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Introduction

A finite element method (abbreviated as FEM) is a numerical technique to obtain an approximate solutioa class of problems governed by elliptic partial differential equations. Such problems are called as boun

value problems as they consist of a partial differential equation and the boundary conditions. he felement method converts the elliptic partial differential equation into a set of algebraic equations !hicheasy to solve. he initial value problems !hich consist of a parabolic or hyperbolic differential equation the initial conditions (besides the boundary conditions) can not be completely solved by the finite elemmethod. he parabolic or hyperbolic differential equations contain the time as one of the indepenvariables. o convert the time or temporal derivatives into algebraic e"pressions# another numerical technli$e the finite difference method (F%M) is required. hus# to solve an initial value problem# one needs bothfinite element method as !ell as the finite difference method !here the spatial derivatives are converted algebraic e"pressions by FEM and the temporal derivatives are converted into algebraic equations by F%M

&istorical 'ac$ground

he !ords finite element method !as first used by *lough in his paper in the +roceedings of ,nd AS*E (American Society of *ivil Engineering) conference on Electronic *omputation in -/0.*lough e"tended the matri" method of structural analysis# used essentially for frame1li$estructures# to t!o1dimensional continuum domains by dividing the domain into triangular elementsand obtaining the stiffness matrices of these elements from the strain energy e"pressions byassuming a linear variation for the displacements over the element. *lough called this method asthe finite element method because the domain !as divided into elements of finite si2e. (Anelement of infinitesimal si2e is used !hen a physical statement of some balance la! needs to be

converted into a mathematical equation# usually a differential equation).

Argyris# around the same time# developed similar technique in 3ermany. 'ut# the idea of dividingthe domain into a number of finite elements for the purpose of structural analysis is older. It !asfirst used by *ourant in -45 !hile solving the problem of the torsion of non1circular shafts.*ourant used the integral form of the balance la!# namely the e"pression for the total potentialenergy instead of the differential form (i.e.# the equilibrium equation). &e divided the shaft cross1section into triangular elements and assumed a linear variation for the primary variable (i.e.# thestress function) over the domain. he un$no!n constants in the linear variation !ere obtained byminimi2ing the total potential energy e"pression. he *ourant6s technique is called as appliedmathematician6s version of FEM !here as that of *lough and Argyris is called as engineer6sversion of FEM.

From -/0 to -78# the FEM !as developed in the follo!ing directions9

(-) FEM !as e"tended from a static# small deformation# elastic problems to

dynamic (i.e.# vibration and transient) problems#

small deformation fracture# contact and elastic :plastic problems#

non1structural problems li$e fluid flo! and heat transfer problems.

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(,) In structural problems# the integral form of the balance la! namely the total potential energye"pression is used to develop the finite element equations. For solving non1structural problemsli$e the fluid flo! and heat transfer problems# the integral form of the balance la! !as developedusing the !eighted residual method.

(5) FEM pac$ages li$e ;AS?S etc.

'asic Steps

he finite element method involves the follo!ing steps.

First# the governing differential equation of the problem is converted into an integral form.

hese are t!o techniques to achieve this 9 (i) Bariational echnique and (ii) Ceighted

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In this step# over a typical element# a suitable appro"imation is chosen for the primary

variable of the problem using interpolation functions (also called as shape functions) andthe un$no!n values of the primary variable at some pre1selected points of the element#called as the nodes. ?sually polynomials are chosen as the shape functions. For -1%elements# there are at least , nodes placed at the end1points. Additional nodes areplaced in the interior of the element. For ,1% and 51% elements# the nodes are placed at

the vertices (minimum 5 nodes for triangles# minimum 4 nodes for rectangles#quadrilaterals and tetrahedral and minimum D nodes for parallelepiped shaped elements).Additional nodes are placed either on the boundaries or in the interior. he values of theprimary variable at the nodes are called as the degrees of freedom.

o get the e"act solution# the e"pression for the primary variable must contain a complete set ofpolynomials (i.e.# infinite terms) or if it contains only the finite number of terms# then the number ofelements must be infinite. In either case# it results into an infinite set of algebraic equations. oma$e the problem tractable# only a finite number of elements and an e"pression !ith only finitenumber of terms are used. hen# !e get only an appro"imate solution. (herefore# the e"pressionfor the primary variable chosen to obtain an appro"imate solution is called an appro"imation). heaccuracy of the appro"imate solution# ho!ever# can be improved either by increasing the numberof terms in the appro"imation or the number of elements.

In the fourth step# the appro"imation for the primary variable is substituted into the

integral form. If the integral form is of variational type# it is minimi2ed to get the algebraicequations for the un$no!n nodal values of the primary variable. If the integral form is ofthe !eighted residual type# it is set to 2ero to obtain the algebraic equations. In eachcase# the algebraic equations are obtained element !ise first (called as the elementequations) and then they are assembled over all the elements to obtain the algebraicequations for the !hole domain (called as the global equations).

In this step# the algebraic equations are modified to ta$e care of the boundary conditions

on the primary variable. he modified algebraic equations are solved to find the nodalvalues of the primary variable.

In the last step# the post1processing of the solution is done. hat is# first the secondaryvariables of the problem are calculated from the solution. hen# the nodal values of theprimary and secondary variables are used to construct their graphical variation over thedomain either in the form of graphs (for -1% problems) or ,1%51% contours as the casemay be.

Advantages of the finite element method over other numerical methods are as follo!s9

he method can be used for any irregular:shaped domain and all types of boundary

conditions.

%omains consisting of more than one material can be easily analy2ed.

Accuracy of the solution can be improved either by proper refinement of the mesh or by

choosing appro"imation of higher degree polynomials.

he algebraic equations can be easily generated and solved on a computer. In fact# ageneral purpose code can be developed for the analysis of a large class of problems.

bectives of the *ourse

he obectives of the course are as follo!s 9

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o develop the finite element formulation for a model one1dimensional problem li$e a"ially

loaded bar for the case of simplest appro"imation (i.e.# linear appro"imation).

o discuss the possible refinements of the simplest appro"imation.

o develop the frame !or$ of a finite element code to solve the one1 dimensional

problem.

o e"tend the finite element formulation to other one1dimensional problems li$e the beam

problem. o develop the t!o1dimensional finite element formulation for a model ,1% problem li$e ,1

% steady1state heat conduction problem.

Introduction

In this lecture# integral formulations of a boundary value problem are developed. here are t!otypes of integral formulations 9

Cea$ or Ceighted residual formulation and s

Bariational formulation

In finite element method# the solution of a boundary value problem is obtained by using one ofthese t!o integral formulations. Chen it is difficult to solve the differential equation of a boundaryvalue problem# this method provides an alternative !ay to obtain the solution. 'ut# usually# it is anappro"imate solution

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Model 'oundary Balue +roblem

o illustrate the development of integral formulations# the follo!ing model boundary valueproblem is considered. It represents the a"ial e"tension (or compression) of a bar sho!n in Fig.,.-.

