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 n different choices of the weighting functions, derive an equation for a by requiring that 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 over each 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)=x2 in 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 solving and 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 and α small 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.)

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., Natural BC’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.)

2.2: Calculus of Variations (cont.) Notes: There are no general procedures for finding the operator for a given set of equations However, is known for many of the more common finite element analysis problems. Special case for which can always be found: “Self-adjoint” equations

2.2: Calculus of Variations (cont.) Notes: For self-adjoint equations, and can be shown to be: (Depending on problem details, may be necessary to integrate by parts before taking variation.)

2.2: Calculus of Variations (cont.) Example: axial deformation of fixed rod with axial load – Can re-write governing equations as:

2.2: Calculus of Variations (cont.) Example: Functional is then calculated as follows: Euler equations for this functional:

2.2: Calculus of Variations (cont.) So, what’s all of this have to do with finite elements? Have a set of equations and boundary conditions to solve for . Have a functional related to and via the Euler equations on . Finite element analysis attempts to find the best approximate solution to Weak form of governing equations!

2.2: Calculus of Variations (cont.) Look more closely at 1D version: Suppose we make “usual” approximation – A “space” of trial functions Must belong to same “space”

2.2: Calculus of Variations (cont.) Plug in approximations (ignoring boundary terms for now) – Since each ak is arbitrary, best approximation comes from Function of a1, a2, …, an  Get n equations for n constants!

2.2: Calculus of Variations (cont.) Notice the following: Galerkin’s Method and Calculus of Variations give same equations when “proper” is used! Galerkin’s method!

2.2: Calculus of Variations (cont.) Notice something else:

2.2: Calculus of Variations (cont.) Integrate 2nd term by parts (and ignore boundary terms again): Rayleigh-Ritz Method on  gives same equations as J = 0 !

2.2: Calculus of Variations (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 the calculus of variations on an appropriate variational principle along with the approximate solution , estimate the displacement of the rod.

2.2: Calculus of Variations (cont.) Solution: Find appropriate variational principle: Problem: there is a nonzero boundary condition – Needs to be integrated by parts!

2.2: Calculus of Variations (cont.) Solution: Doing this gives: Require the first variation to equal zero:

2.2: Calculus of Variations (cont.) Solution: Using the given approximate function: After some integrating, result is: =0 =0

2.2: Calculus of Variations (cont.) Solution: Solve the two equations to get: Same as Galerkin’s method solution!

2.2: Calculus of Variations (cont.) What if we had forgotten about the BC? Functional becomes: So the first variation becomes: Force is cut in half!

2.2: Calculus of Variations (cont.) Solution: Solution becomes: