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.
03.08. The final finite element equations in matrix-vector form - II
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The Finite Element Method for Problems in Physics
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From the lesson
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In this unit, you will write the finite-dimensional weak form in a matrix-vector form. You also will be introduced to coding in the deal.ii framework.
Meet the Instructors
Krishna Garikipati, Ph.D.Professor of Mechanical Engineering, College of Engineering - Professor of Mathematics, College of Literature, Science and the Arts
Let me pull all this together and write it in a very simplified manner.
Okay, so what we have
here is c transpose Kd
equals c transpose F.
Okay.
This is close to the final form of our
finite element equations in matrix-vector form.
Okay?
But there's one other important step we need to take here.
Which is, to recognize that this vector of weighting function,
degrees of freedom, actually does interpolate the weighting function.
Right. So recall that in each element,
we had wh of e equals sum over A,
NA cAe going back just briefly
momentarily actually to our local
element numbering system for
the c degrees of freedom.
Okay.
Also recall that our weak form,
our statement of the weak form is the following,
right, in much abbreviated,
in a much abbreviated version it was defined
belonging to S such that, St is short for such that.
For all wh belonging to V, and I'm sorry,
these are both, that's Sh and Vh, right, right?
For all wh belonging to Vh the following integral holds,
the following equation holds, right?
So this is our weak form again, yet again.
Okay.
What we've done now is essentially reduced that integral
weak form to the equation at the top of the slide.
Okay? In matrix-vector form.
All right?
So, and let me, yet again, state that here.
All these statements, while, while a little superfluous,
are repetitive definitely, are useful to drive home points.
So we, we make them repeatedly.
All right.
So, that is our matrix-vector weak form, and the important thing
that I want to point out is that this holds for all wh belong to Vh.
My question to you now is that,
what is it in the matrix-vector weak form that represents wh?
Remember, using basis functions, finite dimension basis functions,
we've gone from this integral form at the bottom of the slide,
the last equation on the slide which involved fillings.
To matrix-vector form which involves just degrees of freedom for matrices for
vectors, and we have entries for matrices.
So in that setting, what is it that represents wh?
Okay?
Essentially, what represents wh is the following.
This equation of ours is completely equivalent to
our statement at the top of the slides.
C transpose Kd equals
c transpose F.
When we realize that this has to hold for all c, right?
Belonging to this R nel space.
After all is just a vector with nel entries, okay?
It is when we specify that our matrix-vector weak form must hold for
all c, right?
Because all c belonging to this nel degree of sorry, this nel dimensional space
that we are essentially imposing the same requirement as we do here.
Why is this?
It is because the c of degrees of freedom which interpolate the weighting functions.
If our weak form has to go for all weighting functions belonging to Vh,
we've already fixed the functional form of Vh by choosing particular polynomial basis
functions.
All right? Or by choosing certain basis functions.
That degree of arbitrariness which must still hold within the space Vh
is insured if we require, that the matrix-vector weak form holds for
all c belonging to this nel dimensional space.
Okay, well, but if this is, if this is the case, it is clear therefore
that this can only hold if Kd equals F.
There's a formal way of demonstrating that, and
this just involves moving everything to the left-hand side,
and then considering what forms c can possibly have,
all the forms that c can possibly have and discovering that this must be the case.
Okay. We can make that, that rigorous argument
if you need to, if we need to, but I think it's pretty clear why, why this works.
Okay, so, this is the final form of our final, for, for
our finite element equations, okay?
FE short for finite element.
Right.
Remarks.
Okay.
The first remark I want to make is that the matrix K that you have
in front of you or probably in your notes is symmetric.
It's positive definite.
Okay.
Furthermore, it has abandoned, so-called tridiagonal structure.
With [SOUND] banded,
tridiagonal structure.
Okay.
The symmetry comes from the fact.
Symmetry from the fact that the term of the weak form which
gave rise to our matrix K has this sort of structure.
Right?
Where this is our stress.
Okay.
So if you look at this integral as a as an object called a functional in w and u.
It is bilinear, okay?
Furthermore, it remains unchanged if we just swap the places of w and u, okay?
Right? So the reason this thing is symmetric is
because we have this term in the weak form and
we could very well interchange those two positions, right?
Wh and.
Okay?
This is what makes our final K matrix symmetric as well.
If we were working with a different set of equations which, for some reason,
did not give us this form of an integral where wh and
could be interchanged, and still have the same meaning, right?
And, and still have the same integral,
then we would not have a symmetric matrix K.
Okay? Oh, of course, I should also state that
now K, again for obvious reasons, is called a stiffness matrix.
Okay, and as you might imagine this is related to the fact that
when we look at it in, in the form Kd equals F which you have before you
it is reminiscent of a of a spring.
