reddy_chap3-4

Chapter 3

SECOND-ORDERDiFFERENTIAL EQUATIONSIN ONE DIMENSION: FINITE

ELEMENT MODELS

3.1 BACKGROUNDThe traditional variational methods (e.g., the Ritz, Galerkin, and least-squares) describedin Chapter 2 cease to be effective because of a serious shortcoming, namely, the difficulty in constructing the approximation functions. The approximation functions, apart fromsatisfying continuity, linear independence, completeness, and essential boundary conditions, are arbitrary; the selection becomes even more difficult when the given domain isgeometrically complex. Since the quality of the approximation is directly affected by thechoice of the approximation functions, it is discomforting to know that there exists no systematic procedure to construct them. Because of this shortcoming, despite the simplicityin obtaining approximate solutions, the traditional variational methods of approximationwere never regarded as competitive computarionally when compared with traditional finitedifference schemes. The finite element method overcomes the shortcomings of the traditional variational methods by providing a systematic way of constructing the approximationfunctions.

Ideally speaking, an effective computational method should have the foflowingfeatures:

1. It should have a sound mathematical as well as physical basis (i.e., yield convergentsolutions and be applicable to practical problems).

2. It should not have limitations with regard to the geometry, the physical composition ofthe domain, or the nature of the “loading.”

3. The formulative procedure should be independent of the shape of the domain and thespecific form of the boundary conditions.

4. It should be flexible enough to allow different degrees of approximation without reformulating the entire problem.

5. It should involve a systematic procedure that can be automated for use on digitalcomputers.

CI1AI’ER 3: SacOND-ORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: FINrTE ELEMENT MODELS 105

Table 3.1.1 Steps involved in the finite element analysis of a typical problem.

104 .N EcIRODUCT1ON 50 ThE FINSTE ELEMENT METHOD

The finite element method is a technique in which a given domain is represented as acollection of simple domains, calledfinite elements, so that it is possible to systematicallyconstruct the approximation functions needed in a variational or weighted-residual approximation of the solution of a problem over each element. Thus, the finite element methoddiffers from the traditional Ritz, Galerkin, least-squares, collocation, and other weighted-residual methods in the manner in which the approximation functions are constructed. Butthis difference is responsible for the following three basic features of the finite elementmethod:

1. Division of whole domain into subdomains that enable a systematic derivation of theapproximation functions as well as representation of complex domains.

2. Derivation ofapproxiraationfi.snctions over each element. The approximation functionsare often algebraic polynomials that are derived using interpolation theory. However,approximation functions need not be polynomials (like in meshless form of the finiteelement method).

3. Assembly ofelements is based on continuity of the solution and balance of internal fluxes;the assemblage of elements results in a numerical analog of the mathematical model ofthe problem being analyzed.

These three features, which constitute three major steps of the finite element formulation,are closely related. The geometry of the elements used to represent the domain of a problemshould be such that the approximation functions can be uniquely derived. The approximationfunctions depend not only on the geometry but also on the number and location of points,called nodes, in the element and the quantities to be interpolated (e.g., solution, or solutionand its derivatives). Once the approximation functions have been derived, the procedure toobtain algebraic relations among the unknown coefficients (which give the values of thedependent variable at the nodes) is exactly the same as that used in the Ritz and weighted-residual methods. Hence, a study of Chapter 2, especially the weak-form development andthe Ritz method, makes the present study easier.

The finite element method not only overcomes the shortcomings of the traditionalvariational methods, but it is also endowed with the features of an effective computationaltechnique. The basic steps involved in the finite element analysis of a problem are given inTable 3,1.1.

In the sections that follow, our objective is to introduce many fundamental ideas that formthe basis of the finite element method. In doing so, we postpone some issues of practicaland theoretical complexity to later sections of this chapter and to Chapters 4—14. Thebasic steps of a finite element analysis are introduced via a model second-order differentialequation.

-

3.2 BASIC STEPS OF FINITE ELEMENT ANALYSIS3.2.1 Model Boundary Value Problem

Consider the problem of finding the function u(x) that satisfies the differential equation

__(a)+cu_f=0 for 0<x<L (3.2.1)

I. Discretization (Or representation) of the given domain into a collection of preselected finite elements. (Thisstep can be postponed until the finite element formulation of the equation is completed.)(a) Construct the finite element mesh of preselected elements.(b) Number the nodes and the elements.(c) Generate the geometric peopeeties (e.g., coordinates and cross-sectional areas) needed for the problem.

2. Derivation of element equations for all typical elements in the mesh.(a) Construct the variational formulation of the given differential equation over the typical element.(b) Assume that a typical dependent variable u is of the form

U =

and substitute it into Step 2a to obtain element equations in the form

(c) Select, if already available in the literature, or derive element interpolation functions and compute theelement matrices.

3. Assembly of element equations to obtain the equations of the whole problem.(a) Identify the interelement continuity conditions among the primary variables (relationship between the local

degrees of freedom and the global degrees of freedom—connectivity of elements) by relating element nodesto global nodes.

(Is) Identify the”eqnilibrium” conditions among the secondary variables (relationship between the local sourceor force components and the globally specified source components).

(c) Assemble element equations using Steps 3a and 3b.4. Imposition of the boundary conditions of the problem.

(a) Identify the specified global primary degrees of freedom.(b) Identify the specified global secondary degrees of freedom (if not already done in Step 3b).

5. Solution of the assembled equations.6. Postprocessing of the results.

(a) Compute the gradient of the solution or other desired quantities from the primary degrees of freedomcomputed in Step 5.

(b) Represent the results in tabular and/or graphical form.

and the boundary conditions

/ duu(0)=uo, ja— (3,2.2)

\ dxj x—Lwhere a = a(x), c = c(x), f = f(x), and u0, and Qo are the data (i.e., known quantities) ofthe prnblem. Equation (3.2.1) arises in connection with the analytical description of manyphysical processes. For example, conduction and convection heat transfer in a plane wallor fin [see Fig. 3.2.1(a) j, flow through channels and pipes, transverse deflection of cables,axial deformation of bars [see Fig. 3.2.l(b)j, and many other physical processes are described by Eq. (3.2.1). A list of field problems described by Eq. (3.2.1) when c(x) = 0 arepresented in Table 3.2,1 [see Reddy (2004)1. Thus, if we can develop a numerical procedureby which Eq. (3.2.1) can be solved for all possible boundary conditions, the procedurecan be used to solve all field problems listed in Table 3.2.1, as well as many others. Thisfact provides us with the motivation to use (3.2.1) as the model second-order equation inone dimension. A step-by-step procedure for the formulation and solution of (3.2.1) by the

106 aes erraorxrcnowroma nsrm m.Earrorr surruon CHAPTER 3: SECOND.OIER DnERE?m41. EQUATIONS IN ONE DIMENSION FINITE ELEMENT MODELS 107

Internal heat generation,f(x)Convectiop from lateral

Maintained attemperature,

c(x) u(x)

Exposed to ambienttemperature, u,

(a)

Body force,f(x)Subjected to axial

Specifieddisplaceme___.ø.4

Ia L

(b)

f(x)

______________

(duo(O)=o a— Q

LHx

(c)

Figure 3.2.1 (a) Heat transfer in a fin. (b) Axial deformation of a bar. (c) Mathematical idealization

of the problem in (a) or (b).

finite element method is summarized in Table 3.1.1. The mathematical problem consistsof solving the differential equation (3.2.1) in one-dimensional domain 2 = (0, L) subjectto a suitable set of specified boundary conditions at the boundary points x = 0 and x =

as shown in Fig. 3.2.1(c). As already shown in Chapter 2, the type of boundary conditions associated with a differential equation emerges in a natural way during the weak-form development of the differential equation. A detailed discussion of these ideas ispresented next.

3.2.2 Discretization of the Domain

In the finite element method, the domain Q of the problem [Fig. 3.2.2(a)] is divided into aset of subintervals, i.e., line elements, calledfinite elements. A typical element is denotedS2 and it is located between points A and B with coordinates X0 and xb (i.e., of length

x0). The collection of finite elements in a domain is called the finite elementmesh of the domain [see Fig. 3.2.2(b)].

The reason for dividing a domain into a set of subdomains, i.e., finite elements, is twofold. First, domains of most systems, by design, are a composite of geometrically and/or

Table 3.2.1 Some examples of engineering problems governed by the second-order equation (3.2.1)(see the footnote for the meaning of some parameters*).

