04.07. The matrix-vector equations for quadratic basis functions - I - I

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Del curso dictado por University of Michigan

The Finite Element Method for Problems in Physics

200 calificaciones

University of Michigan

The Finite Element Method for Problems in Physics

200 calificaciones

This course is an introduction to the finite element method as applicable to a range of problems in physics and engineering sciences. The treatment is mathematical, but only for the purpose of clarifying the formulation. The emphasis is on coding up the formulations in a modern, open-source environment that can be expanded to other applications, subsequently. The course includes about 45 hours of lectures covering the material I normally teach in an introductory graduate class at University of Michigan. The treatment is mathematical, which is natural for a topic whose roots lie deep in functional analysis and variational calculus. It is not formal, however, because the main goal of these lectures is to turn the viewer into a competent developer of finite element code. We do spend time in rudimentary functional analysis, and variational calculus, but this is only to highlight the mathematical basis for the methods, which in turn explains why they work so well. Much of the success of the Finite Element Method as a computational framework lies in the rigor of its mathematical foundation, and this needs to be appreciated, even if only in the elementary manner presented here. A background in PDEs and, more importantly, linear algebra, is assumed, although the viewer will find that we develop all the relevant ideas that are needed. The development itself focuses on the classical forms of partial differential equations (PDEs): elliptic, parabolic and hyperbolic. At each stage, however, we make numerous connections to the physical phenomena represented by the PDEs. For clarity we begin with elliptic PDEs in one dimension (linearized elasticity, steady state heat conduction and mass diffusion). We then move on to three dimensional elliptic PDEs in scalar unknowns (heat conduction and mass diffusion), before ending the treatment of elliptic PDEs with three dimensional problems in vector unknowns (linearized elasticity). Parabolic PDEs in three dimensions come next (unsteady heat conduction and mass diffusion), and the lectures end with hyperbolic PDEs in three dimensions (linear elastodynamics). Interspersed among the lectures are responses to questions that arose from a small group of graduate students and post-doctoral scholars who followed the lectures live. At suitable points in the lectures, we interrupt the mathematical development to lay out the code framework, which is entirely open source, and C++ based. Books: There are many books on finite element methods. This class does not have a required textbook. However, we do recommend the following books for more detailed and broader treatments than can be provided in any form of class: The Finite Element Method: Linear Static and Dynamic Finite Element Analysis, T.J.R. Hughes, Dover Publications, 2000. The Finite Element Method: Its Basis and Fundamentals, O.C. Zienkiewicz, R.L. Taylor and J.Z. Zhu, Butterworth-Heinemann, 2005. A First Course in Finite Elements, J. Fish and T. Belytschko, Wiley, 2007. Resources: You can download the deal.ii library at dealii.org. The lectures include coding tutorials where we list other resources that you can use if you are unable to install deal.ii on your own computer. You will need cmake to run deal.ii. It is available at cmake.org.

De la lección

4

This unit develops further details on boundary conditions, higher-order basis functions, and numerical quadrature. You also will learn about the templates for the first coding assignment.

04.01. The pure Dirichlet problem - I 18:14

04.02. The pure Dirichlet problem - II 17:41

04.02c. In-Video Correction 1:00

04.03. Higher polynomial order basis functions - I 23:35

04.03c0. In-Video Correction 0:57

04.03c1. In-Video Correction 0:34

04.04. Higher polynomial order basis functions - I - II 16:38

04.05. Higher polynomial order basis functions - II - I 13:38

04.06. Higher polynomial order basis functions - III 23:23

04.06ct. Coding assignment 1 (functions: class constructor to “basis_gradient”) 14:40

04.07. The matrix-vector equations for quadratic basis functions - I - I 21:19

04.08. The matrix-vector equations for quadratic basis functions - I - II 11:53

04.09. The matrix-vector equations for quadratic basis functions - II - I 19:09

04.10. The matrix-vector equations for quadratic basis functions - II - II 24:08

04.11. Numerical integration -- Gaussian quadrature 13:57

04.11ct.1. Coding assignment 1 (functions: “generate_mesh” to “setup_system”) 14:21

04.11ct.2. Coding assignment 1 (functions: “assemble_system”) 26:58

Conoce a los instructores

Krishna Garikipati, Ph.D.
Professor of Mechanical Engineering, College of Engineering - Professor of Mathematics, College of Literature, Science and the Arts

Welcome back.

What we worked out in the previous segment was the form of the stiffness matrix for

a general element.

