Finite Element MethodProf. C. S. UpadhyayIndian Institute of Technology, KanpurModule o. ! "#$ecture o. ! "% (Refer Slide Time: 00:23) Till now,we have looked at the one-dimensional finite element problem using linear basisfuntion. So, if !ou had this domain, what we have done is, we had partitioned the domaininto elements, we had orresponding nodes and then alled the element, we went ahead andonstruted appro"imation using these funtions phi i, whih are linear# $lso the %uestion - isthat all we an do or an we do more than heking suh linear appro"imation& The answer is!es.So,what toda! wearegoing to look at,ishigherorderappro"imationfor thefiniteelement solution.So, what is the notation& 'et us first introdue when it is a linear appro"imation, then we willrepresent it b! ( e%ual to ), where ( we all has the order of appro"imation# *ow, what wehaveimpliitl!assumed,+ will assumeit throughout this ourse,unless it is speified,speifiall!, that this order of appro"imation is the same in all the elements, that throughoutthedomain,wearegoingtousethesameorder of appro"imation, oneanalwa!s usedifferent order of appro"imation,in different regions of auto mode,that ase we will nothandle now.(Refer Slide Time: 02:30) So, let us now see, how this order of appro"imation an be inreased. So, the ne"t e"amplethat would we have to do is,( e%ual 2, that is the order of appro"imation is %uadrati.-orthis, let us see, what is it that we have to do, as far as the appro"imation is onerned. So, forthis, !ou take the same domain that we had taken earlier, we send n load p, this is " e%ual tol, this is " e%ual to 0 and some distributor fores f ". -or f ", let us also takes .$ is e%ual toonstant in the domain, that need not be true. So, using this, given this data for the problem,let us now go ahead and do the appro"imation.So,how are we going to do this& /hat weneed to do now, is to first break the domain into element as we have doing till now.0a! be + will take 1 elements here.So,these partitions that + have drawn,using these endvertial lines2 these are our elements. So, this is element ), this is element 2, this is element 3and this element 1. So, in this element, in eah of this element, what + am going to do is addan e"tra point# So,in eah of the element,+ am going to add an e"tra point# This point isloated at the enter of the element.So,now we are going to do this2 the loations of thispoint. So, now, instead of having ), 2, 3, 1, 3 points for the 1 elements that we have drawn,now we have 4 points. So, we will give them name# So, this is point " ), this is " 2, this ispoint " 3, " 1, " 3, " 5, " 6, " 7, " 4. /hat !ou should note, is that these e"tremities of the element are given b! the point " ), " 3,so, these are the e"tremities of element ), e"tremities of the element 2 are the point " 3 and "3, e"tremities of the elements 3 are the points " 3 and " 6, e"tremities of the element 1 are thepoint " 6 and " 4. These points " 2, " 1, " 5, " 7 are generall! alled mid time nodes or midedge nodes. So, we will give a name to them, these are mid side nodes and the points " ), "3, " 3, " 6, " 4 are alled these n verties or simpl! the e"tremities of the elements. So, thispoint beomes end nodes or elements nodes or e"tremities of the elements# *ow, what do wedo ne"t&(Refer Slide Time: 06:06) So,what will do is - over this point heked that we have reated,we are going to use itsomehow to reate our basi funtion, whih are %uadrati in nature. So, + have this point set2+ am going to onstrut the basis funtion,whih now should be %uadrati in the elements.So, first, + am going to draw, then, we will talk about how to onstrut mathematiall!