Problem data SecondaryField variable variableof study a a c f QHeat Temperature Thermal Surface Heat Heat

transfer T — Tm conductance convection generation QkA APfi f

Flow through Fluid head Permeability 0 Infiltraction Point sourceporousmedium cfr f Q

Flow through Pressure Pipe resistance 0 Point sourcepipes P hR 0 Q

Flow of Velocity Viscosity 0 Pressure gradient Shear stressviscous fluids v —dP/dx

Elastic cables Displacement Tension 0 Transverse force Point forcea T f P

Elastic bars Displacement Axial stiffness 0 Axial force Point forcea EA f P

Torsion of Angle of Shear 0 0 Torquebars twist stiffness T

8 GJElectrostatics Electrical Dielectric 0 Charge Electric

potential constant density flux• e p E

= thermal condoctaace; fi = csavecrive film coadeetauce; p = petimesn P = pressure or force; T = axobieot tetnperaureof the surroondiug fluid medians; R = l2SMh/(rrd’) with a being the viocosisy, h the length, and d the diameter of the pipe;P = Young’s modulus; A = seen of cross section; J = poise moment of inertia.

materially different parts, and the solution on these subdomains is represented by differentfunctions that are continuous at the interfaces of these subdomains. Therefore, it is appropriate to seek approximation of the solution over each subdomain. Second, approximationof the solution over each element of the mesh is simpler than its approximation over theentire domain. Approximation of the geometry of the, domain in the present case is not a

L

Hx(a)

End points of an interval Element L2 = (x, x)

3ABDN

XrO x2 x3 /j4_j\ XNXNft’L

>X Xe=Xa Xe÷1X5

(b)

Figure 3.2.2 (a) Whole domain. (b) Finite element discretization (mesh).

lOS AN INTRODUCOON TO THE FINITE ELEEND(r METHOD

concern, since it is a straight line, We must, however, seek a suitable approximation of thesolution over each subdomain (i.e., finite element).

The number of elements into which the total domain is divided in a problem depends mainly on the geometry of the domain and on the desired accuracy of the solution. In the one-dimensional problem at hand, geometry is simple enough to representit exactly. Since the exact solution is not known a priori, we may begin with a number of elements that are considered to be reasonable. Most often, the analyst has knowledge of the qualitative behavior of the solution, and this helps to choose a starting mesh.Whenever a problem is solved by the finite element method for the first time, we arerequired to investigate the convergence of the finite element solution by gradually refining the mesh (i.e., increasing the number of elements) and comparing the solution withthose obtained by higher-order elements. The order of an element refers to the degree ofpolynomial used to represent the solution over the element. This is made clearer in thesequel.

3.2.3 Derivation of Element Equations

Next, we develop the algebraic equations among the unknown parameters, much the sameway as we did in the Ritz and Galerkin methods discussed in Chapter 2 [see Eq. (2.5.5)].The main difference here is that we work with a finite element (i.e., subdomain) as opposedto the total domain. This step results in a matrix equation of the form [Kd]{ce} =

which is called the finite element model of the original equation. Since the element isphysically connected to its neighbors, the resulting algebraic equations will contain moreunknowns (ces as well as Fes are unknown in the element equations) than the number ofalgebraic equations. Then it becomes necessary to put the elements together (i.e., assembly)to eliminate the extra unknowns. This process is discussed in Section 3.2.5.

The derivation of finite element equations, i.e., algebraic equations among the —

unknown parameters of the finite element approximation, involves the following threesteps:

1. Construct the weighted-residual or weak form of the differential equation.

2. Assume the form of the approximate solution over a typical finite element.3. Derive the finite element equations by substituting the approximate solution into the

weighted-residual or weak form.

A typical element f2 = (.Xa, xb) [see Fig. 3.2.3(a)], whose endpoints have the coordinates x = X5 and x = xb, is isolated from the mesh. We seek an approximate solution to thegoverning differential equation over the element. In principle, any method that allows thederivation of necessary algebraic relations among the nodal values of the dependent variablecan be used. In this book we develop the algebraic equations using the Ritz method, whichis based on the weak form of the differential equation. Other methods, such as the least-squares method, may also be used to construct the finite element equations. Apart from themethod used to derive the algebraic equations, the steps presented in this book for the Ritz(or weak-form Galerkin) finite element models are the same for other methods. The threesteps in the derivation of finite element equations associated with the model differentialequation over a typical element of the mesh are discussed next.

h

XHja

=O 1h,

(a)

q(x)

e-( du) (duQ1 =—a— a—j.i._...-. 2

x=x

(b)

Figure 3.2.3 Finite element discretization of a one-dimensional domain for the model problem in(3.2.1). (a) A typical finite element from the finite element mesh. (b) A typical element,with the definition of the primary Cu) and secondary (Q) variables at the element nodes.

Step 1. Weak Form and Minimum of a Quadratic Functional

In the finite element method, we seek an approximate solution to (3.2.1) over each finiteelement. The polynomial approximation of the solution within a typical finite element l2is assumed to be of the form

u = Eu(x) (3.2.3)

where u are the values of the solution u(x) at the nodes of the finite element Q, and arethe approximation functions over the element. The particular form in (3.2.3) will be derivedin the next section. Note that the approximation in (3.2.3) differs from the one used in theRitz method in that (x) is now replaced with (ando = 0), and ta plays the role ofundetermined parameters and the role of approximation functions. As we shall see later,writing the approximation in terms of the nodal values of the solution is necessitated by thefact that the continuity of u(x) between elements can be readily imposed. The coefficientsu are determined such that (3.2.1) is satisfied in a weighted-integral sense. As discussedin Chapter 2, the necessary and sufficient number of algebraic relations among the canbe obtained by recasting the differential equation (3.2.1) in a weighted-integral form:

o=f w [- (ar) +cu - f] dx (3.2.4)

where w (x) denotes the weight function and l2 = (Xe, XE) 5 the domain of a typical element[see Fig. 3.2.3(a)]. For u u and each independent choice of w, we obtain an independent

0SAPER 3: SECQNI3-OPDER D[FFERFMITAL EQUATiONS IN ONE DIMENSION: FINTEE ELEMENT MODELS 109

Element e= (Xe, Xb)

I

r110 AN INO1ODUCrION1DT1IEFINrrE ELEMErMEm0D i ChAPTER 3; SECOND ORDERDIFFERESIflAL EQUATIONS IN ONE DIMENSION; FINOtELEMENE MODELS 111

algebraic equation relating all u of the element. A total of n independent equations are If we select u (x) in (3.2.3) such that it automatically satisfies the end conditions U(Xa) =required to solve for the parameters u j = 1 2 n When w is selected to be u and u (xb) = i4 then it remains that we include the remaining conditionsand (3.2.4) is used to obtain the ith equation of the required n equations, the resulting/finite element model (i.e., system of algebraic equations among the nodal values) is termed

Qi = ( —a , = ( a (3.2.6)the Galerkin finite element model. Since (3.2.4) contains the second derivative of u, the \. dtJ, \ dx)

approximation functions i,fr must be twice differentiable In addition if the secondaryin the weak form (3 25) With the notation in (3 26) the weak form becomesvariables are to be included in the model, li must be at least cubic. Similar arguments

apply for cases of the weighted-residual methods discussed in Chapter 2. For details of the / dw duweighted residual finite element models see Chapter 14 and Reddy (1986) 0=j + cwu — wfj dx

— w(xa)Qi — w(x5)Q2 (32 i)

To weaken the continuity required of the functions 1fre (x) we trade the differentiationin (3.2.4) from it tow such that both u and w are differentiated equally—once each in the This completes the three-step procedure of constructing the weak form of the modelpresent case. The resulting integral form is termed the weak form of (3.2.1). This form is equation (3.2.1). The finite element model based on the weak form (3.2.7) is called thenot oniy equivalent to (3.2.1) but it also contains the natural boundary conditions of the weak form Galerkin finite element model. It is clear that the weak form (3.2.7) admitsproblem. The three-step procedure of constructing the weak form of (3.2.1) was presented approximation functions that are lower order than the weighted-residual statement (3.2.4).in Chapter 2 and is revisited in the next few paragraphs Students ofengineering recognize that Fig 3 2 3(b)is thefree body diagram of a typical

The first step is to multiply the governing differential equation with a weight function element. For axial deformation of bars, it denotes displacement, du/dx is the strain e, Erw and integrate over a typical element, as given in (3.2.4). The second step is to trade is the stress o and A denotes the force, where E is Young’s modulus and A is thedifferentiation from u tow using integration by parts: area of cross section of the bar; hence, Q = EA(du/dx) = a(du/dx) has the meaning of

force. The quantities Q and Q are the reaction forces at the left and right ends of the/ dw du du] member; Q is a compressive force while Q is a tensile force [algebraically, both are0= j a-- + cwu — wf) dx — [waj (3.2.5)

positive, as shown in Fig. 3.2.3(b)]. For heat conduction problems, it denotes temperature,‘ du/dx is the temperature gradient, —k(du/dx) is the heat flux q, and Aq denotes the heat,The third and last step is to identify the primary and secondary variables of the weak where k is the thermal conductivity and A is the area of cross section of the bar; hence,form. This requires us to classify the boundary conditions of each differential equation into Q kA(du/dx) = a(du/dx) has the meaning of heat; Q —kA(du/dx) is the heatessential (or geometric) and natural (or force) boundary conditions. The classification is input at node 1, while Q = kA(du/dx)b lenote the heat input at node 2. Thus, the arrowmade uniquely by examining the boundary term appearing in the weak form (3.2.5), on the second node should be reversed for heat transfer problems. For additional details on

I du] heat transfer, see Section 3.3.1.Lw_i The weak form in (3.2.7) contains two types of expressions: those containing both w.