Figure 2.1

he bar has a variable area of cross1section !hich is denoted by the functionA(x). he length of

the bar is L . he =oung6s modules of the bar material is E . he bar is fi"ed at the endxG 0. he

forces acting on the bar are (i) a distributed force f(x)# !hich varies !ithxand (ii) a point force P

at the endxG L . he a"ial displacement of a cross1section at x # denoted by u(x)# is governed bythe follo!ing boundary value problem consisting of a differential equation (%E) and t!o boundaryconditions ('*)9

%E9 0H"H@ (,.-a)

'*9 (i) u G 0 at "G0 (,.-b)

(ii) EA(")at "G@ (,.-c)

he differential equation represents the equilibrium of a small element of the bar e"pressed interms of the displacement using the stress1strain and strain1displacement relations. heboundary condition (,.-b) is a geometric or $inematic boundary condition. Since# it is a condition

on the primary variable u(x)# it is called as Dirichlet boundary condition. he second boundarycondition (condition -c) is a force boundary condition# or a condition on the secondary variable(i.e.# a"ial force). Since# it is a condition on a derivative of the primary variable it is called as theNeumann boundary condition.

Cea$ or Ceighted

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(,.-a)# it !ill not be equal to f (x ). In this case# the difference is called as residue or error and isdenoted by R (x ). hus#

(,.,)

In Ceighted

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E"pression (,.8b) is also called as the Weak Formulation of the boundary value problem (,.-a)#(,.-b) and (,.-c) because the solution given by the formulation is required to satisfy !ea$ersmoothness conditions compared to that of the solution of the original boundary value problem.

%epending on the choice of w # various special forms of the !eighted residual method e"ist. heyare 9(i) 3aler$in Method#

(ii) +etrov 3aler$in Method and(iii) @east Square Method.If the condition of smoothness on w (i.e. the 5rdcondition) is rela"ed# t!o more special formsemerge9(i) Sub1domain Method and(ii) *ollocation Method.In this lecture# only the 3aler$in version of the !eighted residual method !ill be developed. Forthe details of other version# see the boo$ by &uebener..

Birtual Cor$ Formulation

+hysical interpretation of equation (,.8b) e"ists if ! is interpreted as a virtual displacement. heconcept of a virtual displacement can be defined as follo!s. ;ote that# because of the forcesacting on the bar# it !ill have some real displacement. Even though it is un$no!n# a typical realdisplacement can be represented by a solid curve sho!n in Fig. ,.,. ;o! imagine that the barundergoes some additional small displacement# !hich is not real but imaginary. Such a

displacement is called as virtual displacement and is denoted by the symbol . he dotted line

of Fig. ,., represents the graph of . hus is a function !hich represents a virtual

small change in the value of u(x)at every point of the interval J0# L K.

Fig 2.2

Ce assume that satisfies the same admissibility requirements as that of w . hus#

1. he virtual displacement must be 2ero at the boundary !here u is 2ero. hus# in the

present problem# G 0 atx = 02. he virtual displacement must be unconstrained at the boundary !here the derivative of

u is specified. hus# in the present problem# is unconstrained atx =L

3. he virtual displacement must be smooth enough for an integral of the product of

R(x)and to be finite. After the integration by parts# this integral involves the derivative of

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. hus# in the present problem# must be finite at every point of the intervalJ0# @K.

hus# the virtual displacement is chosen from the same class of admissible functions from!hich !is chosen.

If wis set equal to , equation (2.5b) becomes

(,./)

E"tension of a bar is a one1dimensional problem. herefore# u is independent of the other t!o

coordinatesy andz . Further# all the stress components other than are 2ero. In particular#

and are 2ero. hus#

(,.7a)

(,.7b)

Further#

dV=A dx (,.D)

(,.)

;ote that each term in equation (,.) is an e"pression of the virtual !or$. he left side of equation(,.) represents the virtual !or$ done by the internal forces. he first term on the right side ofequation (,.) represents the virtual !or$ done by the distributed e"ternal force f("). he secondterm on the right side of the equation (,.) represents the virtual !or$ done by the point e"ternalforce +. ogether the right side of equation (,.) represents the total virtual !or$ of the e"ternalforces. Chat equation (,.) represents is that9 for the residue to be 2ero (or for u to be a solutionof the boundary value problem ,.-a# ,.-b and ,.-c in some sense)# the total virtual !or$ must be2ero or the virtual !or$ done by the internal forces must be equal to the virtual !or$ done by thee"ternal forces.

If the symbol in the e"pression is interpreted as an operator some!hat similar to thedifferential operator d# equation (,.) can be transformed to an e"tremi2ation problem of aquantity called as functional. his gives the second integral form of the boundary value problem.'efore it is formulated# it is first necessary to discuss the idea of a functional and its e"tremum.his is discussed in the ne"t section

Functional

Functional is an operator# !hich operates on a function and returns a number. In other !ords#functional is a function !hich is defined over a set of functions and !hose range is a set ofnumbers. he set of functions !hich constitute the domain of a functional is usually required tosatisfy certain conditions on smoothness andor on the values at the end : points of the interval.Such a set is called as a set or a class of admissible functions.

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As an illustration of a functional# consider a set of functions u(") !hich are functions of a single

variable " for 0 " @# !hich have the value 2ero at " G 0and !hich possess a continuous firstderivative at all points of the interval J0# @K. hen# e"amples of functional are9

(,.-0)

(,.--)

(a(") is a $no!n function)# (,.-,)

(b(") is a $no!n function)# (,.-5)

(c is a $no!n number). (,.-4)

In each of the above e"amples# the operator ta$es a function u(")and returns a number either

by integrating a quantity depending on u# and some $no!n functions or by evaluating u atsome point of the interval. he integrand may involve higher derivatives of u or some functions ofuli$e logarithmic# trigonometric# e"ponential etc or the po!er of u# its derivatives or its function. Inclassical boo$s# a functional is often !ritten as9

(,.-8)

!here

(,.-/)

&o!ever# this case does not include a type of functional defined by equation (,.-4).

he branch of calculus# !hich deals !ith the operations performed on functionals# is called as thevariational calculus. Some ideas from the variational calculus need to be discussed ne"t.

Bariation and E"tremum of a Functional

*onsider a set of admissible functions on !hich a functional is defined. If one of the admissibility

conditions is uG 0 atxG 0# then a typical function from this set can be represented as sho!n inFig. ,.5(a).

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Figure 2.3 (a) Figure 2.3 (b)

A small change in the argument of I is called as variation (or first variation) of u and is denoted by

u. It is defined as a small change in the value of u at every point of the interval J0# @K. ;ote that

u is a function of " over the interval J0# @K. Since u is constrained at " G 0# no change ise"pected in u at " G 0. herefore# the variation u ta$es the value 2ero at the end1point of the

interval !here u is specified. he function u is sho!n in Fig. ,.5(b). Since the value of u issmall every!here# it is often e"pressed as

u = (,.-7)

!here

Ga small number and

G an arbitrary function of " !hich satisfies the condition = 0 atx= 0.

his small change in the argument of I induces a change in the value of I. It is called as the

variation (or first variation) of I and is denoted by . o find the e"pressions of # !e consider

the value of the functional at the argument or . ;ote that I( ) is a function

of for given u and . E"panding about G 0# !e get

(,.-D)

he quantity represents the variation . hus#

= = (,.-)

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he functionalIis said to be e"tremum (minimum or ma"imum) !hen or is 2ero.he corresponding e"tremi2ing function u is found from the condition

(,.,0)

for every function such that G 0 at x G 0 and belongs to the set of admissiblefunctions.

