Good morning.

Welcome to our lecture today.

I would first of all like to recapitulate what we did last time.

Last time we introduced the concept of beam theory and today we will continue to develop the formulations for beam elements.

So when you remember our model problem, then we considered a Bernoulli beam which is subjected to a distributed lateral load, lowercase q of x, as well as to a single load, uppercase qL and a bending moment ml.

And this leads to a differential equation of fourth order.

So we have here the flexural stiffness EI times the fourth derivative of W, which is the deflection, minus the lateral distributed load equals zero in the entire domain.

And we introduced directly boundary conditions which is in this case now the deflection at a specific position and the slope at a specific position.

And in addition we introduced Neumann boundary conditions where we prescribed the bending moment and the lateral load.

In the tutorial we will see that there are various combinations of these boundary conditions possible.

This was for a beam that is clamped on the left hand side.

Today we will continue to derive the weak form and to do so we start here with this equation.

Maybe I copy the entire strong form here.

And then we will continue here.

So this is the strong form.

And of course the boundary conditions can be different for a specific problem.

And here our residuum.

So now derive the weak form.

So to derive the weak form we need a residuum.

And in our case I think this is obvious that this is now a function of the deflection end of X, whereby the deflection is a function of X at the end of the deflection.

A function of X itself.

So this is EIW fourth derivative with respect to X minus Q of X.

And of course this should vanish.

So what do we have to introduce beyond this to derive a weak form?

We have to introduce a test function.

A V of a test function space.

And we have to multiply the test function by the residuum and integrate over the entire domain.

I put here that this is a function of X.

So here we get integral from zero to L.

This is the domain of interest here.

So it's V of X, then EIW fourth derivative, minus Q of X, dx equals zero.

And what do I have to state?

Exactly, for all V of the test function space.

Okay.

And let us now focus on the first term in brackets.

So we have here the V times this term here.

And the rest we will see later what to do with that.

So we have here the integral from zero to L, V of X, EIW fourth prime, four primes, dx.

And what can we do with that?

How can we proceed here?

Yes, yes.

So we can do the following.

We can write integral from zero to L.

And then V of X, EIW three primes.

And then we take another prime of the entire term.

And now we have to subtract something to get equality with the left hand side.

What do I have to subtract here?

And how many primes for the W?

Three, yes.

So this is the first step.