.

and it (bilinear form) and those containing only w (linear form):As a rule, the coefficient of the weight function w in the boundary expression is called/ d da secondary variable, and its specification constitutes the natural or Neumann boundary . Be(w, it) = j { a—- + cwu ‘ dxcondition. The dependent unknown it in the same form as the weight function w appearing j, \ dx dx

13 2 8in the boundary expression is termed a primary variable and its specification constitutes fxbthe essential or Dirichlet boundary condition. For the model equation at hand, the primary le(w)

= J w f dx + w (Xa)Qi + W (XE) Q2and secondary variables are

d The weak form can be expressed asu and a—INQ Be(w,u)=le(w) (3.2.9)

The primary and secondary variables at the nodes are shown on the typical element in which is called the variational problem associated with (3.2.1). As will be seen later, theFig. 3.2.3(b). . bilinear form results directly in the element coefficient matrix, and the linear form leads to theIn writing the final form of the weighted integral statement (i e weak form) we wish right hatid side column vector of the finite element equations Derivation of the variationalto use, we must address the fate of the boundary terms. For a typical line element, the end problem of the type in (3.2.9) is possible for all problems described by differential equations.points (nodes 1 and 2) are the boundary points. At these points we have the following four However, the bilinear form Be(w, a) may not be linear in u, and it may not be symmetricconditions (none of them specified at the moment) in its arguments w and it.

I du\ / du\ Those who have abackgroundin applied mathematics or solid and structural mechanicsu(x0)= u _a1_) = Q, u(x) = u ja1-) = will appreciate the fact that the variational problem (329) when Be(w a) is symmetricXX, x=xb Be(w, u) = Be(u, w) and lt(w) is linear in w, is the same as the statemeritoftheuiinisnum

112 AN INTR000CHON TO THE FOTTE ELEMENT MErHOD CHAPTER): SECONDORJ)ER OIPFERENTL4j. EQUATIONS IN ONE DIMENSION: HAn ELEMENT MoDs 113

ij[(d)2]dx

-f ufdx-u(xo)Ql-u(xb)Qs (32.10)

Thus, the relationship between the weak form and the minimum of quadratic functional P

is obvious [cf. (3.2.9)]:0=8p=Be(w,u)_le(w), w=lu

The statement IP = 0 in solid and structural mechanics is also known as the principle

of minimum total potential energy. When (3.2.1) describes the axial deformation of a bar,p(u, u) represents the elastic strain energy stored in the bar element, le(u) represents the

work done by applied forces, andIt(u) is the total potential energy (flu) of the bar element.Thus, the finite element model can be developed using either the statement of the principle ofminimum total potential energy of an element or the weak form of the governing equations

of an element. However, this choice is restricted to those problems where the minimumof a quadratic finctional Ie(u) corresponds to the governing equations. On the other hand,we can always construct a weak form of any set of differential equations, linear or not,of order 2 and higher. Finite element formulations do not require the existence of thefunctional P(u); they only need weighted-integral Statements or weak forms. However,when the functional P(u) exists with an extremum (i.e., minimum or maximum principle),existence and uniqueness of solution to the variational problem and its discrete analog canbe established. In all problems discussed in this book, the variational problem is derivablefrom a quadratic functional.

Step 2. Approximate Solution

Recall that in the traditional variational methods, approximate solutions are sought over thetotal domain Q = (0, L) at once. Consequently, the approximate solution [u(x) UN (x) =

: cJct)J(x) ± (x)] is required to satisfy the boundary conditions of the problem. Thisplaces severe restrictions on the derivation of the approximation functions 4, (x) and 4(x),especially when discontinuities exist in the geometry, material properties, and/or loadingof the problem (see Chapter 2 for details). The finite element method overcomes this shortcoming by seeking approximate solution (3.2.3) over each element Obviously, geometryof the element should be simpler than that of the whole domain, and the geometry shouldallow a systematic derivation of the approximation functions, as we shall see shortly.

To put the elements back together into their original positions, i.e., connect the approximate solution from each element to form a continuous solution over the whole domain,we require the solution to be the same at points common to the elements. Therefore, weidentify the end points of each line element as the element nodes, which play the roleof interpolation points (or base points) in constructing the approximation functions over

an element. Depending on the degree of polynomial approximation used to represent the

solution, additional nodes may be identified inside the element

Since the weak form over an element is equivalent to the differential equation and thenatural boundary conditions (3.2,6), i.e., conditions on Q of the element, the approximatesolution u of (3.2.3) is required to satisfy onlythe end conditions i4(Xa) = u and u (Xb) =u. We seek the approximate solution in the form of algebraic polynomials, although this isnot always the case. The reason for this choice is two-fold: First, the interpolation theory ofnumerical analysis can be used to develop the approximation functions systematically overan element; second, numerical evaluation of integrals of algebraic polynomials is easy.

As in classical variational methods, the approximation solution u must fulfill certainconditions in order that it be convergent to the actual solution u as the number of elementsis increased. These are:

1. It should be continuous over the element and differentiable, as required by the weak form.2. It should be a complete polynomial, i.e., include all lower-order terms up to the highest

order used.

3. It should be an interpolant of the primary variables at the nodes of the finite element (atleast the nodes on the boundary of the element so that the continuity of the solution canbe imposed across the interelement boundary).

The reason for the first condition is obvious; it ensures a nonzero coefficient matrix. Thesecond condition is necessary in order to capture all possible states, i.e., constant, linear, andsoon, of the actual solution. For example, if a linear polynomial without the constant term isused to represent the temperature distribution in a one-dimensional system, the approximatesolution can never be able to represent a uniform state of temperature in the element shouldsuch a state occur. The third condition is necessary in order to enforce continuity of theprimary variables at points common to the elements.

For the weak form in (3.2.7), the minimum polynomial order of u is linear. A completelinear polynomial is of the form

u(x)=c±cx (3.2.11)where c and c are constants. The phrase “complete polynomial” refers to the inclusion ofall terms up to the order desired; Omission of c would make it an incomplete linear polynomial, Similarly, c + cx2 is an incomplete quadratic polynomial because the linear termis missing. The expression in (3.2.11) meets the first two conditions of an approximation.The third condition is satisfied if c and c meet the conditions

u(xa) C +CXa 14(xb) C +CXbi4 (3.2.12)Equation (3.2.12) provides two relations between (ci, c) and (ut, u). which can be expressed in matrix form as

u I x c= e (3.2.13)

U2 1 x c2

= -(uxb -ux) (au ± au)

=

of the quadratic functional P(u), IP = 0, where

Ie(u) = Be(u,u)_le(u)

Inverting (3.2.13), we obtain

(3.2.14)

114 AN !NrRooUcrrON TO ThE F ITE asMorer METHOD OLkPTER 3: SECOND.ORDER DIFFERENTIAL EUATI0NS IN ONE DIMENSION: FINITh ELERSENT MODELS 115

where he Xb —x0 and( =xb,a =—x, =1, and /3 =—1)

cz=(—1)”x, =(—1)i; X=Xa, 4xb (3.2.15)

In (3.2.15), i and j permute in a natural order:

ifi=1 thenj=2; ifi=2 thenj=1

The a and /3 are introduced to show the typical form of the interpolation functions.Substituting c and c from (3.2.15) into (3.2.11), we obtain

where

u(x) = —[(au +au) + (8u +u)x]

= -(a + x)u + i:(a + x)u

= 4(x)u + 4(x)u=

4(x)u (3.2.16)

(3.2.17)(x) (a+$x)Xb-X

XbX5 he XbX5

which are called the linear finite element approximation functions.The approximation functions ij’x) have some interesting properties. First, note that

u so u(xa) fr(x5)u + i14(x)u

implies (x5)= I and (x5)= 0. Similarly,

so u(Xb) = V4(xb)u + I/4(xb)u

gives ‘(xo) = 0 and,2e(xb) = 1. In other words, t’f is unity at the ith node and zero at theother node. This property is known as the interpolation property (i.e., u is an interpolantof u(x) through nodes I and 2) of (x) and they are also called interpolation functions.When they are derived to interpolate function values only and not the derivatives of thefunction, they are known a the Ligrange interpolation functions. When the function andits derivatives are interpolated, the resqlticg interpolation functions are known as the Hermitefamily of interpolation functions. These will be discussed in connection with beam finiteelements in Chapter 5.

Another property of (x) is that their sum is unity. To see this, consider a constantstate of u = Co. Then the nodal values u and u should be equal to each other and bothequal to the constant Ca. Hence, we have

u(x)=if4(x)u +sfr(x)u=co -+

This property of tr(x) is known as the partition of unity. In summary, we have

0 if i j= . . (3.2.18a)

lift = j

Ue(X) = c +cx

True solution, u(x) ( =e4r(x) +u2’2(x)

t4x)

Figure 3.2.4 (a) Linear interpolation functions. (b) Linear finite element approximation.

wherex = Xa andx = Xb [seeEq. (3,2.15)1. The functions are showninFig. 3.2.4(a) andu(x) is shown in Fig. 3.2.4(b). Althoughproperties (3.2.18a) and (3.2.18b) are verifiedforthe lineas Lagrange interpolation functions, they hold for Lagrange interpolation functionsof all degrees.