Chile deriving the variational integral form from equation (,./)# some properties of the variational

operator are needed. hey are stated ne"t. ;ote that is an operator similar to the differential

operator d operating on usual functions. So# the operator obeys certain properties of d operator

li$e the product rule etc. Further# operator commutes !ith d operator# integral operator etc. therelevant properties needed in the derivation are9

(i) , (,.,-)

(ii)

, (f is $no!n function or a constant)# (,.,,)

(iii)

, (,.,5)

(iv),

(,.,4)

(v), (,.,8)

(vi), (,.,/)

(vii)(,.,7)

Bariational Formulation

;ote that !hile deriving equation (,./)# it has been assumed that the function usatisfies both the

boundary conditions (,.-b) and (,.-c). Additionally# if it is assumed that u also satisfies thedifferential equation (thereby ma$ing u the solution of the boundary value problem (,.-a)# (,.-b)and (,.-c)# equation (,./) becomes9

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;o!# applying the properties of the variational operator # the first term becomes9

(+roperty iii)

(+roperty i)

(+roperty ii)

(+roperty iv)

Similarly# the second term becomes

,

,(+roperty ii) #

(+roperty iv)

Finally# the last terms becomes

,

, (+roperty ii)

(+roperty iv)

Substitution of equations (,.,1,.5-) into equation (,.,D) leads to

?sing property (vi)# this becomes

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(,.5,)

!here

(,.55)

hus# !e have proved the follo!ing result. he solution u of the boundary value problem (,.-a)#(,.-b) and (,.-c) e"tremi2es the functional I given by equation (,.55). he functional I is calledthe variational functional of the boundary value problem (,.-a)# (,.-b) and (,.-c). ;ote that thefunctional (,.55) provides an integral form of the boundary value problem. he integral needs tobe e"tremi2ed to obtain the solution of the boundary value problem. his formulation is called asthe Variational Formulation . Since it is derived from the virtual !or$ formulation (equation ,./)#it is equivalent to the principle of virtual !or$. @i$e equation (,./)# this formulation also has aphysical interpretation. ;ote that the quality I (e"pression ,.55) represents the total potentialenergy of the bar. herefore# the variational formulation states the follo!ing principle. he solutionof the boundary value problem (,.-a)# (,.-b)# and (,.-c) e"tremises the total potential energy ofthe bar. his is called as the Principal of the tationary Value of the !otal Potential "nergy .

In the variational formulation# solution of the boundary value problem is obtained by e"tremi2ingthe corresponding functional. Chile doing so# a set of admissible functions is chosen. heconditions !hich the e"tremi2ing function u is supposed to satisfy are similar to the conditions!hich the !eight function ! is e"pected to satisfy. hese conditions are

-. he function u must satisfy the geometric or $inematic boundary conditions (equation -b).

hus u G 0 at x G 0. Further# the variation u must be 0 at this point.,. he function u and its variation u must be unconstrained !here the force boundary

condition is specified. hus# u and d u are unconstrained at xG @.5. he function u must be smooth enough to ma$e the functional I finite. hus# u must be

such that dud" is finite at every point of the interval (0# @).

If our starting point is the integral form given by the variational functional (,.55)# rather than thedifferential form given by the boundary value problem (,.-a)# (,.-b) and (,.-c)# then it can besho!n that the function u !hich e"tremi2es the functional I (equation ,.55) actually satisfies thedifferential equation (,.-a). hus# equation (,.-a) is called the "uler "#uation of the functional I(equation ,.55). Chile e"tremi2ing the functional# the set of admissible functions is assumed tosatisfy the geometric (or $inematic) boundary condition (equation ,.-b). hus# equation (,.-b) iscalled as the "ssential $oundary %ondition . he condition at the other boundary (equation,.-c) appears naturally !hile e"tremi2ing the functional I. herefore# equation (,.-c) is called asthe Natural $oundary %ondition .

Euler Equation and ;atural 'oundary *ondition

o derive the Euler equation corresponding to the functional I (,.55)# !e proceed as follo!s. @et ube an admissible function !hich e"tremi2es the functional. hen u satisfies the condition that its

value is 2ero at " G 0. Since v is equal to # at "G 0. ;ote that the e"tremi2ation

condition (,.,0) involves the e"pression for the functional I at . ?sing e"pression (,.55)#!e get

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(,.54)

E"panding the square term and separating the terms in the po!ers of # !e get9

.

(,.58)

Since# the first term on the right side of equation (,.58) is nothing but I(u)# !e get

(,.5/)

a$ing the limit as # !e get

(,.57)

Since the function u e"tremi2es the functionalI# IG 0 . herefore# !e get

(,.5D)

for every !hich satisfies the condition G 0 at " G 0. he e"tremi2ing function u is the solution ofabove equation.

o simplify the above equation# the first term is integrated by parts. hen# equation (,.5D)becomes

(,.5)

Since G0 at "G 0# the second term of equation (,.5) is 2ero. *ombining the remaining t!oboundary terms and also the integral terms# !e get

(,.40)

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;ote that the above equation must be 2ero for a set of infinite functions !hich satisfy thecondition G 0 at "G 0 but other!ise are arbitrary. @et us first choose the follo!ing subset of this

set# namely# the set of functions !hich are also 2ero at "G @. For such functions ( G 0 atxG @)#equation (,.40) becomes

(,.4-)

;ote that# equation (,.4-) is true for every !hich is 2ero at "G 0 and @ but other!ise is arbitrary.here are still infinite numbers of such functions. he only !ay in !hich equation (,.4-) can besatisfied for such an infinite set of functions is that its integrand must be 2ero at every point of theinterval J0# @K. In other !ords# the e"tremi2ing function u must satisfy the differential equation9

(,.4,)

!hich is same as equation (,.-a). *ombining equations (,.40) and (,.4-)# the condition satisfied

by u reduces to

(,.45)

Since# this condition is also true for an infinite number of functions# namely# a set of functions!hich satisfy the condition G 0 at "G 0 but other!ise are arbitrary# the only solution of equation(,.45) is

at x= L.(,.44)

his condition is same as the boundary condition (,.-c).

Chat !e have sho!n is as follo!s. An admissible function u (i.e. a function u that satisfies the

condition u G 0 at "G 0) !hich e"tremi2es the functional I(e"pression ,.55) satisfies the %E(,.-a) (*alled as the Euler equation) and the '* (,.-c) (called as the natural boundary condition).

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$ilinear and &uadratic Forms

he t!o integral formulations discussed above# namely (,.8b) or (,./) and (,.55) can bee"pressed conveniently if the follo!ing notation is introduced. For any t!o functions g(") and h(")defined over the interval J0# @K# !e define

'inear Form (

(,.48)

$ilinear Form (

(,.4/)

&uadratic Form (

(,.47)

&ere# @ is the length of the bar# f is the distributed force acting on the bar# + is the point forceacting on the bar at the end "G @# E is the =oung6s modulus of the bar and A is the area of thecross1section of the bar.If u is the a"ial displacement and ! is the !eight function# then the integral formulation (,.8b) isobtained by replacing the functions g and h by u and ! and equating the e"pressions (,.48) and(,.4/)# hus# the !ea$ or !eighted residual formulation is given by

B(u,w)=F(w) (,.4D)

Further# if h is replaced by the virtual displacement rather than by !# then !e get the integral

form (,./) corresponding to the virtual !or$ formulation9(,.4)

If the quadratic form (,.47) is used instead of the bilinear form (,.4/)# then the integral form(,.55) corresponding to the variational formulation is e"pressed as

(,.80)

;ote that the form (,.80) represents an e"pression# !hich needs to be e"tremi2ed to get thesolution. n the other hand# the forms (,.4D) or (,.4) represent equations# !hich need to besolved to get the solution.

