ADVANCED MATHEMATICS APPLIED

TO CIVIL ENGINEERING

UNIVERSIDAD INDUSTRIAL DE

SANTANDER

Prof. LUIS E. ZAPATA ORDUZ, PhD

Facultad de Ingenierías Físico-Mecánicas

Escuela de Ingeniería Civil

Bucaramanga, I semestre 2016

II.

MATHEMATICS BEHIND THE APPLIED

MECHANICS

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Introduction

There are quite a few engineering problems for which we cannot obtain

exact solutions or the exact solution may be too expensive.

Inability to obtain an exact solution:

The complex nature of governing DEs

Difficulties from dealing with the boundary and initial conditions

¡¡¡ We have to use numerical approximations !!!

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In engineering problems

There are three clases of numerical methods:

(i) Finite difference method (FDM)

(ii) Finite volumen method (FVM)

(iii) Finite element method (FEM)

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Finite Difference Method:

Arithmetic operations to determine derivatives

Derivatives are replaced by difference equations

Useful in simple problems (geometrics)

Very difficult to apply to complex geometries or complex BVPs

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Finite Volume Method:

Based on application of conservation principles over each control volume

Especially use as a numerical method for solving PDE

This method do not requiere a structured mesh (advant. over FEM)

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Finite Element Method:

Uses integral formulations rather tan difference equations

A continuous function is assumed to represent the aprox. solution

This is an aprox. solution to the exact DE

Very useful in civil engineering applications: complex problems

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We will focus on the mathematics behind the Finite Element Analysis

(FEA)

FEA/FEM characteristics:

Powerful numerical technique to solve DEs with BVPs

A continuous domain is discretized to simple geometrical elements: FEs

The elements are related to each other through nodal points

The governing eqs, loading and BCs system of eqs aprox. solution

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The field of mechanics can be divided into three mayor areas:

(i) Theoretical Mechanics fundamental laws and principles of mechanics

(ii) Applied Mechanics scientific/engineering applications of Theore. Mec.

(iii) Computational Mechanics simulation through numerical methods

Computational mechanics can be applied to nanomechanics,

micromechanics, continuum mechanics, and other mechanical systems.

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Nanomechanics branch of nanoscience studying fundamental

mechanical properties of physical systems at the nanometer scale

Micromechanics analysis of composite or heterogeneous materials on

the level of their individual materials constituents

Continuum mechanics studies bodies at the macroscopic level, using

continuum models in which the microstructure is homogenized by

phenomenological averages

In the next we will focus our attention to those mathematical problems related

to FEM/FEA and continuum mechanics civil engineering applications

(restricted to static and linear-elastic problems)

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When we are using the FEM in Civil Eng. applications we are solving

the whole elasticity field (15 eqs to determine 15 unknowns)

If stress = 0 and strain = 0 FEM = 0

Unknown I Displacement field (3 unk)

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Unknown II State of Strain (6 unk)

Unknown III State of Stress (6 unk)

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Equations I Equilibrium Equations 2D (3 eqs)

These eqs must be satisfied at all point inside the body

The eqs were derived from Newton´s law of motion

bx, by, and bz are body forces per unit volume b acting body´s centroid

Body forces: gravity forces, inertial forces, electric – magnetic forces

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Equations II Stress-Strain Relationships (6 eqs)

Called constitutive laws or constitutive relationships

These relate the stress and strain components

For linear-elastic isotropic materials, we have:

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Equations III Linear Strain-Displacement Relationships (6 eqs)

Are derived from purely geometric considerations

Define the strain components used for the characterization of deformation

The strains associated with the displacement field (Green-Lagrange strains)

These strains are expressed in terms of the displacement gradients

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The strains are computed from the Green-Lagrange (G-L) tensor

Stresses corresponding to G-L strains second Piola-Kirchhoff stresses

PK2 stresses linearly related to the Cauchy stresses as follows:

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prestresses Cauchy stresses

PK2 stresses

|Jacobian| F = Deformation gradient matrix

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Assuming isochoric deformation (volumen-preserving) J = 1.0

Assuming the prestressed state (reference configuration) So = 0

We restrict our analysis small deformations small strains

S ≈ σ

In the mathematical models for the linear elastic field:

We will always have 15 unk although we may never see the 15 eqs

The 15 eqs are imbedded into the integral formulation based for the FEM

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Typical Elasticity Field for a Continuous Deformable Body

Statically admisible stress field S(x,y,z) if it satisfies the equil. eqs at all

points in the domain and the surface equil. eqs at all points on the surface:

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Kinematically admisible displacement field d(x,y,z) if it is continuous

and diferenciable at all points in the domain and the geometric BCs at all

points on the surface:

Strain field compatible E(x,y,z) if it is derived from a kinematically

admissible displacement field through the strain-displacement relationships

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Newton´s Laws

First law A particle remains at rest, or continues to move in a straight

line with uniform velocity, if there is no unbalanced force acting on it

Second law The acceleration of a particle is proportional to the resultant

force acting on the particle and in the direction of this force (the constant

of proporcionality is the mass of the particle)

Third law The force of action and reaction between interacting bodies

are equal in magnitude, opposite in direction, and collinear

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FEM/FEA calculations: Positive Sign Convention

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Aproximations to DEs

There are four ways to solve any DE problem (related to FEA):

(i) Exact solution

(ii) Minimization of the quadratic functional

(iii) Strong form (alternative integral form of the DE to obtain the solution)

(iv) Weak form (alternative bilinear representation of the integral formulation

of the DE to obtain the solution)

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Recall:

¿What is a DE?

An equation involving one or more variables and its derivatives with

respective to one or more independent variables.

Homogeneous DE:

Nonhomogeneous DE:

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Example:

Solve the following DE using the exact form:

u´´- u = 1 0 < x < 1 u = u(x)

s.t.: u(0) = 1, u´(1) = 0

Ans.

The exact solution homogeneous solution and a particular solution:

u(0) = uh(0) + unh(0) = 1; uh(0) = 0 and unh(0) = 1

The total exact solution is: u(x) = -1+ c1ex + c2e

-x

Using the BCs: c1 = 2/(1 + e2) and c2 = c1e2

Therefore, we have:

u(x) = -1 + 2ex/(1 + e2) + 2e2-x/(1 + e2)

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Approximate Solution to DE

In many cases we may not be able to solve the DEs exactly (close form

solution)

In fact, although the close form exist, in practice it may be too costly

An alternative approximate the DE using either its integral form or its

finite difference:

NOTE: The solution to the integral formulation is an approximation which is

a linear representation of the unknown parameters cj:

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In this approximation:

cj are the unknown parameters

ϕj are known as trial functions (functions of position x in the domain Ω)

The problem is reduced to find cj such that the approximate solution satisfies

the equations governing the actual solution, u(x).

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Integral Approximation: Strong Formulation

In the process of satisfying the governing eqs through approximations, we

obtain (not accidentally but by planning) n-algebraic relations among the n

parameters c1, c2, …, cn.

One way is to use the integration formulation: strong formulation an

approximate solution over the entire domain Ω = (0,1):

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Nonhomogeneous solution Homogeneous solution

Known trial functions

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The function ϕo(x) is chosen it satisfies the nonhomogeneous essential

(actual) BC

For those problems, for which essential BCs are zero ϕo(x) = 0

The trial functions ϕj are chosen satisfy the homogeneous essential

(specified) boundary conditions

The trial functions ϕj should be linearly independent and the given series

must be able to approximate the solution to any given accuracy

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We know that

Therefore, the choice of the trial functions ϕj(x) must be zero for all j.

Let us choose the approximate solution of the form (i.e. using j = 2):

Let us choose the trial function of the form (for now it is magic):

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¿The above trial functions satisfy the homogeneous specified BCs?:

In fact, happens to satisfy the forced BCs (in most cases it is not neccesary):

Now, the choice of ϕo(x) is such that: we know

In summary, the following BCs must be satisfied for the strong formulation:

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Hence, for our example the trial functions are:

Now, we choose c1 and c2 such that they satisfy the original DE:

Since the above equation must be zero for any value of x the

coefficients of the Powers of x must be zero:

Therefore, we can conclude that there is no set of solutions to the above

equation by using the strong formulation.