The element interpolation functions in (3.2.17) were derived in terms of the globalcoordinate x [i.e., the coordinate appearing in the governing differential equation (3.2.1)),but they are defined only on the element domain = (xe, Xb). If we choose to express them in terms of a coordinate . (convenient in evaluating integrals of ifr), withthe origin fixed at node 1 of the element cle, the functions frf of (3.2.17) in terms of

fr(I) = 1— -, t(i) = (3.2.19)

The coordinate is termed the local or element coordinate [see Fig. 3.2.3(a)].The interpolation functions are derived systematically: Starting with an assumed

degree of a complete algebraic polynomial for the primary variable u and determiningthe coefficients c of the polynomial in terms of thevalues u of the primary variable uat the element nodes, the primary variable is finally expressed as a linear combination ofapproximation functions (x) and the nodal values u, which are called the element nodaldegrees offreedom. The key point in this procedure is the selection of the number and thelocation of nodes in the element so that the geometry of the element is uniquely defined andinterelement continuity may be easily imposed. The number of nodes must be sufficient toallow the rewriting of the coefficients in the assumed polynomial in terms of the primaryvariables. For a linear polynomial approximation, two nodes with one primary variable pernode are sufficient to rewrite the polynomial in terms of the values of the primary variableat the two nodes and also define the geometry of the element, provided the two nodes arethe end points of the element.

The degree (or order) of the polynomial approximation can be increased to improve theaccuracy (see Fig. 3.2.5). To ifiustrate the derivation of the interpolation functions of higherorder, we consider the quadratic approximation of u(x)

‘‘1i

he(a) (b)

z=1

(3.2.18b)u (x) = c + cx + cx2 (3.2.20)

116 AN DOD’.ODUCTION TOT EFNITEELEMENT METHOD CNAPTER 3: SECOND-ORDER DIFFERENTh4[, EQUATIONS INONE DIMENSION: FINITE ELEMENT MODELS 117

Since there are three parameters c (i = 1 2, 3), we must identify three nodes in the elementso that the three parameters can be expressed in terms of the three nodal values u. Two

of the nodes are identified as the endpoints of the element to define its geometry, and thethird node is taken interior to the element. In theory, the third node can be placed at anyinterior point. However, the midpoint of the element, being equidistant from the end nodes,

is the best choice. Other choices are dictated by special considerations (e.g., to have acertain degree of singularity in the derivative of the solution), and we will not discuss suchspecial cases here. Thus, we identify three nodes, two at the ends and one in the middle,

of the element of length h [see Fig. 3,2.5(b)]. Following the procedure outlined for thelinear polynomial, we eliminate c by rewriting t4(x) in terms of the three nodal values,(ut, 4,34). The three relations among q and u are

34 u(4)rc +cx +c(4)2 (3.2.21a)

34 34(4) = c + ‘44 + ‘4(4)2

34 1 4 (4)2 ‘4= 1 x. (4)2 ‘4 (3.2.21b)

34 1 4 (4)2 ‘4where 4 (i = 1, 2, 3) is the global coordinate of the ith node of the element (4 =4 =X + 0.5he, and4 =xb). Inverting (3.2.21b), we obtain

‘4= 4=x(4)2_4(x)2

‘4= = (x5)- (4)2 (3.2.22)

c=_yjCu, y=—(x1—.4), D=a

where DC denotes the determinant of the matrix in (3.2.21b), and a, , and y are defined- in (3.2.22) in terms of the nodal coordinates. The subscripts used in (3.2.22) permute in a

natural order;

if i=1, then j=2 and k=3if i = 2, then j = 3 and k = 1 (3.2.23)if 1=3, then j=l and k=2

For example, ‘4, , and y are given by

‘4=4(4)2 Xe(Xe)2 = 2—

4)2, y=4 —4Note that if 4 is very close to either 4 or 4 (i.e., node 2 is placed close to node 1 ornode 3), making the rows of the coefficient matrix in (3.2.21) nearly the same, which makesthe matrix nearly singular and not invertable.

Substituting for q from (3.2.22) into (3.2.20) and collecting the coefficients of 34,34,and 34 separately, we obtain

34(x) =(x)u + (x)’4 + 4(x)34 = J(x)u (32.24)

where are the quadratic Lagrange interpolation functions,

(x) = -(a + x + yx2) (i = 1,2,3) (3.2.25)

Once again, the quadratic interpolation functions can be expressed in terms of the localcoordinate i. When the interior node, node 2, is placed at a distance .1 = ah, 0< a < 1,the quadratic interpolation functions are given by

= a(l — a)(i (3.2.26)

For a = , i.e., when node 2 is placed at the midpoint of the element, the interpolationfunctions in (3.2.26) become

= (i— ) ( —

= 4 ( —

(3.2.27)

Plots of the quadratic interpolation functions are given in Fig. 3.2.6. The function is equalto 1 at node 1 and zero at the other two nodes, but varies quadratically between the nodes.

u(x) u(x)u(x)

UI I.E1 X2 X3

(a) (b)

Figure 3.2.5 (a) Linear approximation. (b) Quadratic approximation.

or, in matrix form,

118 AN INTEODUCIION TO TEE FINrrE EMENTMETEOD CHAFFER 3: SECOND-ORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: FINITE ELEMENT MODELS 119

Lagrange interpolation of the variables, i.e., all problems for which the primary variablesare the dependent unknowns—not their derivatives.

Step 3. Finite Element Model

The weak form (3.2.7) or (3.2.9) is equivalent to the differential equation (3.2.1) overthe element , and it also contains the natural boundary conditions (3,2.6). Further, thefinite element approximations (3.2.16) and (3.2.24) are the interpolants of the solution. Thesubstitution of (3.2.16) or (3.2.24) into (3.2.7) will give the necessary algebraic equationsamong the nodal values u and Q of the element 2e. In order to develop the finite elementmodel based on the weak form (3.2.7), it is not necessary to decide a priori the degree ofapproximation of u. The finite element model can be developed for an arbitrary degree ofinterpolation:

_____________________

uu (3.2.28)

bwhere frJ are the Lagrange interpolation function of degree a — 1. For n >2, the weakform in (3.2.7) must be modified to include nonzero secondary variables, if any, at interior

Figure 3.2.6 One-dimensional Lagrange quadratic element and its interpolation functions: nodes. This modification is discussed next.

(a) geometry of the element; (b) interpolation functions. The integration by parts in Step 2 [see Eq. (3.2.5)] of the weak-form development foran element with interior nodes is carried out by intervals (xi, x), (x, xi),..., (x_1,x):

The interpolation properties (3.2.18a) and (3.2.l8b) can be used to construct theLagrange interpolation functions of any degree. For example, the quadratic interpolationfunctions (3.2.27) can be derived using the interpolation property (3.2.18a). Since 1t)must vanish at nodes 2 and 3, i.e., at i = and x =h, respectively, is of theform

- he)( he)

The constant C1 is determined such that is equal to 1 at i = 0:

1=Ci(0he)(0he) or C1=2/h

This gives

= (.t — he)(. — h) = (1 — ;) (1 —which is identical to that in (3.2.27). The other two interpolations functions can be derivedin a similar manner.

Although a detailed discussion is presented here on how to construct the Lagrangeinterpolation functions for one-dimensional elements, they are readily available in bookson numerical analysis, and their derivation is independent of the physics of the problem tobe solved. Their derivation depends only on the geometry of the element and the numberand location of the nodes. The number of nodes must be equal to the number of terms inthe polynomial. Thus, the interpolation functions derived above are useful not only in thefinite element approximation of the problem at hand but also in all problems that admit

(a)

n—I ( / dw du ) r du 10=1 f (a——+cwu-wf dx-jw(x)a—I

\ dx dx L dxJ j

= I (a——+cwu—wf di—w(x) —a— —w(4)(a—lfx / dwdu ) du / du

J \ dx dx ( dx) \ dxJ

e / du—w(x2)

x

/ du

\ dx

/ du\ /du_w(x_i)(,a_) _w(x)),,a_) -

dx dx

fx / dw du du

dx dx)dx_w(x)(_aj=j ça——+cwu—wf

/ du\ / du\ 1 e du\ / du\ 1-w4) [la_( + _a_) I -w3)Ra_)+ (-a—)

+ dx dxj+jdx,’-

dx j

du\ du\ 1 / du\w(x1) ad ( dxJ+j—) + —a—)

fx / dw du

dx dx ) dx - w(x)Q - w(x)Q - w(x)Q=j (,a——+cwu—wf

— . .. — w(x1) Q-I — w(x)Q (32.29a)

120 AN INTRODUCTION TO THE FINITE ELEMENT METHODCHAI’TERI: SECONDORDER DIFFERENTIAl, EQUAI1ONS IN ONE DIMENSION: FINITE ELEMENT MODELS 121

where x and x denote the left and right sides respectively of node and The ith algebraic equation of the system of n equations can be written as

= (_a) Q= (a) +(_a) (t=l 2 n) (3231a)‘ dx dx dx 1=1

where(32 29b) fXb / dd

K = B?ifr) I (a__L+cr1)dx/ du” / du” / du” j \ dx dx /

Q_1= (a_) +(_aT) Q=(a_)dx x x

= f f dx (3.2.31b)

Thus, Q, i = 2, 3 n — 1, denotes the jump in the value of the secondar3’ variable in Note that the interpolation property (3.2.18a) is used to writegoing from the left to the right of the ith node. This value is zero if no external source isappliedat the node. Thus, for an element with n nodes, the weak form becomes

= (3.2.32)Xb dwdu j=I

0= f (aT — + dx

- f wq dx-

w()Q (3.2.30) Equations (3.2.31a) can be expressed in terms of the coefficients K, j, and Q as

K1u + K2u+ ‘ . ‘ + Ku = f + QNote that Q = Q and Q = Q represent the unknown pomt sources at the end nodes,andallother Q (i =2,3,...,n— l)arethespecifiedpointsources,ifany,attheinteriornodes.