)it* +ethod( Part ,

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(,.57) for I# substituting uN for

u and multiplying all the terms by N # !e get

(5.-)

;ote that# using the above equation is equivalent to using the integral form corresponding to thevirtual !or$ formulation (equation ,./) or the !eighted residual (!ea$) formulation (equation,.8b). Chile using the !eighted residual formulations# ho!ever# one needs to choose anappropriate form for the !eight function ! also. his point is made clear in section 5./

As stated in the introduction# in

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(5.--)

Substituting the e"pression (5.D) for in equation (5.--) and changing the summation inde"

from ito# !e get

( )"

(5.-,)

;o! define the follo!ing quantities 9

(5.-5)

(5.-4)

hen# equation (5.-,) becomes

(5.-8)

In matri" form# this can be !ritten as

(5.-/)

;ote that# if the functions are non1dimensional# then the dimensions of the vectors and

turn out to be that of the displacement and force respectively. hen# the dimension of

becomes that of stiffness (i.e. force per unit length). herefore# is called as the stiffness

matri"# is called the displacement vector and is called as the force vector.

he matri" has the follo!ing properties.

E"pression (5.-5) sho!s that

(5.-7)

hus# the matri" is symmetric. ;ote that

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(5.-D)

hus#

for any non1trivial vector .(5.-)

herefore# is a positive definite matri". his implies that is invertible or e"ists.&ence#

(5.,0)

hus#

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Figure 5.-

Ce choose the follo!ing numerical values of various geometric# material and force parameters.

EA = 1, L = 1, P =10. (5.,4)

Ce choose a t!o1term appro"imation for the solution. hus#

(5.,8)

!here the basis functions are chosen to be the polynomials 9

, . (5.,/)

hen# the derivatives of the basis functions are given by

(5.,7)

Substituting the e"pressions (5.,/) and values from equation (5.,4)# the e"pression for

becomes

=

= ,

=

(5.,D)

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Substituting the e"pressions (5.,5) and (5.,/) and values from equation (5.,4)# the e"pression forOFP ta$es the form 9

#$% = ,

= .

(5.,)

herefore# the matri" form of the algebraic equations (equation 5.-/) becomes

(5.50)

he solution of this equation is given by

(5.5-)

herefore# the appro"imate solution (equation 5.,8) becomes

(5.5,)

o study the improvement in the accuracy of the solution# !e consider the 51term appro"imationne"t. hus#

(5.55)

!here#

(5.54)

he derivatives are given by

(5.58)

Substituting the e"pressions (5.58) and values from (5.,4)# the matri" becomes

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(5.5/)

Further# using the e"pressions (5.,5) and (5.54) and the values from equation (5.,4)# the vectorOFP becomes

(5.57)

hen# the matri" form (equation 5.-/) of the algebraic equations become

(5.5D)

he solution of above equation is

(5.5)

herefore# the appro"imate solution (equation 5.55) is given by

(5.40)

For comparison purpose# the e"act solution is needed. It can be found as follo!s. Substituting thee"pression for f from equation (5.,5) and the values of EA # L and P from equation (5.,4)# theboundary value problem (equations ,.-a# ,-b and ,-c) becomes

0 HxH- (5.4-a)

BC : at x= 0, (5.4-b)

(ii) atx= 1 (5.4-c)

he solution of this problem is given by

(5.4,)

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hus# it is seen that the 51term appro"imation gives the e"act solution. Substituting " G- inequations (5.5,) and (5.4,)# !e get

(5.45)

(5.44)

hus# the ,1term solution# although not e"act# gives the e"act value of the ma"imumdisplacement

+odule - (

'ecture $ ( )it* +ethod ( Part 2

/ntegral Forms %orresponding to a Point Force in the /nterior of the Domain

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)it* +ethod %orresponding to a Point Force in the /nterior of the Domain

As# before# choose an N 1term appro"imation to the solution u (equation 5.D) containing the un$n

coefficients and the $no!n function . he $no!n functions must be linearly independent

must satisfy the condition G 0 at x G 0. In

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Ce choose the same numerical values of the parameters EA # L and P as in e"ample -. hosevalues are given by equation (5.,4). Ce choose the value of force F as

F = 20. (5.84)

First# !e chose the t!o1term appro"imation for the solution !hich is the same as in e"ample -. It

is given by equations (5.,8) and (5.,/). Since the value of EA and the e"pressions for # i G -# ,

are the same as in e"ample -# the coefficient matri" is the same as before (equation 5.,D).For the evaluation of the right side vector# !e use e"pression (5.85). Since# there is no distributed

force# i.e. f G 0# and the force F acts at the midpoint # i.e. # equation (5.85) becomes

(5.88)

Substituting the e"pressions for (equation 5.,/) and the values of P and F (equations 5.,4 and5.84 )# !e get

,

= .

(5.8/)

;o!# the matri" form of the algebraic equations (equation 5.-/) becomes

(5.87)

he solution of this equation is given by

(5.8D)

herefore# the appro"imate solution (equation 5.,8) becomes

(5.8)

;e"t# !e choose the three1term appro"imation for this solution given by equations (5.55) and

(5.54). hen the coefficient matri" is given by equation (5.5/). Substituting the e"pression for

(equation 5.54) and the values of P and F (equations 5.,4 and 5.84)# the e"pression (5.88) forO F P becomes9

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(5./0)

;o!# the matri" form of the algebraic equations (equation 5.-/) becomes

(5./-)

he solution of this equation is given by

(5./,)

herefore# the appro"imate solution (equation 5.55) becomes

(5./5)

hus# there is no improvement in the solution !hen the third term is added.

For comparison purpose# !e find the e"act solution. Substituting # f G 0 and the values ofEA # L # P and F (equations 5.,4 and 5.84)# the boundary value problem (equations 5.48a# 5.48b#5.48c# 5.48d) becomes

%E9 0 H xH 0.8#0.8 HxH-& (5./4a)

'*9 at xG 0# (5./4b)

(ii)

atxG -, (5./4c)

(iii)

(5./4d)

he solution of this problem is given by

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(5./8)

he graphical representation of the three solutions is sho!n in

Fig. 5.8

Figure 5.8

Chereas the e"act solution is piece!ise linear# the appro"imate solution is quadratic over the!hole domain. he ma"imum value of the displacement (i.e. u at x G -) predicted by theappro"imate solution matches !ith the e"act solution. &o!ever# the ma"imum error of theappro"imate solution is 1D.55Q !hich occurs at x G 0.8 (the point of application of F ).