#So,whiledrawing,what hadwesaidabout thelinear funtion,what wewantedthesefuntions to do, what should have a value ) at one of the nodes and 0 at all other nodes, thesame priniple we are going to use as far as onstrution of the %uadrati funtions. So, let ussa! the funtion phi ) that is orresponding to the nodes " ) or the point " )# The phi ) will be) at the point " ) and 0 at all other points, so, if + want to make that phi ) as this funtion.Similarl!, + would like to make phi 2, phi 2 will be a funtion, whih is ) at this point and thepoint " 2,and 0 at all other points# So2 what we see is that we have defined it in a loalregion, that is, phi 2 is onl! defined in the first element.So, if + an draw that element here, this is element + ), this is element + 2, this is element + 3and this is element + 1, then + an draw phi 3# $ording, phi 3 should be ) at this point and 0ever!where else.So,b! the same token that + have done phi 3 like this,then phi 1.phi 1should be ) here, 0 at all other points phi 1. Similarl!, phi phi phi phi will be ) here, 0 willbe where are8 Similarl!2 if + want to use, phi 5 will be this, phi 6 will be this, phi 7 will bethis and phi 4 will be this.So, what we got is phi 2 is here, so naming wise2 this is phi 3, this is m! phi 1, this m! phi 3phi, sorr!, + made a mistake here. So, + have to go bak and hange m! naming onvention.So, before phi 2, this is going to be phi 3, this olumn is going to be phi 1, this one is phi 3,this one is phi 5, phi 6, phi 7 and then phi 4.(Refer Slide Time: )2:27) So,here,+ have the 4 phi,whih + have onstruted, the! look %uadrati,we have to alsogive an e"pression to them mathematiall!, wh! are the! %uadrati& 9eause the! have avalue ) at one point, and 0 at two other points, and then we e"tend it as 0 ever!where. So,the! should be %uadrati, that means the! look like that. So, in some will this funtion, howam + going to write m! 300 ml solution. So, u - . of ", we will use our earlier notation, is itis going to be, u is the 4-term solution u 4 of "# This is e%ual to sigma i is e%ual to ) to 4 u iphi i. So, this is a pi"el optial. So2 now, what is this, that the funtion has to satisf! in order to beadmissible basis funtion& The first thing that we have done till now is that, this funtion phii should be,it should be linearl! independent, seond thing that the! have to satisf! is thatthe! should be omplete, and third thing that we have done b! onstrution is the! shouldloal support,that is,the! are onl! defined in one or two elements and in the rest of theelement there, 0.So,b! onstrution,we see that this propert! has satisfied linear independene, we do notknow this,we do not know.So,let us first make this funtion in suh a wa! that the! are%uadrati, linearl! independent and omplete. So, how are we going to do that& So to do that,let usnowonsult this funtion element b! element.So,if+lookatelement - agenerielement + k - if !ou look at the generi elements + k, what are the nodes that this elements hasor the point it will be& " 2 k minus ) " 2 k " 2 k plus ). Similarl!, whih is the phi, whih isnon-:ero here& So, phi whih are non-:ero, here are phi 2 k minus ) phi 2 k plus ).(Refer Slide Time: )3:)0) So, let us now go and transform this to our element notation that we had introdued earlier.So, from the element notation point of view, in the element, + will all the first node as " ) ofan element k, the seond one, " 2 of an element k, and the third one is " 3 of element k.Similarl!, the phi 2 k minus ), in this element ould beome, would be alled * ) of elementk and phi 2 k will be alled * 2 of k. Similarl!, phi 2 k plus ) will be alled n 3 of k# $nd likein the ase of linear appro"imation,in this we want this funtion * ) of k to be to have avalue ) at the point " ) of k and 0 at all other points. * 2 of k should have a value ) at thepoint " 2 of k 0, at all other points, * 3 of k should have a value ) at the point " 3 k and 0 atall other points. So, what we an write in short is, * i k as the point " i k is e%ual to ) and atall other points ; it is e%ual to 0# Set propert! is what we want to have for this funtion. So, ifwe want to onstrut * ) k with this propert! suh that * ) k is 0 at the point " 2 k and " 3 k.So, how will + form the funtion * ) k& This should minus at the point " 2 k, ertainl!, thisrepresentation should have this form,something into is there,then at " 2 k this e"pressiongoes to 0,it should also vanish at the point " 3 k# So,it should also have this part sittingthere#