Next, we develop the finite element model of Eq. (3.2.l)when the (n — l)st-degree (3.2.33a)Lagrange polynomials are used to approximate u(x). Following the Ritz procedure devel-

+ K8 + + K8 8 — +oped in Section 2.5.2, we substitute (3,2.28) for u and w = ifr, w = w = into nlUI ,2U2 — Jfl fl

the weak form (3.2.30) to obtain n algebraic equations: In matrix notation, the linear algebraic equations (3.2.33a) can be written as

[Xb difr / e di/iJ “ e tV’’ e d I[Kel{uel {fe} + {Qt} or K = f8 + Q8 (3.2.33b)

0= j

a-a-— u-—,1 + c’1 uIr(x) — f x — (x)Q The matrix K8 is called the coefficient matriz or stiffiiess matrix in structural mechan\j=I j1 j=I

ics applications. The column vector f is the source vecto or force vector in structural(1st equation) mechanics problems. Note that Eq. (3.2.33) contains 2n unknowns: (u, u u) and

X e / “ / I (Q Q Q),calledprimaryandsecondaryelementnodn1degreesoffreedom;hence,0

= f f u’L ) + c’ ( ui,!i(x)) — tf dx — (x)Q I it cannot be solved without having an additional n conditions. Some of these are provided, dx dx / \‘ / j=I by the boundary conditions and the remainder by “balance” or equilibrium of the secondary

(2 d ation) I variables Q at nodes common to several elements. This balance can be implemented byn equ I putting the elements together (i.e., assembling the element equations). Upon assembly andimposition of boundary conditions, we shall obtain exactly the same number of algebraicequations as the number of unknown primary and secondary degrees of freedom. The ideas

0 — 1 iI ( u8 ¶1”\ + cl,!’8 (“u’(x)’\ — i/if dx — i8(x8)Q underlying the assembly procedure are discussed in the next section.—

adx “ dx ) -‘ ) “ Thecoefficientmatrix[K8],whichissymmetric,andsourcevector(f8)canbeevaluated

for a given element and data (a, c, and f). For elementwise-constant values of a, c, and f(Ith equation) (say, a8,c8, andf8) the coefficients K and ,f8 can easily be evaluated for a typical element,

as discussed next,

Linear Element. For a mesh of linear elements, the element S2 is located between the0

= fx[a (E + c (E ui/4(x)) — f] dx — (x)Q global nodes xa =x8 and xb =x81 [see Fig. 3.2.2(b)]. Hence,

.8 j—1 J—i j—l X,4 dl/J dl,!r(nth equation) K

= f + ce1/1j1J) dx, f= j f8ifr dx

122 AN lWtRODUCT1ONTRE Fntar.4ENEMEflIOD CHM’tEIO 3: SECOND-ORDER DW RANTIAIEQUAI1ONS IN ONE DIMERSION: FINITE ENT MODELS 123

= f [ac () () +C (i ) (i )] d

=

1h, / / .\t]K2=j

= —? + cehe=Kj (by symmetry)

çh./ t a 1=

a — — + 5— J d.t = +

Jo \ he h0 eie / nO

ph / .\ fh 1fi°jf0(1)d.=feh0,foj fdfehe

he) 2 h 2

Thus, for constant the total source fh0 is equally distributed to the two nodes. Thecoefficient matrix and column vector are

[K9 [‘ —] Cehe [ ] {fe}fe’! { } (3.2.34)

If a = ax and c = c, the coefflciçnt matrix [K9 can be evaluated as

[K0]=(+X1’1F 1 —l 2 1

h0’ 2 1 6 1 2

The reader should verify this. Note that when a is a linear function of x, this is equivalent to replacing a in the coefficient matrix with its average value [compare (3.2.34) with(3.2.35)]:

aavg (x0 +xe+i)a0

A(x) =Ao[2 — (xli)] Area ofcross section, A0

Xe X

IL

—A A0j

A0 =Aj2 — (x0lL)], A1 =A0[2 — (x1/L)]

or, in the local coordinate system f,

=j (ae + ceii!) d., .t=

fe di

wberex=4+i and

- di ddx=dx, —=——-

dx dx

The irf can be expressed in terms of I as [see (3,2,19)]

i/4(i)=1 —I/h0, rfr(I)=I/h0

We can compute K and je by evaluating the integrals. We have

Similarly,

Figure 3.2.7 Approximation of an element with linearly varying cross section by an equivalent element of constant cross section,

For example, in the study of bars with linearly varying cross section (see Fig. 3.2.7)

f Ae1Aea=EA(x)=E Ae+

hX

this amounts to replacing the linearly varying cross section with a constant cross sectionwithin each element, the cross-sectional area of the constant section being the average areaof the linearly varying element. Here A0 denotes the cross-sectional area at x0 and A0+ isthe area atx =x,

Quadratic Element. For a mesh of quadratic elements, element f2 is located betweenglobal nodes x5 =x_1 and xb = x.1.Hence,

x24 ds/i difrK=J (ae_1_L+cei,;) dx

dx dx

= fh(aeLI +CeiI1i) di (32.36a)

fe

jfrfXj i/iffedi (3.2.36b)

0

where ,f(I) (i = 1,2,3) are the Lagrange quadratic interpolation functions in (3.2.27).Evaluating the integrals in (3.2.36a) and (3.2.36b), we obtain

(he / 3 4,\/ 3 4KI=J a+)l--+

7ae 2=

(3.2.35)

CHAPTER 3; SECONDORVER DIFFERENI1AL EQUATIONS IN ONE DIMENSION: FINifE ELEMENT MODELS 125

In summary, for elementwise-constant values of the data a, c, andf, the element matricesof a linear and quadratic elements are

Linear Element

(ae [1 —1] Cehe [ ]) {:} { } + {} (3,2,39)

Quadratic Element

7 —8 ii(ad I— —8 16 —81

1 —8 7J

124 AN INTRODUCTION iO TNT FINITE ELEMENT METhOD

eçh, f’ 3 4.C1f4 8.1

K12=j

d.1

8ae 2=

—-- +

h. /4 8.1’\/4 8.1Kn=j

+ c[4(1_)][4(1_y[)]) d.1

7a 2=

and so on. Similarly,

fffe[1_+2()]

= jh.

fe [4(1_i)] d.1 =

f = f (by symmetry)

Thus, the element coefficient matrix and source vector for a quadratic element are

[K94 [_8 16 4]Cehe

16 2] (3.2.37a)

(f9 = 4 (3.2.37b)

Note that, for quadratic elements, the total value of the source fh is not distributedequally between the three nodes. The distribution is not equivalent to that of two linear elements of length he. The computation of f should be based on the interpolation functionsof that element. The sum of f for any element should always be equal to the integral off(x) over the element

fe_ff(x)dx - (3.2.38)

r 4 2 —ll\ Iu

216 2Illud301 hi IL1 2 4j L3 ifli

IQ1= 14 1 + I (32.40)

611(ii Q

When the coefficient c = 0, the corresponding contribution to the above equations shouldbe omitted.

When a, c, and f are algebraic polynomials in x, the evaluation of K and fJ isstraightforward. When they are complicated flmctions of x, the integrals in [K9 and (fd}

are evaluated using numerical integration. A complete discussion of numerical integrationwifi be presented in Chapter 7.

3.2.4 Connectivity of Elements

In deriving the element equations, we isolated a typical element (the eth element) fromthe mesh and formulated the variational problem (or weak form) and developed its finiteelement modeL The finite element model of atypical element contains n equations among 2nunknowns, (ut, u, . .., u) and (Q, Q,..., Q). Hence, they cannot be solved withoutusing the equations from other elements to get rid of extra unknowns. From a physical pointof view, this makes sense because we should not be able to solve the element equationswithout considering the complete set of elements and the boundary conditions of the totalproblem.

To obtain the finite element equations of the total problem, we must put the elementsback into their original positions (undoing of what was done before formulating the discreteproblem). Inputting the elements with their nodal degrees of freedom back into their originalpositions, we must require that the solution u(x) is uniquely defined (i.e., u is continuous)and their source terms Q are “balanced” at the points where elements are connected toeach other. Of course, if the variable u is not continuous, we do not impose its continuity;but in all problems studied in this book, unless otherwise stated explicitly (like in the caseof an internal hinge in the case of beam bending), the primary variables are assumed to becontinuous. The continuity of the primary variables refers to the single-valued nature of the

(a

+(_a) = -* Q + QC+l = (3.3.45b)

The interelement continuity of the primary variables can be imposed by simply renaming

the variables of all elements connected to the common node. For the connection in Eq.(3.2.41), we simply use the name

(3.2.46)

where I is the global node number at which the three elements are connected. For example,for a mesh of N linear finite elements (n = 2) connected in series [see Fig. 3.2.2(b)], wehave

= j+1+ Q+l

CHAPtER 1: SECOND-ORDER po’PeancnAl. EQUAnONs TN NE DMENSTON: FTNYrE ELEMEND MODELS 127

The balance of secondary variables can be interpreted as the continuity of a(du/dx)[not a(du/dx)j at the node (say, global node I) common to elements 2e and !2e+i whenno change in adu/dx is imposed externally:

/ duY I dUtI

or

(a) + (_a) =0 Q + Q1 =0 (3.3.45a)

If there is a discontinuity of magnitude Qr in a in going from one side of the node to theother side (in the positive x direction), we impose

126 AN 0DUONmnsmoD

solution; balance of secondary variables refers to the equilibrium of point sources. Thus,the assembly of elements is carried out by imposing the following two conditions:

1. If the node i of element 12e is connected to the node j of element f2 and node k ofelement the continuity of the primary variable u requires

= uy = (3.2,41)

When elements are connected m series as shown in Fig 328 the continuity of u requires

u=u’ (3242)

2 For the same three elements the balance of secondary vanables at connecting nodesrequires

+ + Q9)= Qi (3.2.43)

where I is the global node number assigned to the nodal point that is common to thethree elements, and Q is the value of externally applied source, if any (otherwise zero),at this node [the sign of Qi must be consistent with the sign of Q in Fig. 3.2.3(b)]. Forthe case shown in Fig. 3.2.8, we have

0 if no external point source is appliedQ + Q = Q if an external point source of magnitude (3.2.44)

Q is applied

÷ Externally applied Q

(b)

Figure 3.2.8 Assembly of two Lagrange elements: (a) continuity of the primaiy variable; (b) balanceof the secondary variables.