Equations (5.8) and (5./8) give the follo!ing values of the displacement derivatives of the

appro"imate solution and the e"act solution atx G - 9

atxG -,

atxG -. (5.//)

Further# at the point of application of F (i.e.# atx G 0.8) 9

atxG 0.8

at ,

at (5./7)

hus# the appro"imate solution is neither able to provide a good appro"imation to the derivate norable to capture the ump in the derivative.

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Ceighted

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(5.74)

o convert the above equation into a set of N algebraic equations in the un$no!n coefficients #

!e choose N !eight functions as follo!s 9

fo (5.78)

&ere# the functions represent another set of linearly independent functions defined over

the interval J0# L K. ;ote that the functions must satisfy the constraints arising out of the

admissibility conditions of the !eight function . hese conditions are stated in section ,.,.

Chen the functions are different from # the corresponding !eighted residual method

is called as the +etrov13aler$in Method. n the other hand# !hen the functions are the

same as # then the method is called as the 3aler$in Method. he ne"t section describesthe details of the 3aler$in method.

3aler$in Method

In 3aler$in method# !e choose

for .(5.7/)

;ote that the constraints on arising out of the admissibility conditions of are as

follo!s. he first admissibility condition requires that the functions must be 2ero at #

(that is# at the point of application of the %irichlet boundary condition on ). ;ote that this

constraint is identical to the condition (5.7-). he second condition implies that should be

unconstrained at . he third condition requires that the derivative must be finiteover the !hole interval (0# L ).

Substituting the e"pression (5.7/) for in equation (5.74)# !e get

fo (5.77)

;o!# define the follo!ing 9

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,(5.7D)

(5.7)

hen# equation (5.77) becomes

(5.D0)

In matri" form# this can be !ritten as 9

(5.D-)

;ote that# !e get the same set of algebraic equations as in the

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Figure 4.1

&ere also# the e"act solution is piece!ise linear as sho!n in Fig. 4.,.

Figure 0.2

his suggests that piece1ise linear basis functions may be a better choice in simulating theumps in the derivatives or providing a more accurate appro"imation to the derivatives. ne suchset of ( N R-) functions is sho!n in Fig. 4.5. o construct such functions# the domain is partitioned

into N parts (Fig. 4.5) !ith the intermediate points . he functions (e"cept (") and

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(")) are 2ero out side the interval . Cithin the interval# they are

piece!ise linear !ith the ma"imum value of unity at .

Figure 0.

;ote that these functions are linearly independent and satisfy all the constraints arising out of thethree admissibility conditions on u("). In particular# they satisfy the constraint that the derivatives

are finite. For to be finite# the functions need not be piece1!ise linear.hey only have to be piece!ise differentiable !ith umps in the derivatives as sho!n in Fig. 4.4.

Figure 0.0

olution Procedure

As before# an appro"imate solution is assumed as a series of ;R - terms

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(4.-)

&o!ever# here# the un$no!n coefficients are denoted by rather than as (as in equation 5.,).;ote that# the %irichlet boundary condition (equation ,.-b) implies

atx= 0(4.,)

&o!ever# all e"cept are 2ero atx G 0 (Fig. 4.5). herefore# to satisfy this condition#the first coefficient must be 2ero 9

(4.5)

hen# the series (4.,) starts from i G - rather than from i G 0. herefore#

(4.4)

Chen the e"pression (4.4) is substituted in the integral form (equation 5.8,)# !e get

, fo i = 1,2,, N (4.8)

!here the e"pressions for the stiffness matri" and the force vector are given by equations

(5.-5) and (5.85). In matri" form# equation (4.8) can be !ritten as

(4./)

"xample -

In the e"ample# the bar of Fig. 5.4 is considered9

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Figure 4.8

he numerical values of various parameters are as before9

EA = 1 , L = 1 , P = 10 , F = 20. (4.7)

A t!o1term appro"imation is chosen for the solution 9

(4.D)

!here the functions and are defined as follo!s 9

for

for(4.)

for

for (4.-0)

hese functions# along !ith are sho!n in Fig. 4./.

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Figure 4./

;ote that the derivatives of the basis functions are given by 9

for

for (4.--)

for

for (4.-,)

Substituting the e"pressions (4.--) and (4.-,) for and the values of EI and L fromequation (4.7) in equation (5.-5)# the elements of the stiffness matri" J K K become 9

(4.-5)

(4.-4)

(4.-8)

(4.-/)

hus

(4.-7)

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Substituting f G 0 and a G L ,# the e"pression (5.85) for becomes 9

(4.-D)

Substituting the values of P # F and L from equation (4.7) and the e"pressions (4.) and (4.-0) for

in the above equation# the elements of the force vector become 9

(4.-)

(4.,0)

hus#

(4.,-)

hus# the algebraic equations (equation 4./) become 9

(4.,,)

he solution of the above equation is9

(4.,5)

hen# the appro"imate solution (equation 4.D) becomes 9

for

for (4.,4)

his is the same as the e"act solution (equation 5./8).

he e"ample in the ne"t section sho!s that# even for a distributed load# the piece!ise linear basisfunctions can appro"imate the e"act solution to any desired level of accuracy.

"xample 2

In this e"ample# the bar of Fig. 5.-. is considered.

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Figure 4.7

he distributed force is given by

(4.,8)

he values of other parameters are the same as before 9

EA = 1 , L = 1 , P = 10. (4.,/)

Again# a t!o term appro"imation is chosen (equation 4.D) !here the basis functions and theirderivatives are given by equations (4.)1(4.-,). he basis functions are sho!n in Fig. 4./. Sincethe basis functions and the values of EA and L are the same as in e"ample -# the J K K matri" is

the same as before (equation 4.-7). Cith f (x ) Gx # F G 0 and L G -# the e"pression (5.85) forbecomes9

(4.,7)

Substituting the e"pressions (4.)1(4.-0) for and the value of P from equation (4.,/)# theelements of the force vector become 9

(4.,D)

(4.,)

hus#

(4.50)

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hen# the algebraic equations (equation 4./) become 9

(4.5-)

he solution of the above equation is

(4.5,)

hen# the appro"imate solution (equation 4.D) becomes 9

for

for . (4.55)

he e"act solution of the problem is given by equation (5.4,). he graphical representation of the

t!o solutions ( (") and (")) is sho!n in Fig. 4.D.

Figure 4.D

If a four1term appro"imation is chosen

(4.54)

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!here the basis functions are as sho!n in Fig. 4.#

Figure 4.

then the appro"imate solution loo$s li$e as sho!n in Fig. 4.-0.

Figure 4.-0

Figure 4.-0 sho!s that as !e ma$e the appro"imation finer and finer (i.e.# as !e add moreterms)# the solution becomes more and more accurate. hus# as the number (;) of terms in the

appro"imation increases# converges to the e"act solution.

/ntroduction

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In the last lecture# t!o e"amples !ere solved using the hat1shaped# piece!ise linear basisfunctions to illustrate the effectiveness of these basis functions in providing a betterappro"imation to the derivatives. nly a fe! basis functions !ere used in solving these e"amples.In this lecture# a general formulation is proposed using (;R-) piece!ise linear basis functions.'efore that# the basis functions and the corresponding division of the domain into the parts (calledas the mesh) are discussed in detail.

