Finite Element MethodProf. C. S. UpadhyayDepartment of
Mechanical Engineering Indian Institute of Technology, anpurModule
! " #ecture ! $In the previous lecture, we had looked at the finite
element method and we had done a simpleimplementationof that
methodusingthehat-shapedfunctions or thelinear functions toaproblem
of a bar.(Refer Slide Time: :!!"If I can draw that problem again,
this is the bar with an end load # and sub$ected to a
uniforml%distributed load f. So, in that case, we had built up the
element e&uations. 'e introduced theidea of element
calculations and then we had discussed how to take these element
e&uations andput them in the global matrices( that is, the
global stiffness matri) * and the global load vector +.The element
calculations which we had written as * of l, element l and the
matri), and the loadvector from the element + due to the element l
- how these e&uations will be put in these
globale&uationsintermsof*and+,'ehadsaidthat thiswill
finall%afterassembl%leadtoane&uation of this form: * -
e&ual to +. 'hat we had done in the last class was we had
developedthis matri) *, we had developed this vector +, but we had
not taken into account the effect of theboundar% conditions. .ow do
we account for the boundar% conditions, 'e had taken care of
thisend force # here b% sa%ing that if we consider the domain is
broken into / elements as we haddone into 0 elements with nodes )1,
)2, )!, )3, )0, )/. )/ is e&ual to 4 and )1 is e&ual to
.(Refer Slide Time: 2:0!"'e had said that in the entr% +/, we
updated it e&ual to +/ plus #. That is in the final entr% of
thevector +, we added the value of this end load # which is applied
and wh% we did it was discussedin the last class. So this part was
taken care of. 'e had left at was a little bit more complicatedpart
of an application of boundar% conditions, which is how do we impose
the 5irichlet or theessential boundar% conditions, This is where we
had trailed off in the last lecture. 'hat do wehave for our
problem, 'e know that for our problem, u si) is the appro)imation
that we havemade. This is e&ual to sum of i e&ual to 1 to /
uiphii) and we know that u si) atfor thisproblem has to be
e&ual to . 6t ) e&ual towhich is the node )1onl% phi1is
e&ual to 1. 6llotherphi(s"are.Sothisise&ualtou1(ReferSlide
Time: 3:17".Thisconditionhastobee)plicitl% imposed in thesolution.
.owdowee)plicitl% imposeit, 4etus go to thes%steme&uations that
we have. If %ou remember, what was our global stiffness matri), 4et
us start withthe s%stem e&uations that we had written.(Refer
Slide Time: 3:0!"'hat did we have earlier, I will write it in a
little bit general wa%. +irst e&uation was plus 86 b%h1, minus
86 b% h1 and , , , . Second one was minus 86 b% h1, here I will
get, because of theassembl%, 86 b% h1 plus 86 b% h2. Third one will
be minus 86 b% h2, , , . This wa%, I willcontinue till the si)th
e&uation which is , , ,minus 86 b% h0 and here I will have plus
86 b%h0. This is the global matri) *. This into the u was e&ual
to the +. .ere we seethat nowherehave we reall% taken careofthe
value of u1(thatis,wehave notimposed the fact that the value of u1
has to be e&ual to specified value. 4et us now be a little
bitfurther general. 'e will sa% that u atis given the value u bar.
+rom what we have done earlier,this will be e&ual to u1. .ow do
we impose this in our solution, The idea is ver% simple. 'e
willtake the first e&uation and what I will do is I will
replace this with 1 (Refer Slide Time: 9:10min". The first entr%,
that is the diagonal entr%, corresponding to the first row, that I
am going tomake as 1. The second entr% I am going set it ( that is,
all other entries in this row, the first rowcorresponds to u1. The
diagonal entr% of the first row, I am going to make it 1. 6ll other
entries, Iam going to make it e&ual to . 6t the same time, what
am I going to do to the +1, I am going toput +1 is e&ual to u
bar. :ust look at the first e&uation.(Refer Slide Time: 7:2"In
the first e&uation, what do we end up getting, The first
e&uation becomes 1 into u1 is e&ual tou bar. So this b%
modif%ing the first row of the stiffness matri), I have enforced
the fact that u1hastocomeout tobee&ual toubar.
The&uestionisthat whenweenforcethis, theothere&uation has
to change also.(Refer Slide Time: 7:!1"If I go back to the previous
e&uation, in the second e&uation, this is the part (Refer
Slide
Time:7:!;"correspondingtou1.Thisistheentr%intheseconde&uationcorrespondingtou1.'eknow
that u1 is e&ual to u bar. .ow do we remove this from this side
and put it into the knownpart, odified +2 will be e&ual to, b%
thistoken, what we had earlier as +2 minus of, we had *i1 that is
the first column of the second row(it will be minus of 86 b% h1into
ubar. This will become the modified right hand side for thesecond
e&uation. 4et us rewrite it( what did we have as our +2, +2 was
f b% 2 into h1 plus h2( soto that, I am going to add 86 u bar
divided b% h1. This becomes the so-called corrected or themodified
right hand side for our problem. ?nce we had transferred this
information to the righthand side, then we are going to set *i1is
e&ual to . That is we are going to blank out all
theentriesinthefirst column, forall therows. Thisisfori e&ual
to2to, forourproblem, /.@ertainl%, we are not going to do this to
the first row because then it makes no sense. +rom thefirst row, we
want u1 is e&ual to u bar and from the second row onwards, we
want to take care ofthe known values of u b% suitabl% modif%ing our
load vector.(Refer Slide Time: 12:3/"If I now use this and write
the s%stem, I will end up getting this s%stem( I will call it
*modified intodisplacement vector - is e&ual to the +modified.
?nce I have these two things, then I can solve for-. .ow do I solve
for -, I can solve for - b% taking the inverse of *. - will be
e&ual to *>inverse into +>. .ere, we are not going to
talk too much about what kind of solvers. 'hat Iwouldre&uest
thestudent shoulddois togouse some commerciall%available code.
+ore)ample, we can use >atlab or >athematica to solve this
problem. That is we can feed to theseprograms, the matri) *, the
vector + and ask it to return back to the vector u. ?nce %ou know
thecomponents of u, then %ou know %our solutions u si) because this
will be e&ual to sigma ui phii). This is how we will construct
the finite element solution to a problem.Aow the &uestion
arises - wh% did we not ignore the first e&uation right
awa%,