23Interior nodes

(a)

c:ie Q e+l

Qt123 n-In 123 m-lm

Figure 3.2.9 shows the connected finite element solution, uh, composed of the elementsolutions u. From the connected solution, we can identify the global interpolationfinctions(p1, which can be defined in terms of the element interpolation functions correspondingto the global node I, as shown in Fig. 3.2.9.

To enforce balance of the secondary variables, itis clear that we can set [see Eq. (3.2.44))+ Q + Q equal to zero or to a specified value Qi only if we have such an expres

sion in the finite element equations. To obtain such expressions, it is clear that we must addthe ith equation of the element f2e, the jth equation of element and the kth equationof element 2e• For the case shown in Fig. 3.2.8, the nth equation of element 2e must beadded to the first equation of the element i.e., we add

and

(Ku + K’u) = f+ je+l

+ (Q +

= f + + Qi (3.2.47)

This process reduces the number of equations from 2N to N + 1 in a mesh of N linear

elements. The first equation of the first element and the last equation of the last element

KU,_1 + (K + K)U + KU+i = fNl + f[ +Qr.l

+ QN

KUN + KUN+I = f + Q (unchanged)

(3,2,49)

These are called the assembled equations. They contain the sum of coefficients and sourceterms at nodes common to two elements. Note that the numbering of the global equationscorresponds to the numbering of the global primary degrees of freedom, U1.This correspondence carries .the symmetry of element matrices to the global matrix. Equations (3.2.49)can be expressed in matrix form as

128 AN INTPODUCflON TO ThE HNITE ELEMENT METhOD

Ut = S

CHAPTER 3: SECOND ORDER DIFEERENflAI. EQUAtiONS IN ONE DIMENSION: FINTTE ELEMENT MODELS 129

will remain unchanged, except for renaming of the primary variables. The left-hand side of(3.2.47) can be written in terms of the global nodal values as

(Ic1 u + K2u+ .. + Ku) + (Ku’ + Ku’ + . . . + Kuj

= (KUN + K2UN+i +‘‘ + KUN+,,-i)

(ve+1,, ve+iii i I ye+i)-rTV 11 VN+n_IT 12 VN4.nT... T “N+2n—2

= K1UN + K2UN+j + ... + K(_i)UN+fl_2 + (K + K’)UN÷,1_l

+ K’UN+ +. . . + KUN+n_2 (3.2.48)

where N =Qi — 1)e ± 1. For a mesh of N linear elements (n =2), we have

K1U1 + KU2 = f1’ + Q (unchanged)

K1U1 ÷(K + K1)U2+ KU3 = f + f ± Q + QK1U2+(4 + K5)U3+ KU4 = + + + Q

w’1() (x)

- x

i-i / I / i+it11

N4-I

u(x)YU1j(x)

1=1

Ir’t(x), x1_1XX1

iO,

(x) X1XXjj

(a)

(b)

Figure 3.2.9 Global interpolation ftmctions for the (a) linear elements and (b) quadratic elements.

to obtain

K111 12

K1 K2+K1 K 0

K1 K222+K1

0 K+K K

vN vN21 22

Ui

(12

U3

UN

fii

:2J2 TJl

:2 :3J2 TJ1

+

f2N fN

Q + QQ + Q

(3.2.50)

132 AN NTRODUCfl0NTO TSR FINrFE ELEMENT METHOD

K24 0 because global nodes 2 and 4 do not belong to the same elementK33 = K2 + K + K1

and so on. The problem can also be solved by numbering global nodes 1 and 2 to be thesame node. This leads to a 3 x 3 global system of equations.

In summary, assembly of finite elements is carried out by imposing interelement continuity of primary variables and balance of secondary variables, Renaming the primaryvariables of an element in terms of the global primary variables and using the correspondence between the local and global nodes allows the assembly. When certain primary nodalvalues are not required to be continuous across elements (as dictated by physics or the variational formulation of the problem), such variables may be condensed out at the elementlevel before assembling the elenients.

3,2.5 linposition of Boundary Conditions

The discussion in Section 3.2.1—3.2.4 is valid for any differential equation that is a specialcase of the model equation (3.2.1). Each problem differs from the other in the specificationof the data (a, c, f) and boundary conditions on the primary and secondary variables (u, Q).Here we discuss how to impose the boundary conditions of a problem on the assembled setof finite element (algebraic) equations. To this end, we use the problem in Fig. 3.2.10. Theboundary conditions of the problem are evident, at least for an engineering student, fromthe structure shown. The known primary degrees of freedom (i.e., displacements) are

u=U1=0, u=U2=0, u=U4=0 (3.3.54a)

The known secondary degrees of freedom (i.e., forces) are

Q + Q + Q = 2P (3.3.54b)

The forces Q , Q, and Q are unknown (reaction forces), and they can be determined afterthe primary degrees of freedom are determined.

Imposing the boundary conditions (3.2.54a) and (3.2.54b) on the assembled system ofequations (3.2.53) with ,t = 0, we obtain

rKl, 0 K2 0 1 ‘QH0 K2 0 JU2=0IjQIK1 K1 K+K+K? K3it 12 1 U3 12P I

[o 0 K2j [u4=oJ [QJ

3.2.6 Solution of Equations

As a standard procedure in finite element analysis, the unknown primary degrees of freedomare determined first by considering the algebraic equations corresponding to the unknownprimary variables. Thus, in the present case, we consider the third equation in (3.2.55) tosolve for U3:

CHAf’TER 3: SECOND-ORDER DIFFERENTIAL EQUA11ON5 IN ONE DIMENSION: FINifE ELEMENT MODELS 133

(K2 + K + K1)U3= 2P — (K1U1 + K1 U2 + KU4) (3.2.56)

Equation (3.2.56) is called the condensed equation for the unknown U3. The term in parentheses on the right-hand side is zero because all specified displacements are zero in thepresent problem. Hence, the solution is given by

U3 = (3.2.57)K + K + K1

The unknown secondary variables are determined by considering the remaining equationsof (3.2.55), i.e., those that contain the unknown secondary variables:

K2U3

= KU3

because U1, U2, and U4 are zero.It is possible, although not common with commercial finite element programs, to move

all the unknowns to the left-hand side of (3.2.53) and solve for them all at once. But thisprocess destroys the symmetry of the coefficient matrix and requires more computationaltime in practical problems.

To obtain numerical values, we use the geometric and material data shown in Fig. 3.2.10.We obtain

EbAb (14 x l06)(ird/4)= 87, 693 lb-in.K1=———=

8xl2

2 E3A, — (30 x l06)(rd/4)= 122,718 lb-in.K11=—----—

4x12

EaAa (10 x l06)(rd/4)= 109,083 lb-in.

6xl2 -

The displacement of node 3 is U3 =0.01878 in., reaction at node 1 is QI = —1,647 lbs,Q = —2, 305 lbs and reaction at node 4 is Q = —2, 049 lbs. Hence, the stresses in elements 1,2, and3 are = 2,739 psi, 2 = 11,739 psi, ando3 = —2,609 psi, respectively.

In general, the assembled finite element equations can be partitioned according to thesets of specified and unspecified displacements into the following form:

K” K’2 U’

- K2’ K U2 = F2(3.2.59)

where U’ is the column of known (i.e., specified) primary variables, U2 is the column ofunknown primary variables, F’ is the column of unknown secondary variables, and F2 is thecolumn of known secondary variables. Writing (3.2.59) as two matrix equations, we obtain

KU’ +K’2U2F’ (3.2.60a)

or

Q K,’,

= 0

Q 0

0 K,’2 0

K,’, K,’ 0

0 K, K

I

(3.2.58)

(3.2.55)

This contains four equations in four unknowns: U3, Q, Q, and Q.

