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Okay, so good morning everyone. So this is really the last class and you won't have any more torture from me, isn't it?

Okay, so in the last class we were doing, we had just started the discretization of the weak form.

Okay, so we defined our increments delta r as delta r little i n i, similarly for delta theta,

which was delta theta i n i and same way for del r and del theta. Okay, so you see these are discrete approximations of these four functions.

So what we have to do is we just plug in the discrete approximations in the weak form and then it just becomes the discretized weak form.

So for example our G static that we had defined, so when we plug in the discrete approximation it becomes discrete,

so we use a superscript H to define that it's discrete. Okay, so this becomes, let me just write down the actual G,

which was N m dot e transpose del r del theta minus the term due to distributed load,

and the boundary condition. Okay, and this N p and M p are the prescribed forces and moments at the ends.

So this is the weak form without being discretized, so I should not put this H here.

Now we have to plug in these approximations over here, isn't it? And you see we have this operator e,

so we have to also discretize that operator, so that e transpose was defined as the identity d over ds,

then here we had this little r prime and it's skew symmetric, then the zero tensor and identity d over ds, so we have to discretize this.

So you can note this for example when it acts on this del r del theta vector, then this becomes del little r prime plus r prime cross del theta

and del theta prime. Okay, so the whole idea was to write this using this operator e transpose, isn't it?

And now when we discretize this, you see what's going to happen. Let's just plug in del r prime with this discrete approximation,

so here we say it's approximate, and del little r prime then becomes del little r i N i prime,

see the prime shifts onto the shape functions, because the del little r i they are constants, right?

They do, they are not functions of s, all s dependence has now shifted onto the shape functions. Okay,

so we take the prime onto the shape functions plus little r prime, so little r prime remains as it is, okay,

because we have discretized just the increments and the variations cross del theta,

so that also becomes cross N i del theta i, and then this del theta prime then becomes del theta i N i prime, okay,

and then we can write this again using the discrete operator, you'll see how to get that discrete operator e transpose.

So, if I write this as N i prime identity, then N i little r prime cross 0 N i prime identity,

times del little r i and del theta i, okay, so you see you have now gotten your discrete operator,

the discrete form of that e transpose, so this is simply you can define it as e transpose with a subscript i, okay,

you could also put in some x to say that it is a discrete form, but let's not burden it with so many symbols,

if you have the subscript i because you see in this matrix you also have this N i's,

so therefore this comes with a subscript i, okay, and then this gets multiplied with del little r i del theta i, isn't it?

Alright, so we are going to use that of course, so then our G s t H, the discrete H is then 0 to L,

N m dot e i transpose, where e i transpose is that number over there, that matrix del little r i del theta i, okay,

and you see what about these one, this N and m, they are simply your integrands which have to be evaluated,

this is also getting integrated, right, ds, so you have to integrate this from 0 to L,

so you would have the current states, right, what is little r and little theta,

so corresponding to that current state you find out what is the current strain,

v and k and then from constitutive law you get your N and m, okay,

so these depend on the current state little r little theta, remember this was a function of little r little theta

and this del r del theta, so little r little theta over defining your current state,

so find out the strains from using this current state and get your N and m and then you integrate, okay,

so that is just the first term, then we talk about the second term, minus 0 to L and the second term is quite easy,

n hat and m hat are the distributed load and distributed couple, usually they are prescribed,

so depends on how you have prescribed it, you can put that function here

and then you dot this with the approximation, so del r you put it as ni del ri,

so ni would just come out and this becomes ni del ri del theta i, isn't it,

and then the last term, can you guess what happens to the last term,

see we have to evaluate that at the boundary, so what is del little r of L,

so let us just do that side calculation, del little r of L that simply ni at L times del little ri,

but what is ni of L, remember your shape functions, if this is your rod and you have going it from 0 to L

and you discretize this as we have done last time with nodes 0, 1, 2, n minus 1 and the last node is n,