Since n=2, have two subdomains. One must be at x = L (= As). Other must be 0 < x < L (= V). • Equation #1: integrate residual of E(u) over V: • Equation #2: evaluate residual of B(u)at x = L:
2.1: Weighted residual methods (cont.) • Solution: • Subdomain Method -- • Solve simultaneous equations: • Plot results:
2.1: Weighted residual methods (cont.) • Solution: • Least Squares Method -- • For dimensional equality, take in ILS(a). Once again, the “integral” over As is just evaluation at x = L. • Equation #1: take derivative with respect to a1:
2.1: Weighted residual methods (cont.) • Solution: • Least Squares Method -- • Equation #2: take derivative with respect to a2: • Solve equations: Same as subdomain method!
2.1: Weighted residual methods (cont.) • Solution: • Galerkin’s Method -- • Weighting functions are • Integrate general expression for volume integral by parts first: Set this equal to zero for k = 1,2!
2.1: Weighted residual methods (cont.) • Solution: • Galerkin’s Method -- • Equation #1 uses N1(x)=x in previous: • Equation #2 uses N2(x)=x2in previous:
2.1: Weighted residual methods (cont.) • Solution: • Galerkin’s Method -- • Solve simultaneous equations: • Plot results:
Section 2: Finite Element Analysis Theory • Method of Weighted Residuals • Calculus of Variations Two distinct ways to develop the underlying equations of FEA!
Section 2.2: Calculus of Variations • A formal technique for associating minimum or maximum principles with weak form equations that can be solved approximately. • A more physically motivated approach than weighted residuals. • Not all problems amenable to this technique.
2.2: Calculus of Variations (cont.) • Minimum/Maximum Principles (Variational Principles) involve the following: • A set of equations and boundary conditions to solve for . • A scalar quantity “related” to E and B (called a functional). • A variational principle states that solvingand is equivalent to finding the function that gives a maximum or minimum value. • Requires -- “First variation of must be zero (stationarity)”.
2.2: Calculus of Variations (cont.) • What is a functional? • A function takes a point in space as input and returns a scalar number as output. (Vector-valued function gives vector as output.) • A functional takes a function as input and returns a scalar number as output. Arc-length of f(x) from x=0 to x=a.
2.2: Calculus of Variations (cont.) • A few examples: • Recall that straight line is shortest distance between two points. How do we prove that?
2.2: Calculus of Variations (cont.) • For a given function , consider a Taylor series expansion of arc-length formula in terms of α: = some number β; Assume β> 0.
2.2: Calculus of Variations (cont.) • Suppose that α is small and negative: • Same problem if β < 0 andαsmall and positive.So, must have β = 0! Can’t happen!!!
2.2: Calculus of Variations (cont.) • Integrate by parts: • But and Must equal zero!!!
2.2: Calculus of Variations (cont.) • Key ideas in this “proof”: • Considered an arbitrary increment of the input function. • Derivative of the functional forced to be zero. • This implies a certain equation must equal zero. • Calculus of Variations gives you a “direct” way of performing these calculations!
2.2: Calculus of Variations (cont.) • Some definitions: • General form of a functional is • A variation of is • Note: if must satisfy some boundary conditions, so must .
2.2: Calculus of Variations (cont.) • Some properties of the variation of : • Derivatives and variations can interchange. • Integrals and variations can interchange. • Variation of sum is sum of variations. • Variation of product obeys “product rule”.
2.2: Calculus of Variations (cont.) • Some properties of the variation of : • “Chain rule” applies to dependent variables only!
2.2: Calculus of Variations (cont.) • Let’s go back to arc-length example: = = Thus, we see that Just like before! =
2.2: Calculus of Variations (cont.) • Minimum/Maximum Principles (Variational Principles) involve the following: • A set of equations and boundary conditions to solve for . • A scalar quantity “related” to E and B (called a functional). What is the relation?
2.2: Calculus of Variations (cont.) • Let’s consider a 1D version of this: • Want to minimize J(u), so require δJ(u) = 0:
2.2: Calculus of Variations (cont.) • Integrate 2nd term by parts: • involves the boundary conditions! • Essential BC’s: E.g., • NaturalBC’s: E.g, • Other BC’s: E.g.,
2.2: Calculus of Variations (cont.) • Integrate 3rd term by parts twice:
2.2: Calculus of Variations (cont.) • Pull all of this together:
2.2: Calculus of Variations (cont.) • Assuming all boundary conditions are either essential or natural, end up with: for any choice of The Euler equation for
2.2: Calculus of Variations (cont.) • The “relation” between being minimum and is as follows: If you can find an operator such that then solving is the same as solving .
2.2: Calculus of Variations (cont.) • Some notes: • If you have boundary conditions that neither essential nor natural, then must explicitly include a “boundary term” in the functional. • As number of dependent variables increases (e.g., 2D), one functional will produce multiple Euler equations: (See Slide #10 for general statement of this idea.)