We proceed now with getting together the other parts of our

matrix vector version of the finite dimensional weak form.

And the first thing we have to do is carry out a similar equation for

the forcing function.

That is, the, the, the distributed body force.

All right?

So let's start with that.

So, we will next.

Consider.

The following term.

Integral over omega e

W,h,f,A,d,x, all right.

No gradients in here so

this actually becomes somewhat simpler integral to calculate.

So integral over omical e.

For wh we have sum over a.

Na, Cae.

And we'll put parenthesis around here to remind ourselves.

This is what wh is represented as.

We have f, A dx now.

Because we know that N A ultimately is represented in terms of C,

that is the coordinate in the by unit pairing domain.

And because we know that we have availability,

we have available to us this map.

Right? For the geometry, isoparametric map.

What this tells us is we can very well write this as an integral

now by changing variables as an integral omega X C.

Sum over A, C A E.

The reason we can do that is, what?

Reason we can pull C A E out of the integral because of course it is just

a degree of a freedom that is used to build our Representation of the waiting

function, it does not depend upon C, right, or does not depend upon position,

so we are able to pull it out, just as like, just as we've been doing all along.

All right and remember that this sum runs A going from one to

three, right, for three nodes in the element.

All right. We have C A e.

N, A, f, A.

Now, the integral over

dx can now be written as dx, dz, dz.

And of course, we remember, from before,

something we calculated a couple of units ago.

dx, dc is he over 2.

All right?

We're going to go ahead and build this integral.

And in order to build it, let us just, just in order to be able to fix ideas,

and get something that we can compute.

Let us consider the situation where f and

A are

uniform, right?

They're uniform over Omega E right?

Over the element of interest.

What this allows us to do is to pull f and A as well out of the integral.

And h e is of course the element width, which is independent of.

Right?

So what that tells us is that what we're trying to compute from up here, right?

I've represented it in here in dots with ellipses.

This is sum A equals

one to three CA e

Times FAHE over 2,

integral minus 1

to 1, NA, DZ.

Right.

Now, just as we did before, in the case of linear basis functions, and just as we did

in the previous segment in order to compute the left-hand side integral.

Right? We go now to matrix factor notation.

All right.

So, what we're trying to compute on the left hand side, right?

The integral that comes from the top of the slide

is now represented as c1e, c2e, c3e,

using vector notation for representing our degrees

of freedom of our waiting function over the element.

Times F A H E over 2.

Now, we get a vector here.

Right?

And this is integral.

The first component here is integral minus 1 to 1.

N 1, right?

And, 1 however, we already know,

is one half C times C minus 1 d C.

And, 2, the, the integral from N 2 is integral minus 1 to 1.

Or the integral from, ga.

The integral for component 2 is the integral minus 1 to 1.

Of n 2.

Which is one minus xi squared d xi.

All right?

And the third component

is integral minus 1 to 1

one-half C times C plus 1 dC.

All right?

Okay.

We go ahead now and compute those integrals.

They are relatively straight forward to compute.

Let's do that.

We get our C vector for the element.

We have F A h, e over two because f and

a, we are assuming are uniformed and

for our integrals we get here first,

we get one half of c cubed

over 3 limits minus 1 to 1.

For the second one we get

c Minus Z cubed, over 3,

minus 1, 2, 1.

And for the third,

we get one-half, Z cubed,

over 3, also minus 1 to 1.

Right.

When we compute result,

we see that we are left with c1e,

c2e, c3e, fAhe over 2 and

here, we get 1 over 3,

4 over 3, 1 over 3.

Right.

Okay.

So this is our general representation of the contribution from

the right-hand side forcing function term, for a general element, okay?

What we're going to do now is go to assembly, right?

So, assembly.

Of global matrix vector equations

All right.

Now, it's worthwhile just in order to position ourselves to recall that

what we're attempting to do here is actually carry out an integral

over the entire domain.

Right, over our entire 1D domain.

And, we've used this partition into elements to write that as a sum over

element integrals.

Of that term, that element integral.

Equal to the sum over elements of that element integral.

Plus a term which arises from the fact that we have a,

what kind of a boundary value problem do we have?

We have a, there [INAUDIBLE] boundary value problem.

Right? So we are WHL times T which is

the attraction times A.

Okay?

So, this is our, finite dimensional weak form

expressed as a sum over element integrals.