K,’,Ul + K,’,U2+(K + K + K,’,)U3+ K,’2U4=2PK2’U’ +K22U2=F2 (3.2.60b)

130 .*.i nrraooucnONromeflNrra En.mrrn1OD CHAFrER 3 SECOND-ORDER DREN31AL EQUAIIONS IN ONE Dfl,4ENSION: NITE ELEMENT MODELS 131

h1=8ft.,h2=4ft, h3=6ft.,

,O.5 ilL,d1=O.875in.,

d5=l.Om.,E3=30x10°psi,

E=l4xlO6psi,E3=lOxlO6ps

P=3000lb -

Steel (E3,A3) ‘ Rigid member(constrained

r 3

h2

r/’ Brass(E53A0) 0Aluminum (E0,A3) Global node number

h1

Note that the discussion of assembly in Eqs. (3.2.49) and (3.2.50) is based on the

assumption that elements are connected in series. In general, several elements can be con

nected at a global node, and the elements do not have to be consecutively numbered. In that

case, the coefficients coming from all elements connected at that global node will add up.

For example, consider the structure consisting of three bar elements shown in Fig. 3,2.10.

Suppose that the connecting bar is rigid (i.e., not deformable) and is constrained to remain

horizontal at all times. Then the continuity and force balance conditions for the structure

are

u=uunsU3, Q+Q+Q2P (3.2.51)

To enforce these conditions, we must add the second equation of element 1, the first equation

of element 3, and the second equation of element 2:

(K?iu + K42u) + (K?iu + K?au) + (K1u+ K2u)

(3.2.52)

We note the following correspondence of local and global nodal values (see Fig.

3,2.10):

u=U1, u=U2, t4=u=u?=U3, u?=U4

K?1U1 + K?1U2 + (K + K?1 + K)U3+ KU4

Hence, (3.2.52) becomes

The other equations remain unchanged, except for renaming of the primary variables. Theassembled equations are

K?1 0 K?2 0 U1

0 K?1 K?2 0 U2 — fi2 +

K?1 K1 k K?2 U3— f+f?+f Q+Q+Q

0 0 K?1 K (14 f Q(3.2.53)

where k = K?2 + K?1 + K. Note that all ,f are zero for the problem in Fig. 3.2.10 becausethere is no distributed axial force (f = 0 for all elements).

The coefficients of the assembled matrix can be obtained directly. We note that theglobal coefficient K11 is a physical property of the system, relating global node Ito global

node J. For axial deformation of bars, K11 denotes the force required at node Ito induce aunit displacement at node J, while the displacements at all other nodes are zero. Therefore,K11 is equal to the sum of all K for which I corresponds to I and j corresponds to J, and

I and j are the local nodes of the element Thus, if we have a correspondence betweenelement node numbers and global node numbers, then the assembled global coefficients can

readily be written in terms of the element coefficients. The correspondence can be expressedthrough a matrix [B], called the connectivity matrix, whose coefficient has the followingmeaning: b1 is the global node number corresponding to the jth node of element i. For

example, for the structure shown in Fig. 3.2.10, the matrix [B] is of order 3 x 2(3 elements

and 2 nodes per element):

13

[B]= 2 3

34

This array can be used in a variety of ways—not only for assembly, but also in the computer implementation of finite element computations. The matrix [B] is used to assemble

coefficient matrices as follows:

K?1 = K11 because local node 1 of element I corresponds to global node 1.

K?2 = K13 because local nodes 1 and 2 of element 1 correspond to global nodes Iand 3, respectively

and so on. When more than one element is connected at a global node, the element coefficients are to be added. For example, global node 3 appears in all three rows (i.e., elements)

of the matrix [B], implying that all three elements are connected to a global node 3. More

specifically, it indicates that node 2 of element 1, node 2 of element 2, and node I of element

3 are the same as global node 3. Hence,

K2 + K + K?1 = K33

Assembly by hand can be carried out by examining the finite element mesh of the

problem. For the mesh shown in Fig. 3.2.10, we have

K23 = K because global node 2 is the same as node 1 and globalnode 3 is the same as node 2 of element 2.

Figure 3.2.10 The geometry and finite lement mesh of a three-bar structure.

=f+f?+f+2P

134 AN IMrRODUCTIONTO THE FINITE ELEMEND METHOI) CHAPTER 3: SECOND-ORDER DFERENTIAL EQUATIONS IN ONE DIMENSION: FINITE ELE34ENE MODELS 135

From (3.2.60b), we have

U2 = (K) (F2 — K21U’) (3.2.60c)

Once U2 is known from (3.2.60c), the vector of unknown secondary variables, F’, can becomputed using Eq. (3.2.60a).

In most finite element computer programs, element matrices are assembled as soonas they are generated and they are not stored in the memory of the computer. Thus,element equations are not available for postcomputation of the secondary variables. Also,due to the fact that the assembled coefficient matrix is modified (when operations areperformed to invert a matrix) during the solution of equations, Eq. (3.2.60a) cannot beused. Therefore, secondary variables can only be computed using their definitions, asdiscussed next.

3.2.7 Postcomputation of the Solution

The solution of the finite element equations gives the nodal values U1 of the primaryunknowns (e.g., displacement, velocity, or temperature). Once the nodal values of the primary variables are known, we can use the finite element approximation u (x) to compute the desired quantities. The process of computing desired quantities in numerical formor graphical form from the known finite element solution is termed postcomputation orpostprocessing; these phrases are meant to indicate that further computations are madeafter obtaining the solution of the finite element equations for the nodal values of theprimary variables.

Postprocessing of the solution includes one or more of the following tasks:

1. Computation of the primary and secondary variables at points of interest; primary variables are known at nodal points.

2. Interpretation of the results to check whether the solution makes sense (an understandingof the physical process and experience are the guides when other solutions are notavailable for comparison).

3. Tabular and/or graphical presentation of the results,

To determine the solution u as a continuous function of position x, we return to the approximation (3.2.28) over each element:

I4(x)=u)sfr)(x)

u(x) uii(x)i=l (3.2.61)

where N is the number of elements in the mesh. Depending on the value of x, the corresponding element equation from (3.2.61) is used. The derivative of the splution is obtainedby differentiating (3.2.61):

du-, ,di/4

dx‘‘

dx

d2 dik2

dx ‘dx (3.2.62)

du” VNd

dx i dx

Note that the derivative du/dx of the finite element approximation u based onLagrange interpolation is discontinuous, for any order element, at the nodes connectingdifferent elements because the continuity of the derivative of the finite element solution atthe connecting nodes is not imposed. In the case of linear elements, the derivative of thesolution is constant within each element (see Fig. 3.2.11).

The secondary variables can be computed in two different ways. In Eq. (3.2.58),we determined the unknown secondary variables QI Q’ and Q from the assembledequations of the problem in Fig. 3.2.10. Since the assembled equations often representthe equilibrium relations of a system, the Q computed from them will be denoted by(Q)ii. The Q can also be determined using the definitions in (3.2.6), replacing u withU. We shall denote Q computed in this way by (Q). Since (Q) are calculatedusing the approximate solution u, theyare not as accurate as (Q)j. However, in finiteelement computer codes, (Q) are calculated instead of This is primarily dueto computational reasons. Recall that, in aniving at the results in (3.2.58), we used partof the assembled coefficient matrix. In the numerical solution of simultaneous algebraicequations in a computer, the original assembled coefficient matrix is often modified, andtherefore the coefficients needed for the determination of the secondary variables are notavailable, unless they are saved in an additional arriy. For the problem in Fig. 3.2.10, we

du7’5

dxDiscontinuity

j he

du

dx

u(x)

u”(x) = u’(x)

Figure 3.2.11 Gradient of the finite element Solution.

—-4

1_

•a’

E‘4-

4-4

-n

-O

—.

IDQ

..

4-•

-.

0 8•n

--

---

-:.

;--

..

-:-

.-u,.

-:-

-II

=I

--

--

--.

--

—._

_ID

-4•

-—--

4-

--

.

-ii

--.

-.-.

ID-.

-.

g-.

,.

-g--

-.

-8

-—

I-I

---ID

IIII

--

-‘

-,

-

--

—‘—

.ID

-II

--I

II—

.‘

-a

ID--

--

IIII

___

g-

iiI

gII

-

•--4

-1--

.-

C

ID.

‘in

till--.

a’—

rx—

--

i—I

8>

1’g

I.

__

ID

a.

-a

IDa

-4

--,

L•

t—

II-

—a’

-

—-:-

•H

-1

.-

-.

.o

4-’

-In

-It

It-

ItID

--

•0

IDID

..

a-

—•

—-

--

_-

.=

______

-p

.—

—.

----‘

•.—

_____

8—

I-4

g-

r=

-.

-

:‘

IIII

II4.—

-—

.--

‘-

2-

I’

9—

a’

It4

..-.1

--a

4-

—.

8.

IIH

.I

-4-

4-

Ii

II,

±,±

4..,-g

’-

.-4

-II

-‘ii

Ii

-

II

-

I-

I:..i_

_______

2-.

—.-

-.

4.

--f-

-•

1I

-—

4-—

-

8-.+

-±

_

p:•

—1

‘‘-.1

-••

•

ID

n

________

-I

--•

-It

.1

--

.-

..

..

4.4

_____

-.

-.

.±

.--

.-

--

—•.

--

—.:-

•.-.

-.

-,_

:c,t

-•.

-•

--

••

.w•

-•

-.

•--

-•

••

-.

•--:.

•-

•-

--

-2

.•

•

.•

•--..

•-

---

-4-

-.

-.

•-

--

C-

C--

--.

