Finite Element MethodProf C. S. UpadhyayDepartment of Mechanical EngineeringIndian Institute of Technology, anpurModule ! " #ecture ! $In the previous lectures, we had talked about how to compute the stiffness matrix and the load vector foragivenmodel problemusingthefiniteelement method. What weretheessential featureswhat wediscussed in the previous lecture? We had defined these global basis functions phii, which were given asfunctions of this type.(Refer lide !ime" ##"$#%!hat is, the functions which at the node i had a value & and at all other nodes they had a value #. !heseare so'called piecewise linear functions that we have formed( they were continuous you can see that theyare non )ero only in the elements i minus & and element i. !hese were functions with local support. *nething that we should answer before going further is why did we choose these functions?If you remember in the previous lectures, we had talked about the problem with point loads. We saw inthat casethat theproblemwithpoint loads was that it has piecewisesmoothfunctions whicharecontinuous at the points and the derivatives have +umps at those points. o that was the motivation forusing such functions in the Rayleigh'Rit) ,ethod. -ow, it is not always obvious what kind of functionswe have to use for the given problem of interest? We should have some kind of general mechanism bywhich you can see what is the minimum smoothness re.uirements on the function phii that we are usingto form our global basis functions, before we can define global basis functions. /or different differentiale.uations you will have different phii0s. o let us see how to do it.If you remember that we had made our weak formulation of this type" integral x is e.ual to # to 1 23 duby dx and into dv by dx, where I will be using v or w as my weight functions and this on the right handside is e.ual to integral x is e.ual to # to 1 f v dx plus as we had put an end load 4 v evaluated at x e.ualto 1. o this was our weak form for the model problem that we have taken. o now let us look at thispart. If I look at this part here you see that this term and this term are sitting in this integral. o we wantthis integral to be finite. o the integral should be finite. -ow what does it mean? If this integral has to befinite then each of these terms du by dx dv by dx have to be defined in the domain of interest. o whatwe want is du by dx and dv by dx, they should both be defined in domain omega or I.(Refer lide !ime" #$"5#%o what are the roughest or the least smooth of these functions u and v for which du by dx and dv by dxwill be defined. !he answer is very simple that is, if we look at this domain we can have du by dx and dvby dx defined means that they can be like this. !hat is at particular points I will call point xa, point xb, andpoint xc. 3t these particular points the derivatives can have +umps. !his is if you remember I am writing itas du by dx dv by dx. o if this is the function that we have, then in that case the integral on the left handside in the previous page is defined. 3nything worse than that in this is if I have in a way worse than the+ump means that there will direct delta at these points data derivatives that is in fact even the u can bediscontinuous. In that case this integral will be infinite( it will not be defined. In order to have theseintegrals finite we ask the .uestion what is the worst type of functions that will satisfy that re.uirementand the answer here isthe functions forwhich derivatives have +umps at specificpoints. !hatis thereason why we say that if the derivative has +umps at specific points, then u and v are continuous in thedomain. In such way that at certain points slopes change. o with this re.uirement we ask, what are thefunctions that can be used in our series representation? 6ecause if you remember, in this series also weneed functions which can capture this kind of behavior.(Refer lide !ime" #7"#7%o the phii0s that we have should also satisfy this minimum re.uirement of the derivative. o we want phifor this particular problem, because we want the left hand side to be defined( should also be continuousor we say it lies in 8 )ero. !hat is the reason why we took phii0s as the hat shaped functions that we aretaking. Wecouldhavetakenit tobesmoother, but thesmoother functions wouldnot satisfythisminimum condition and you remember that whenever you are developing a method that is a methodwhich will translate into a computer program you do not want to write the program for one particularproblem( you want to write it for a class of problems.8lass of problems in this case is defined by what? ' 6y different types of boundary conditions, differenttypes of body force, and different types of materials. It turns out that in order to look at a big class ofproblems of practical interest we have to talk about the point loads. o in this case whenever we aretrying to write a program or whenever we trying to define a finite element method we would like to coverthe solutions for all these types of problems that are of interest and for the point load this phii with +umpinderivativeswasneededorwiththematerial interfacesweneededanditcomesoutalsofromourminimum re.uirement for the left hand side to be finite. o that is the reason why one chooses one typeof basis functions for a given differential e.uation. We will see later on when we go to the beam problemthat these basis functions, the definitions of these are going to change. !here again we will see what isthe minimum re.uirements for which we have to define the basis functions.o do not think that the whole process is arbitrary in nature. 2verything comes for a particular reason andthe reason is we have to go back our weak formulation and look at the left hand side, look at the righthand side and ask the .uestion what will make this left hand side and right hand side finite. o this is onething that I wanted to harp on and that is a very important reason, very important thing that we have tokeep in mind when we are doing the finite element method. !hird thing that we obtained was, becausethis function phii were locally defined that is phii was non )ero in only elements i minus & and element i,so in that case what happened is the integral were not all non )ero. !hat is the stiffness matrix was sparse.-ot only sparse, it was banded that is some of the elements around the diagonal that is either side of thediagonal were only non )ero everything else in the matrix was )ero. !oday let us go further and see howdo we go and do everything from a computer based orientation.(Refer lide !ime" "$7%/inally, we are not going to solve this problem by hand. We would like to write a computer programwhich can give us the answer to this problem. o from this point of you we would like to go head. !ostart off, let us if you remember that I take the i+ th entry of the stiffness matrix. o what does it mean?I am looking at the ith row and the +th column of the stiffness matrix and what do we get for this entry?!here are few possibilities that we have. !hat is for + e.ual to i minus &, what do we have? that for + e.ualto i minus & I will have in the element x i minus & to xi, if you remember it will be e.ual to 23 phii prime phiiminus & prime dx for + is e.ual to i minus &. !his is all that is only in the element i minus & this part is non')ero everywhere else it is )ero. Why? because if you see here phiiis non )ero in elements i minus &elements and element i while phii minus & is non )ero in only element i minus & but it is )ero in element i. oelement i does not contribute. imilarly let us look at the second possibility. econd possibility is, when +is e.ual to I, so in that case we will get integral over the element i minus & phii prime phii prime dx plusintegral over element i 23 phii prime d s.uare x for + is e.ual to i.!hird case is when + is e.ual to i plus &. !hen this integral becomes xi to xi plus & 23 phii prime phii plus &prime dx for + is e.ual to i plus &. /or all other cases it is )ero for all other +0s. 9ou see that only thesethree entries are non )ero in the ith row everything else is )ero. 9ou see that the ith row i minus Ithcolumn entry the ith column entry i plus Ith column entry are non )ero. !his is the structure that we willhave for all i0s. imilarly what will the /i be?(Refer lide !ime" &$":#%!hat is the load vector it will be e.ual to because phiiis non )ero in the element i minus & and theelement i so xi minus & xi f phii dx plus integral xi to xi plus & f phii dx( plus we have 4 phii at x e.ual to 1 forour model problem. 1et us ignore this part for the time limit. !hat is for the time being, let us forget thisthing. We will see why. I mean you should be able to know the reason why. 9ou note that, for all phi i0s,because we have taken a six element mesh, that is for i is e.ual to five element mesh plus or for themodel problem, i is e.ual to ; for that case only the phii at x e.ual to 1 is &( all other phii0s are )ero here.o this part 4 into phiiis only going to contribute to f;that is in the general case f n plus &. !o all other fphii0s it is not going to contribute that is why let us keep it out for the time being, we will handle thislater. o now, we have this part of the load vector and if I go to the previous page (Refer lide !ime"&;"&