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DE´s Residual (R)

R is obtained by setting the DE to zero:

Now, the R is multiply by a weighted function v(x) and integrated over Ω:

We may obtain as many linearly independent eqs as there are linearly

independent functions v(x).

In our example, v(x) must be restricted to n = 2 because we have two

coefficients, cj (c1 and c2)

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To overcome this problem approximate the solution to satisfy the DE in

the weighted-integral sense (called weak formulation).

Let us take (¿arbitrarily?)

Gauss-Jordan:

c1 = 324/347

c2 = - 40/347

Hence, the solution to the weak formulation becomes:

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Strong Form: Methods of Weigthted Residual (MWR)

In MWR asume that a solution may be approximated analytically or

piecewise analytically.

In general, a solution to a DE is expressed as a superposition of a base set

of functions that satisfy initial and BC of the problem:

where,

ϕj(x) called trial functions chosen to satisfy the specified BC

ϕo(x) function that is chosen to satisfy the essential nonhomogeneous BC

cj generalized coordinates (are constants determined by a chosen method)

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Key points about the MWR

The assumed solution is not exact the goal is to choose the coefficients

cj such that the residual R becomes small (in fact zero) over Ω

The procedure consists in premultiplying the DE by a weighted function Φi

and integrating over Ω

From theoretical studies the exact solution always satisfies R = 0 if the

weight functions are analytic (even in subdomains)

The number of weight functions Φi should be equal to the number of unk

constants cj the result is a set of n-algebraic eqs for the unk cj

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Four categories

There are four main categories of weight functions applied in MWR in order to

determine the unknown missing coefficients:

1. Collocation Method

In this method R is forced to zero n-times (n = unk coefficients cj)

The selection of collocation points is arbitrary

The weighting functions family of Dirac functions in the Ω

Dirac function in this method: Φi = δ(x – xi)

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2. Subdodmain Method

The integral of the error function is force to zero over selected subintervals

The Ω is divided into n-subdomains Ωn (size free)

The weighted function is chosen such that in each subdomain:

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3. Least-Squares Method

Error is minimized respect to the unk. coefficients in the assumed solution

Hence, the weight function is:

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4. Galerkin Method

We choose the weight functions identical to the trial functions

The method requires the error to be orthogonal to the weight functions

NOTE: In general, for methods 1 – 4 we are solving a matrix equation in order

to obtain the unknown coefficients cj

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Exercise:

Solve the following DE:

u´´- u = 1 0 < x < 1 u = u(x) s.t.: u(0) = 1, u´(1) = 0

1. Determine the weighted functions (use j = 2)

u(x) = -1+ c1ex + c2e

-x

unh(0) = 1 and uh(0) = 0

u´nh(1) = 0 and u´h(1) = 0

Hence, we have:

ϕo(0) = 1 and ϕ´o(1) = 0 nonhomogeneous (actual) BCs

ϕj(0) = 0 and ϕ´j(1) = 0 homogeneous form of the specified BCs

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Solving for the nonhomogeneous term:

BCs(hom) since BCs =2 let us the polynomial: ϕo(x) = a + bx

Solving for the BCs: ϕo(0) = 1 a = 1 and ϕ´o(1) = 0 b = 0

The polynomial is ϕo(x) = 1

Solving for the first homogeneous term (j = 1):

BCs(nonh) = 2 let us the polynomial: ϕ1(x) = a + bx + cx2

Solving for the BCs: ϕ1(0) = 0 a = 0 and ϕ´1(1) = 0 b + 2c = 0

The polynomial is ϕ1(x) = c(x2 – 2x) (but c = 1 absorbed into c1)

The trial function is ϕ1(x) = (x2 – 2x)

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Solving for the second homogeneous term (j = 2):

BCs(nonh) = 2 let us the polynomial: ϕ2(x) = a + bx + dx2

Solving for the BCs: ϕ2(0) = 0 a = 0 and ϕ´2(1) = 0 b + 3d = 0

The polynomial is ϕ2(x) = d(x3 – 3x) (but d = 1 absorbed into c2)

The trial function is ϕ2(x) = (x3 – 3x)

Hence, the trial functions are:

ϕo(x) = 1 ϕ1(x) = (x2 – 2x) ϕ2(x) = (x3 – 3x)

The complete approximate solution:

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2. Determine the DE´s residual (for j = 2)

3. Approximate the DE by using collocation method (use j = 2)

Rta.

c1 = 666/731

c2 = -81/731

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5. Approximate the DE by using subdomain method (use j = 2)

Rta.

c1 = 195/209 and c2 = -24/209

3. Approximate the DE by using least-squares method (use j = 2)

Rta.

c1 = 5730/6143 and c2 = -72800/632729

6. Approximate the DE by using Galerkin method (use j = 2)

Rta.

c1 = 5010/5537 and c2 = -80/791

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Weak Formulation

The formulation previously described is known as strong formulation

We needed to evaluate the following integral (highest order derivative)

This implies that v´´ Є (-∞, 0) or (0, ∞) meaningful approx. solution

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The Strategy:

Reducing the order of differentiability through mathematical techniques

By integrating by parts: strong form weak form

The key is “weakened” the order of continuity of primary variable

The major advantage is when choosing the trial functions

For an ODE: strong form (2m) weak form (up to m-order)

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Test Function (v)

Any cont & diff function vanishes at points of the BC (essential)

Weak formulation test functions obtained directly from trial functions

Test Function: (approximate solution) – (exact solution) very small

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Weak Form

The goal is to have the same order of derivatives for the primary variables

and the test function v(x).

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Therefore, the following eq is called the weak form of the DE

The test function must be such that:

Or symbolically:

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Quadratic Functionals

If B(u,v) = B(v,u) (symmetric bilinear term) the functional for the given

DE can be written as the following quadratic functional:

vs.

The Euler-Lagrange eq. for this quadratic functional is same as the DE. In

our case the quadratic functional is:

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Galerkin Approximation

The Galerkin approach (weak form) is one of the most often used

approximate methods to determine numerical solutions of ODEs and PDEs

ϕo(x) = 0 satisfies the nonhomogeneous essential BCs

ϕj(x) = 0 trial functions: satisfy the homogeneous essential BCs

cj constants called generalized coordinates

NOTE: Galerkin method the trial functions satisfy both BCs essential/natural

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When using Galerkin approx.:

Regardless of using the weak or strong formulation accuracy of the

aprox. solution depends upon the choice of trial functions (TFs)

To assume a proper TF for the unknown exact solution is not an easy task

Specially true when the exact solution to have a large variation over Ω

To overcome TFs are expressed using piecewise continuous functions

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Assuming the following approximate function:

Hence, TFs must satisfy:

(i) Be linearly independent

(ii) Satisfy both essential and natural BCs

(iii) Form a complete set (no missing terms)

(iv) Test Functions = Trial Functions v(x) = Φi(x) for 1 ≤ i ≤ n

NOTE: in Galerkin you can use both the strong or weak form to get answers.

NOTE: in the following, the approximation solution losses the upper bar.

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Galerkin: Strong Form

The strong form is:

Recall: Test Functions = Trial Functions v(x) = Φi(x) for 1 ≤ i ≤ n

In symbolic form, the i-th eq (1 ≤ i ≤ n) is written as:

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In the above equation:

Kij the stiffness coefficient

pi The generalized forces corresponding to the i-th generalized coordinate

In matrix notation, the set of equations can be written as:

K c = P

K known as the stiffness matrix

c vector of generalized coordinates

P vector of generalized forces

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Exercise:

Solve the following DE by using the strong form of the Galerkin method:

u´´- u = 1 0 < x < 1 u = u(x) s.t.: u(0) = 1, u´(1) = 0

Rta.

Same trial functions: ϕ0(x) = 1, ϕ1(x) = (x2 – 2x) and ϕ2(x) = (x3 – 3x)

The approximate sol.: usg(x) = ϕ0(x) + c1ϕ1(x) + c2ϕ2(x)

usg(x) = 1 + c1 (x2 – 2x) + c2 (x3 – 3x)

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The strong form (for i = 1, 2):

In matrix form: K c = P

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The generalized stiffness coefficients (for i = 1, 2):

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The generalized force coefficients (for i = 1, 2):

Solving the system:

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