Mesh and 3lobal 'asis Functions

As mentioned in section 4.-# the piece!ise linear basis functions are generated by dividing the

domain into many parts (Fig. 8.-). Since this is a one1dimensional problem# the domain isidentical to an interval of length @.

Figure .-

o generate (N R-) basis functions# !e divide the domain into N parts using the (N 1-)intermediate points and the t!o end points. hese points are called as nodes . hus# !e have (N R-) nodes. Ce number these nodes sequentially as sho!n in Fig. 8.-. In Fig. 4.5 !e had

denoted the coordinate of the first node as . 'ut# no! !e label it as . hus# the coordinate of

the ith node is denoted by . Ce place the nodes at the points of

discontinuities in the geometry#

discontinuities in the material properties# and

discontinuities in the loading (!hich also includes the points of application of the

concentrated forces)

ther!ise the nodes are equally spread.he parts in !hich the domain is divided by nodes are called as the elements . hese elements

are also numbered sequentially as sho!n in Fig.8.-. hus# i thelement# denoted by # is definedby

fori = 1,2,., N (8.-)

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If the nodes are equally spaced# then the length of each element is equal. @et this length bedenoted by h . hen

h = fori = 1,2,, N . (8.,)

he division of the domain into the elements and the nodes is called as the mesh . Chen all theelements are of the same si2e# it is called as the uniform mesh. he quantity h is called as the

(uniform) mesh si*e .

he piece!ise linear basis functions are defined as follo!s. he functions (e"cept

and ) are 2ero outside the interval . Cithin the interval# they are piece!ise

linear !ith the ma"imum value of unity at . hus# can be e"pressed as (

for ,

for ,

G 0 for belonging to other elements#

fori = 1,2,., N . (8.5)

;ote that the function has only a local support over the interval .

3alerkin +ethod

Ce shall use the 3aler$in method to obtain an appro"imate solution. herefore# the integral formto be used is given by equation (5./D). his equation is rearranged as

(8.4)

As before# !e assume an appro"imate solution as a series of ( N R-) terms 9

(8.8)

!here are the un$no!n coefficients and are the basis functions described in theprevious section. ;ote that# no! the basis functions are numbered from one rather than from 2ero(as done in @ecture 4). herefore the summation inde" also starts from one# and not from 2ero.

Another departure !e ma$e# from the methodology adopted in @ecture 4# is that !e do not applythe %irichlet boundary condition in the beginning. It !ill be applied later.

For 3aler$in method# !e choose the basis functions as the !eight functions. hus#

for i=1,2,...,!*1 (8./)

;ote that these functions are linearly independent and satisfy all the constraints arising out of the

three admissibility conditions on . Substituting the e"pression (8./) for in equation(8.4)# !e get

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he last term in the e"pression (8.-,) for can be simplified as follo!s. ;ote that# at the end

pointx = L # all are 2ero e"cept for i G N R- (Fig. 8.-). herefore# the last term is non2eroonly for i G N R-. hus# the e"pression (8.-,) becomes

fori = 1,2,, N &

fori = N *1. (8.-8)

he e"pressions (8.--) and (8.-8) sho! that the stiffness matri" and the force vector are tobe evaluated as the sums of the integrals over ; elements. he systematic procedure for thisevaluation !ill be discussed in the ne"t @ecture. &o!ever# for given i # the contribution of manyelements is 2ero as sho!n in the ne"t paragraph.

*onsider the e"pressions (8.--) and (8.-8) for and . ;ote that the functions # e"cept

for i G - and N R-# are 2ero outside the interval (Fig. 8.-). herefore# the sums (8.--)

and (8.-8) for and receive the contributions only from the t!o elements 9 element i 1- and i#

that is# from for i G ,#5#..# N . For i G -# the function is

non 2ero only for the first element . hus# for i G -# and receive the

contributions only from the first element. For i G N R-# the function is non12ero only for the

last# i.e.# Nthelement . hus# for i G N R-# and receive the contributions

only from the ;thelement. hus # the e"pressions (8.--) and (8.-8) for and become 9

fori = 1,j = 1,2,., N *1, (-stelement)

(-stelement)

fori = 2., N ,j = 1,2,., N *1,

((i1-)thelement) (ithelement)

fori = !*1,j = 1,2,.,!*1&(8.-/)

(;thelement)

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fori = 1,

(-stelement)

fori = 2.,!,

fori = N *1. (8.-7)

It is observed that the stiffness matri" in equation (8.-5) is a sparse matrix . More precisely# itis a banded matrix . his can be sho!n as follo!s.

;ote that all are 2ero over the ne"t interval e"cept !henj G i1- and i . Similarly#

all are 2ero over the ne"t interval e"cept !henj G i and i R- (Fig. 8.-). herefore#

the only non12ero contributions to the sum on the left side of the equation (8.-5) are from j G i1-# i# i R-. hus# ith equation in the set (8.-5) becomes

(8.-D)

!here# for i G ,#5#..# N #

forj = ,

forj = i,

forj = i *1,

for remainingj (8.-)

Further# for iG -# the only non12ero contributions to the sum on the left side of equation (8.-5)come fromG- and ,. hus

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(8.,0)

!here

forj= 1,

forj = 2,

for remainingj . (8.,-)

Similarly# for i G N R-# the only non12ero contributions to the sum on the left side of equation(8.-5) come fromj G N and N R-. hus#

(8.,,)

!here

forj = N ,

forj = N *1,

for remainingj . (8.,5)

hus# ithequation has non12ero elements only in the columnsj G i1-# i # and i R-. he remaining

columns are 2ero. Further# the -stequation has non12ero elements only in the -stand columns

. Similarly# the last equation (i.e.# equation) has non12ero elements only in the and

columns. hus# the matri" is a banded matri".

/ntroduction

In the last lecture# general e"pressions !ere developed for the stiffness matri" and the forcevector using the piece!ise linear basis functions. hese e"pressions are to be evaluated asthe sums of the integrals over all the elements. his lecture discusses the systematicprocedure of this evaluation. First# the integrals are evaluated element !ise and then theyare assembled over all the elements. For the convenience of the evaluation of the integralsover the elements# local or element notation is introduced.

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@ocal or Element ;otation

In the last lecture# !e introduced a numbering scheme for the elements# nodes# nodal coordinatesand basis functions. his is called as the global numbering scheme . It is sho!n in Fig. /.-.

