Yeah, welcome to the Introduction to the Finite Element Method. First of all, I would like

to make an announcement concerning our schedule. We are today here, June the 11th, and in two

weeks' time we won't have neither lecture nor tutorial, so this is the June the 25th.

We are quite well on time, so I'm not sure whether it is really necessary to have another

lecture or tutorial towards the end, but this is something I would like to discuss maybe later

if you would like to have another discussion about topics of lecture or tutorial. This is

something we have still to see. I will upload this updated schedule to Stutton. Okay, and today I

would like to continue with the two-dimensional description of the finite element method.

Particularly today, we will revisit the conductivity matrix and heat flow vector. We will also

consider numerical integration in two dimensions and also in three dimensions, and then we will

solve the system of equations. So let me first of all recapitulate the formulations for the

element conductivity matrix and the element heat flow vector, which we derived last time. You can

also find them in either the lecture notes or the recapitulations. So this is the formulation for

the element conductivity matrix, and here the element heat flow vector consists of two parts.

The first part, this is based on the heat flux entering or leaving the system by the surface,

so this is a surface integral, and the second part, this is energy or power leaving or entering

the system via distributed heat sources throughout the volume. Today, I would like to start with some

considerations concerning numerical integration, because here we still have the integrals, but

typically when you implement that and a computer evaluates these terms, then you apply numerical

integration. And to this end, let me just copy this figure here. So we have here for the two-dimensional

case given two cases, and of course I still have to add the section title, which is 4-4-2, numerical integration.

And on the left-hand side, you see a quadrilateral element here, for example with four integration

points or Gauss points, and then the right-hand side a triangular reference element with a single

Gauss point. So let us start with the quadrilateral elements. As for the shape functions, they are closely

to the one-dimensional elements, so we just extend the number of dimensions, and let me do that for the

two-dimensional case. In the three-dimensional case, the quadrilateral element would be a hexahedral

element, but for the sake of simplicity here and in the interest of time, we just consider the

two-dimensional case. So what do we have here? In general, we have to solve an integral over the physical

domain. So here the argument is a function of x, x is the position vector, which is in two dimensions, of course a

two-dimensional vector. Then we translate that into the integration over the reference domain of the element,

and we do that by substituting dVe by the determinant of the Jacobian times the volume element in the

reference configuration. I still have here dVe, we are working now in 2D, we will see how this translates,

because in 2D we have to keep in mind that our system is actually a three-dimensional system, but we assume a

constant thickness t in the third direction. So here the volume element looks like this, we have a thickness t in the third

direction, and this dVe squared, this is what I have sketched here, and this is substituted by this arbitrary thickness

times the increments in both directions. So for instance we have here dxi1 and here dxi2,

and you see as in the one-dimensional case in each coordinate direction we integrate from minus one to one,

so in the two-dimensional case we get two integrals from minus one to one, one for the integration over xi1 and one for the integration over xi2.

Then, very similar as in the one-dimensional case, we substitute this integration in this reference interval from minus one to one

by the summation over quadrature points and we evaluate this function f only at the quadrature points,

the same for the determinant of the Jacobian, and we put here associated weights for the integration points.

The thickness t is just constant, so we put it here in front of the summation.

So this is more or less what one has to do, and since we still have these integrals separately in the two dimensions,

we can use the same weights and quadrature point positions as in the one-dimensional case,

so this is just an extension to the two or three-dimensional case.

In the three-dimensional case, by the way, we would have three integrals and of course three summations,

but this is straightforward to be extended.

A little bit different from the one-dimensional case are tetrahedral or triangular elements,

and this is the next I would like to discuss.

In 2D, we would have a triangular element,