Right, K being the spring stiffness, d being the displacement of the spring and
f being the force on the spring.
And indeed, in the early days of finite elements when there was a lot of
structural mechanics work done with finite elements, this was the,
well this was the right interpretation, okay?
But now if you, if you imagine that you're doing a heat conduction problem,
you would probably would, you would probably,
properly call it something else maybe the conductivity matrix, right.
But traditionally, it's called a stiffness matrix.
So let me just put that in quotes.
Okay now, the other bit is the positive definiteness.
The positive definiteness comes from the fact
that our in this particular case, our so-called
material constant, E, is greater than 0.
We didn't actually state that but if E and D is greater than 0,
then we have a positive definite system.
All right the particular banded tridiagonal structure
comes from the fact that there are two first order derivatives,
and that we have a linear set of basis functions.
Okay, so.
The bandedness is actually is, is included when I say tridiagonal.
So tridiagonal
from.
Single derivative
on wh and.
All right, in this form that, that's before us in the integral,
okay there's a single derivative on each of them.
And from the fact that we have a linear basis functions.
Okay.
The next remark I want to make, and really it's the final remark of
this segment, it is the fact that our force vector in this case,
I'm just looking back to my notes here to make sure that I have, oh, there we go.
That I have here before me, so
that I don't make any errors in putting in back up here, okay.
Our force vector in this case is the following.
It's fAhe over 2,
times 2 being all the way down to 1,
plus 0 all the way down to tA,
in the last entry.
Plus EA over he, multiplying our
Dirichlet boundary conditions u knot,
and zeros all the way down.
All right, there are a few things I want to point out about this.
If we look at this contribution to the force, to the force term, it comes from
the distributed forcing function that was specified at every point along the bar.
Right, our original function, F.
Right, which we assume is a constant in order to get this final form.
If F were not a constant,
it would be inside this it would be within this the column vector here.
Okay so, so
one thing to note is that if we consider the mesh and
consider a stretch of the mesh with elements of that type.
Okay?
What we're seeing is that the contribution on a node,
which has elements on either side of it, is of the form fAhe.
Okay?
It's only the very last element.
Right?
This is the point L.
Okay?
It's only this very last element which has a contribution, fA.
So at this very last node which has a contribution fAhe over 2.
Okay? And
this is really because an interior node has a contribution fAhe.
It sort of has the entire force on it.
Because it has elements on both sides.
So both those elements contribute force to it.
However, this last node has a contribution only from the element to its left.
Okay?
This is pretty obvious.
This contribution is from the traction and it's no surprise that it appears only
on the very last node, because that's where the traction is indeed applied.
This contribution is also very important one.
You recall that this came about from the fact that Dirichlet boundary condition is
known and therefore that degree of freedom in the d vector in our
trial solution vector could be moved over to the right-hand side.
Dirichlet b.c for boundary condition.
'Kay?
So what this is giving us is the loading, right, on the problem,
obtained by specifying a Dirichlet boundary condition, okay?
So this is what you may call
within quotes, a Dirichlet
driven load, right?
You, we, we all very well know that partial differential equations can be
driven by either the Dirichlet or Nuemann boundary conditions.
This is how the Dirichlet boundary condition itself drives the problem.
Of course, in the kinds of problems, we are considering,
the sort of boundary value problem we're considering, u knot was typically zero.
Okay?
So that actually drops out.
But in the context of elasticity, what it does for us is give us the effect of the,
the, the, the, the load that gets transmitted to the structure.
Because of the fact that on the left end, at the left end of the structure,
we've fixed the displacement to be equal to zero.
Okay?
So let me say just one more thing here.
This Dirichlet driven load is equal to 0
in the considered case.
Okay. So there we have it.
The various contributions to the, to the effective forcing on the problem,
coming from the distributed forcing function from the traction,
the, the Neumann boundary condition, right.
Since I'm calling it traction,
I also ought to say that this is the Neumann boundary condition,
and finally, the effect of the Dirichlet boundary condition as well.
Okay, we will stop this segment here, but
essentially at this point we have completed our
simplest finite element problem in 1d for linear elliptic equations.
Okay.
Well, I should, I should also state one more thing, right.
I've just said it's formulated the problem.
What is the solution?
Since we have Kd equals F,
the solution is d equals K inverse F.
Okay, that is the formal solution.
Of course, you may or may not actually invoke K depending on the size of
your problem, that is an entirely separate question which we don't get into here.
This is just the formulation.
Once we have d, of course we can then go back and reconstruct
in any element as being simply the interpolation over A using
the bases functions, which we know very well, times dAe.
Okay?
So we've recovered the actual field from the solution.
Okay, here's where we definitely do stop the segment.
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