Figure 4.- 3lobal numbering scheme

For the convenience of the element !ise evaluation of the stiffness matri" and the force vector# atthe element level# !e introduce another numbering scheme for the nodes# nodal coordinates andbasis functions. his scheme is called as the local or element numbering scheme. For k1thelement# this numbering scheme is illustrated in Fig. /.,

Figure 4.2 "lemental numbering scheme for k5th element

For k1th element# note that# the global numbers of the t!o end nodes are k and $R- respectively(Fig. /.-). ;o! !e number them as - and , (Fig. /.,). Ce call it as the local or element node

numbering scheme . ;ote that the global notation for the coordinates of the end nodes is

and (Fig. /.-). ;o!# !e denote them as and (Fig. /.,). his is called as the local orelement notation . he subscript here denotes the local node number. For every element# thenodes !ill al!ays be numbered as - and , according to the local numbering scheme. hereforeto identify the element under consideration# a superscript is added to the notation of thecoordinates. he superscript denotes the element number. Figure /.- sho!s that the only non1

2ero basis functions for the 1th element are and . his is the global notation for the

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basis functions. ?nder the local or element notation # !e denote them as and . &ere#the subscript denotes the local number of the node at !hich the basis function attains the valueunity and the superscript represents the element number under consideration. hese functionsare called as the shape or interpolating functions . ?sing equation (8.5)# the e"pressions forthe shape functions can be !ritten as9

(/.-)

!here the e"pression (8.,) for the element si2e becomes

= (/.,)

*onsider the e"pression (8.8) for the appro"imate solution9

(/.5)

;ote that# since is unity at node (Fig. /.-)# the un$no!n coefficient is the un$no!n

displacement at node . he un$no!ns are called as the degrees of freedom of the rod.

Since# the subscript denotes the global number of the node # they are called as global

degrees of freedom . For the 1th element# the only non 2ero basis functions are and

.herefore# the e"pression for over the 1th element becomes

(/.4)

For the convenience of element !ise calculations of the stiffness matri" and force vector# !e

introduce another notation for and . Ce denote them as and . &ere# the subscriptdenotes the local number of the node to !hich the displacement belongs and the superscriptrepresents the element number under consideration. his notation is called as the local or

element notation and ( # ) are called as the local or element degrees of freedom of the1th element. hen the e"pression (/.4) under the element notation becomes

(/.8)

E"pressions for Element Stiffness Matri" and Force Bector

No16 1e shall determine the ro1 and column numbers of the matrix e#uation 7.-08 to

1hich the 5th element makes the contribution. For this purpose6 consider e#uation7.-8(

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(/./)

!here the e"pressions for the stiffness matri" and the force vector are given by equations

(8.--) and (8.-,)9

(/.7)

and

(/.D)

he above e"pression can be !ritten as

(/.)

(/.-0)

and

(/.--)

(/.-,)

;ote th at and represent the contributions of the 1th element to the stiffness matri"

and the force vector . he second term of equation (/.--) is a contribution from the lastelement only. his term needs to be treated separately and therefore not included in the

e"pression for and the present discussion. ?sing the e"pressions (/.-0) and (/.-,)# the

contribution to equation (/./) of the 1th element can be !ritten as

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(/.-5)

;ote that the values of for !hich is non2ero over the 1th element are G and G

. herefore# the only non2ero ro!s of and are G and G . hus# the only

non2ero equations of the set (/.-5) are G and G . hese equations are as follo!s9

,

(/.-4)

(/.-8)

Since# the only values of for !hich is non2ero over the interval are G and

. herefore# the above equations become9

,

(/.-/)

.

(/.-7)

In matri" form# this can be !ritten as

, (/.-D)

!here

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(/.-)

and

(/.,0)

In elemental notation# the above quantities can be !ritten as

(/.,-)

and

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(/.,,)

;ote that# the elements of and are 2ero !hen or and or . hus#

(/.,5)

and

(/.,4)

herefore# these quantities are often e"pressed as

,

(/.,8)

and

(/.,/)

he matri" (/.,8) is called as the element stiffness matrix !hereas the vector (/.,/) is called asthe element force vector . ;ote that the notation for the elemental quantities uses small letters 9small $ for the stiffness matri" and small f for the force vector. he notation for the correspondingfull matri" or full vector uses capital letters 9 capital T for the stiffness matri" and capital F for theforce vector.

he e"pressions (/.,8) and (/.,/) help in the systematic evaluation of the global stiffness

matrix and global force vector of equation (/./). he procedure can be described asfollo!s9

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i. First# the stiffness matri" and the force vector of each element are evaluated using thee"pressions (/.,8) and (/.,/). his can be done using a % loop. hus# for

# the elements stiffness matri" and element force vector areevaluated using

(/.,7)

and

(/.,D)

ii. ;e"t# these quantities are e"panded to the full si2e and . his is doneusing the relationship bet!een the local and global node numbering systems.

iii. Finally# the e"panded matrices and vectors of all the elements are added to get the global

stiffness matri" and global force vector . At this stage# the second term of the

e"pression for (equation /.D) also needs to be added.

he last , steps constitute !hat is called as the global assembly procedure. he details of theprocedure are discussed in section /.4. 'efore that# section /.5 discusses an e"ample onelement calculation

/.5 E"ample on *alculations of Element Stiffness Matri" and Force Bector.

*onsider a typical element . he nodes# the coordinates# the degrees of freedom and the shape

functions of the element (in the local notation) are sho!n in Fig. /.5

Figure 4. 9 typical element k

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Assembly of Stiffness Matri" and Force Bector

After the evaluation of element stiffness matri" and element force vector for all the elements#these quantities need to be assembled to get the global stiffness matri" and global force

vector. As stated at the end of section /.,# this procedure has t!o steps9

i. E"pansion of the element stiffness matri" and element force vector to the full si2e.ii. Addition of the e"panded matrices and vectors over all the elements. At this stage# the

second term of the e"pression for (equation /.D) also needs to be added.

@et us first discuss the first step. ;ote that equations (/.,8) and (/.,/) are the e"pressions for the

element stiffness matri" and the element force vector !hile equations (/.,-) and

(/.,,) are the e"pressions for their e"panded versions and . Chen !e compare

equations (/.,8) !ith (/.,-)# !e observe that (-#-) component of occupies the position

of the e"panded matri" . his is because is the global number of the local node -of the element $. hus# the first step involves9

*hoose the component , ,

Find the global number of the local nodes and of the element . @et they be and

respectively.

hen the component occupies the location in 1th ro! and s 1th column of the

e"panded matri" . hus# the component goes to the location in thee"panded matri".

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(/.54)

he first ro! of the connectivity matri" contains the global numbers of the first and second localnodes of element -. he global numbers corresponding to the first and second local nodes ofelement , are !ritten in the second ro!. *ontinuing in this !ay# the global numbers of the first

and second local nodes of element appear in the 1th ro!. he last ro! contains the global

numbers associated !ith the first and second local nodes of the last# i.e. 1th element. hee"pression (/.54)# in the inde" notation# can be e"pressed as

(/.58)

It means the global number of the local node of the element is obtained as the value of the

component of the connectivity matri" in 1th ro! and 1th column. As an e"ample# consider the

case of G 5 and G ,. he e"pression (/.58) gives . his means 4 is the globalnumber of the second local node of the element 5. his can be verified from Fig. /.-.;o!# the first step of the assembly procedure can be e"presses as follo!s. he e"panded matri"

is obtained from the element stiffness matri" by the relation9

(/.5/)

Similarly# to obtain the e"panded vector from the element force vector # !e use therelation9

(/.57)

hus# !e use the follo!ing procedure9

i. *hoose the component

ii. Find the global number of the local node from the connectivity matri". @et it be .

iii. hen# the component goes to the location of the e"panded matri".

iv.

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,(/.5D)

(/.5)

he matri" corresponds to the second term of equation (/.D). ;ote that# the only basis

function !hich is non2ero at is . Further# it6s value at is -. hus

(/.40)

herefore# the vector O+P can be !ritten as

(/.4-)

"xample on 9ssembly of tiffness +atrix and Force Vector

As an e"ample# consider the mesh of / elements (; G /) and 7 nodes# sho!n in Fig. /.4.