And we've just, over the previous segment and

over the first few minutes of this segment learned how to compute this integral,

in the last segment, and that integral, in this segment.

For general elements.

Okay?

Okay, now, in order to carry out the assembly,

we just need to recall something.

Okay?

We need to recall.

That for the Dirichlet Neumann problem.

Right.

Because we know that u h add 0 equals

some given value which actually is equal to 0 in this case.

Right?

What that implies for us is that w h at 0 is also equal to 0.

All right, because we have a tertiary boundary condition at x equals 0.

We also have a homogenous tertiary condition on the waiting function.

Of course, it turns out that our tertiary condition on the left hand side,

on x equals 0 is itself homogenous,

meaning we are saying that the trial solution there has to be 0.

Okay? Even if it

were something other than 0 not, if it were non-zero, we would still have

a homogeneous Dirichlet condition on the waiting function, right?

And that's just the nature in which we built our waiting function, okay,

the function's base, in fact.

All right, so what does this mean?

What this implies is that for

element 1, okay?

Because w h at 0 is equal to 0, we construct w h e equals 1 as sum A,

not starting with 1, but starting with 2,

to number of nodes in the element, which is 3.

Okay, so we have this.

N A C A e.

Okay.

And in fact, going from local to global

numbering of degrees of freedom,

we will get here here is what we get.

Okay?

Here is what we get, because of the fact that we can go from local to global

numbering of nodes, here's what we get.

When we right out our matrices, right, matrices and

vectors coming from our weak form, here's what we will get.

We will use the fact that

this thing can be written

explicitly as N2 times

c, 2e plus N3,

c 2e plus 1 for e equals 1.

Okay, we're going to use this.

Where do we use it?

Here's what we get.

Right, our

global equations for

the matrix vector

weak form.

Okay?

They take on the following form.

For element e equals 1, we have something a little special, right?

We have here c 2 e, okay?

C 2 e, plus 1.

Okay?

This is for e equals 1.

Okay.

In fact, let me do this.

Let me explicitly use the fact that e equals 1 here.

When I do that, c 2 e becomes c 2, and c 2 e plus 1 becomes c 3.

Okay, we have

2EA over h1.

Okay?

We're assuming here, of course, that E and A are uniform or

not only over each element, but also between elements, right?

Effectively over the entire domain.

There is nothing preventing us from developing the more complicated

formulation where E and A are allowed to vary with position.

Okay? All right.

Now, what we get for the matrix contribution here is

a matrix that is not square but has only two rows.

Okay, and in fact, that matrix

has the form minus 4 over 3,

8 over 3, minus 4 over 3.

1 over 6, minus 4 over 3,

7 over 6, okay?

Multiplying it here, however, is the full d vector for that element.

Okay?

That would be d1 in terms of global,

no numbers, d2 and d3.

Okay. That's all for element one.

Okay, and note that I called that h, h1 or e equals 1.

Thus, plus now, sum E equals 2 to number of elements.

Here things look the same.

2e minus 1, the same as for general element,

c2e minus 1, c2e, c2e plus 1.

2EA over he.

Our matrix, which we developed

in the previous segment which is 7

over 6 minus 4 over 3, 1 over 6,

8 over 3 minus 4 over 3, 7 over 6.

Completing bisymmetry.

All right, all of this multiplying

here, d2e minus 1, d2e,

d2e plus 1, right?

Global numbering of degrees of freedom and nodes.

Okay, so those terms together, that's this one, for the first element and

this for a generic other element give us our left-hand side.

This is now equal to, okay?

The right-hand side.

Again, the same thing happens with the forcing function contribution, okay?

So we get for the c vector, for

the forcing function contribution, again, we get c2, c3.

Okay?

The contribution here

is fAh1 over 2.

And it multiplies a vector which is

just 4 over 3 and 1 over 3, 'kay?

The first element, the first entry, the first component from the general

form of the factor that arises from the forcing function is absent.

Because in element one, we start with that global degree of freedom.

Because the weighting function satisfies the homogenous

Dirichlet boundary condition.

All right, we have this plus,

sum over e equals 2 to number of elements.

C2e minus 1,

c2e, c2e plus 1,

fAhe over 2.

Now we get the full form of this vector

1 over 3, 4 over 3, 1 over 3.

In addition, let's never forget the Neumann boundary, too.

Right.

Which is wh at l.

But that is simply c2,

nel plus 1,

times tA, okay?

That's it.

That's really our weak form written out fully in matrix-vector notation.

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