Figure /.4 Mesh !ith / elements

he connectivity matri" for this mesh can be !ritten as9

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.

(/.4,)

@et

,(/.45)

and

(/.44)

be the element stiffness matri" and the element force vector of the elements G-#,#5#4#8#/.

*onsider the element -# i.e. . ;ote that

(/.48)

hen as per equation (/.5/)# components of the stiffness matri" of the element -# i.e. of #

occupy the follo!ing locations in the e"panded matri" 9

(/.4/)

Similarly# as per equation (/.57)# components of the force vector of the element -# i.e. of #

occupy the follo!ing locations in the e"panded vector 9

(/.47)

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he remaining components of the e"panded matri" and the e"panded vector are

2ero. hus# the matri" becomes9

.

(/.4D)

and the vector becomes9

(/.4)

Similarly# !e obtain the e"panded versions of the element stiffness matri" and the element

force vector for the remaining elements# i.e. for G ,#5#4#8#/. It can easily be verified that#

for the 5 rd element (i.e. for )# the e"panded matri" and the e"panded vectorare9

(/.80a)

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(/.80b)

his completes the first step.

In the ,ndstep# !e add all the e"panded matrices and vectors. hus# equation (/.5D) gives thefollo!ing e"pression for the global stiffness matri"9

(/.8-)

Similarly# the sum of the e"panded force vector becomes9

(/.8,)

&o!ever# before !e get the global force vector # !e need to add the vector to the above

e"pression. Since (no. of elements) G /# the 1th component# i.e. the 71th component of

the vector !ill be . he remaining components !ill be 2ero as per equation (/.4-). hus#

becomes9

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(/.85)

Substituting the e"pressions (/.8,) and (/.85) in equation (/.5)# !e get the follo!ing e"pression

for the global force vector 9

(/.84)

;o!# as in section /.5# assume that and (distributed force) are constant for the entire bar.

Further# assume that the length of each element is constant. @et us denote it by h . hen

(/.88)

hen# equation (/.5,) implies that the element stiffness matri" is identical for each elementand is given by

(/.8/)

Similarly# equation (/.55) implies that the element force vector is identical for eachelement and is given by

(/.87)

Substituting the e"pression (/.8/) in equation (/.8-)# !e get

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(/.8D)

Further# substituting the e"pression (/.87) in equation (/.84)# !e get

(/.8)

In actual calculations# the assembly procedure is appropriately modified to reduce thecomputational time and storage requirements. Chen# the number of elements is large# storing ofthe e"panded matrices and vectors for each element needs a lot of storage requirement.herefore# the process is modified as follo!s9

nce the e"panded version of the element stiffness matri" of the first elementis obtained# the element stiffness matrices of other elements are not e"panded.

Instead# the locations of the components of the stiffness matri" of the element t!o

are determined using equation (/.5/). From the connectivity (/.54)# it is easy to see that

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(/./0)

/ntroduction

%etermination of the element stiffness matri" and the element force vector and their assemblyinto the global stiffness matri" and global force vector !ere discussed in the last lecture. 'eforesolving the (global) finite element equations# the essential or the %irichlet boundary condition (i.e.#the boundary condition on the primary variable) needs to be applied. his lecture discusses theprocedure of the application of the essential boundary condition.

;ote that# in the

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'oundary *onditions at 'oth the Ends

*onsider a bar !hose both ends are fi"ed as sho!n in Fig. 7.5.

Figure 7.3 Bar with essential boundary condition at both the ends

Since# !e have an essential boundary condition at the end also# the boundary conditions (-.-c) becomes

at . (7.-D)

hen the !eighted residual e"pression becomes

(7.-)

If !e develop the finite element equations for this problem using the /1element mesh of Fig. 7.-# then the globalstiffness matri" remains unchanged# but the global force vector becomes

(7.,0)

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;o! the finite element equations become

(7.,-)

or

.

(7.,,)

he essential boundary condition at the left end (x G 0)

(7.,5)

is applied by modifying the matri" and the vector using the e"pressions (7.-4) and (7.-8). ;ote that# here

. he essential boundary condition at the right end ( )

, (7.,4)

is also applied in similar fashion. 'ut no!# it is the nodal displacement of the 7 thnode that is $no!n. herefore# !eapply the boundary condition as follo!s 9

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(7.50)

;o! the finite element equations become

(7.5-)

or

(7.5,)

;ote that by adding all the equations# !e get

(7.55)

Since # !e get

(7.54)

his is the same equation as equation (7.,7).

;ote that# there are no essential boundary conditions. herefore# no modification of either thestiffness matri" or the force vector is needed before !e solve the system of equations (7.5-). 'utthere is a difficulty in solving these equations. It can be easily verified that the determinant of the

stiffness matri" is 2ero or is singular . In this case# the system of equations (7.5-)possesses many solutions or the solution is not uni#ue . his can be seen as follo!s. *onsiderthe vector

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, ( = abita numbe).

(7.58)

It means all the nodal values of the a"ial displacement are identical (and are equal to ). hen#equation (7.-0) becomes

.(7.5/)

his implies that the a"ial displacement of the rod is same at every point (and is equal to ). hisrepresents the rigid translation of the bar. ;ote that

. (7.57)

@et be a solution of the system of equations (7.5-).hen

. (7.5D)

hus# if is a solution of the system of equations (7.5-)# then is also a solution ofthe same system of equations. herefore# solution of the system of equations (7.5-) is notunique.

o ma$e the solution unique# !e have to ma$e sure that there is no vector li$e !hichsatisfies the equation (7.57). In other !ords# !e have to ensure that there is no rigid bodytranslation mode in the deformation of the bar. o ensure this# !e can set any nodal displacement

to any arbitrary value . o apply this condition# !e carry out the modifications similar toequations (7.-4) and (7.-8)9

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forj = 1,..,, / i ,

& (7.5)

ranspose the $no!n terms from the remaining equations to the right side 9

forj = 1,..,, / i,forj = 1,..,, / i (7.40)

hen# the modified becomes non singular and !e can solve the modified system of equations

to get a unique solution. ;ote that the solution gives .

Spring at ne End

*onsider a bar# fi"ed at the left end and supported by a spring at the right end (Fig. 7.8).

Figure 7.8 'ar supported by a spring at one end 7 : pre5compression of the spring8

he above problem can be modeled by replacing the spring !ith a force as sho!n in Fig. 7./ .

Figure 7.6 Equialent !roble"

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he e"pression for the spring force is

(7.4-)

!here is the spring force and is the initial compression in the spring. Since the force iscompressive# the boundary condition (-.-c) becomes

, at .(7.4,)

hen the !eighted residual integral becomes

(7.45)

or

(7.44)

If !e develop the finite element equations for this problem using the /1element mesh of Fig. 7.-#then the global stiffness matri" becomes

(7.48)

Further# the global force vector becomes

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(7.4/)

hen# the finite element equations become

(7.47)

or

(7.4D)

;o! the essential boundary condition

(7.4)

is applied by modifying the matri" and vector as described by equations (7.-4) and

(7.-8). ;ote that here .