•-

--

---

C

--

--

--

---

•--

--

H—

tID

+

S.

I

140 AN INTRODuCTIoN To.IENTMEfflOD CHAPTER): SECOND-ORDER DIFPERENTIAL EQUATIONS IN ONE DIMENSION: PINifE ELEMENT MODELS 141

-U-.,..

JIIII III

--- °-:

0.2- 0.4

Figure 3.2.12 Comparison of the finite elementsc

0.30

-- 0.25- - --

-. 0-. -

- 0.20 -

-- . .

- -

- =. 0.15 -. - - —

- .0.05-

z

,0.00

-

•:; HIIIIIIIjIIIIIUIIIi

Figure 3.2.13 Comparison of the finite element solutions du/dx with the exac

-

- solutions. - . - - - - - -- -

3.3 SOME REMARKS

A few remarks are in order on the steps described for the model equation.

Remark 1. Although the Ritz method was used to set up the element equations, any othermethod, such as a weighted-residual (i.e., the least-squares or Galerkin) method could beused.

Remark 2. Steps 1-6 (see Table 3.1.1) are common for any problem. The derivation ofinterpolation functions depends only on the element geometry and on the number andposition of nodes in the element The number of nodes in the element and the degree ofapproximation used are related,

Remark 3. The finite element equations (3.2.33b) are derived for the linear operatorequation

d / d\A(u)=f, where (3.3,1)

Hence, they are valid for any physical problem that is described by the operator equationA(u) = for its special cases. One need only interpret the quantities appropriately. Examplesof problems described by this operator are listed in Table 3.2.1. Thus, a computer programwritten for the finite element analysis of (3.2.1) can be used to analyze any of the problemsin Table 3.2.1. Also, note that the data a = a(x), c = c(x), and f = f(x) can be differentin different elements.

142 AN INThODUCOONtO THE FINFE ELEMENT METHOD CHAPTER 3: SECOND-ORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: FINiTE ELEMEN’rMoOELS 143

Remark 4. Integration of the element matrices in (3.2.3lb) can be implemented on acomputer using numerical integration. When these integrals are algebraically complicated,we have no other choice but numerical integration (see Chapter 7).

Remark 5. As discussed in (3.2.45a) and (3.2.45b) the point sources at the nodes areincluded in the finite element model via the balance of sources at the nodes. Thus, inconstructing finite element meshes, we should include nodes at the locations of point sources.If a point source does not occur at a node, it is possible to “distribute” it to the elementnodes, consistent with the finite element approximation. Let Q denote a point source ata point .1 =,1o’ I .I , where i is the local coordinate. The point source Qo can berepresented as a “function” with the help of the Dirac delta function

f(i) Qo(.I —

where the Dirac delta function B() is defined byfCO

j FGI)3(I — o) d = F(0)-00

The contribution of the function f(i) to the nodes of the element l2 = (1i, t) is computedfrom [see Eq. (3.2.3lb)]

= j f(.i)i/i(.I) di j Q08(i —i0)(i) di Qofr(io)0 0

where are the interpolation functions of the element 2e’ Thus, the point source Qis distributed to the element node i by the value Qo(.Io). Equation (3.3.4) holds forany element, irrespective of the degree of the interpolation, the nature of the interpolation(i.e., Lagrange or Hermite polynomials), or the dimension (i.e., one-dimensional, two-dimensional, or three-dimensional) of the elements. For one-dimensional linear Lagrangeinterpolation functions, Eq. (3.3.4) yields

fQo(1_) fQ°(?)Note that f10 + f Qo. When .1 =h0/2, we have f = f = Qo/2 as expected.

Remark 6. There are three sources of error that may contribute to the inaccuracy of thefinite element solution of a problem:

1. Domain approximation erro which is due to the approximation of the domain.2. Computational errors, which are due to inexact evaluation of the coefficients and

or are introduced owing to the finite arithmetic in a computer.

3. Approximation erro which is due to approximation of the solution by piecewisepolynomials.

For the structure shown in Fig. 3.2.10, the geometry of the problem is exactly representedand the linear approximation is able to represent the exact solution at the nodes when a is aconstant, c = 0, and f is arbitrary [see Reddy (1986), p. 403; also, see Sec. 14.5]. Therefore,the first and third type of errors are zero. The only error that can be introduced into thefinal numerical results is possibly due to the computer evaluation of the coefficients Kand je and the solution of algebraic equations. However, in general, computational as well

strain, s = elongation/original length

stress, = Young’s modulusx strain

load, P = stress x area of cross section

A0E0 r i 1] Iul foolIi — Ii I

h0[_l ]fuet1Fe12J j, 2)

heat flux, q = conductivity x (—temperature gradient)

heat, Q = heat flux x area of cross section

as approximation errors exist even in one-dimensional problems. Additional discussion ofthe errors in the finite element approximation can be found in Oden and Carey (1983) andReddy (1986); also, see Sec. 14.5.

Remark 7. The approach used in matrix methods of structural analysis to solve the problemin Fig. 3.2.10 is not much different than that presented here. The difference lies only inthe derivation of the element equations (3.2.39) for the case c0 = 0. In matrix methods ofstructural analysis, the element equations are obtained directly from the definitions of stressand strain and their relationship. For example, consider the free-body diagram of a barelement [see Fig. 3.2.3(b)]. From a course on mechanics of deformable solids, we have

(3.3.2)

(3.3.3)

(3.3.4)

The strain defined above is the average (or engineering) strain. Mathematically, strain forone-dimensional problems is defined as s = du/dx, u being displacement, which includesrigid body motion as well as elongation of the bar. Hence, the compressive force at the leftend of the bar element is

e AED0_A e_e_ I 2_ 00(0

1 1eJeSI11e1..e L—‘U1 U2

0 0

where E0 is Young’s modulus of the bar element. Similarly, the force at the right end is

pe_4eE’e(UeUe)

In matrix form, these relations can be expressed as

(3.3.5a)

which is the same as (3.2.39) obtained for a linear finite element with a = A0E0,Ce = 0, and= Q + f. Note that in deriving the element equations, we have ued knowledge-of the

mechanics of materials and the assumption that the strain is constant (or the displacement islinear) over the length of the element. If a higher-order representation of the displacement isrequired, we cannot write the force-displacement relations (3.3.Sa) directly, i.e., the elemeritequations of a quadratic finite elements cannot be derived using the arguments presentedabove.

A similar approach can be used to develop the relations between the temperatures andheats at the ends of an insulated fin. From a course on basic heat transfer, we have

temperature gradient = differencin temperatute/length

Then, if the temperature is assumed to vary linearly between the ends of the fin and thereis no internal heat generation, the heat input at the left end of the fin is

QAeqAekeTlT2.5!0(T1e72e)

144 AN OtR0UUCflONTO t EEINTtEELES’IENT MEO100CHAPtER 3: SECOND-ORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: FINITE ELEMENT MODELS 145

/ du\— I 0 or Qo( dx)Ix \ dx

N(s)

Q + Qo

,‘L1 dvdu \0=1 la———vqjdx-—v(L)Qo

j \ dxdx j

where k is thermal conductivity of the fin. Similarly, the heat input at the right end is

QAeqAeke2‘‘

In matrix form, we have

A€k [1 I = { } (3.3.5b)

which are the same as those in (3.3.5a) with E replaced by k, TC by u, and Ffby Q [and reverse the arrow at node 2 in Fig. 3.2.3(b)].

Equations of the type (3.3.5a) can also be derived for discrete systems (as opposed to acontinuum) consisting of spring elements, pipe-flow elements, electhcal resistor elements,and so on, These elements will be discussed in Chapter 4 using physical principles. However,such a direct approach cannot be used wheii the element data is not constant or when higher-order approximation of the dependent unknowns is used.

Remark 8. Another interpretation of (3.2.39) force = 0 can be given in terms of the finitedifference approximation. The axial force at any point x is given by P(x) = EA du/dx.Using the forward difference approximation, we approximate the derivative du /dx and write

—P P(x)I, = EeAe[u(xe+i) — u(x)]/h (3.3.6a)

P P(X)Ix,.L = E&[t4(x+l) — u(x)]!h (3.3.6b)

which are the same as (3.3.5a), with u = u(x) and u = u(x+i). Note that no explicitapproximation of u(x) itself is assumed in writing (3.3.6a) and (3.3.6b), but the fact thatwe used values of the function from two consecutive points to define its slope implies thatwe assumed a linear approximation of the function. Thus, to compute the value of u at apoint other than the nodes (or mesh points), linear interpolation must be used.

Remark 9. For the model problem considered, the element matrices [K9 in (3.2.31b) aresymmetric: K = KJ. This enables one to compute K (I = 1, 2,..., n) for j i only.In other words, we need compute only the diagonal terms and the upper or lower diagonalterms. Because of the symmetly of the element matrices, the assembled global matrix willalso be synimethc. Thus, we need to store only the upper triangle, including the diagonal, ofthe assembled matrix in a finite element computer program. Another property characteristicof the finite element method is the sparseness of the assembled matrix. Since Kjj = 0, ifglobal nodes I and J do not belong to the same finite element, the global coefficient matrixis banded i.e., all coefficients beyond a certain distance from the diagonal are zero. Themaximum of the distances between the diagonal element, including the latter, of a rowand the last nonzero coefficient in that row is called the ha(f-bandwidth. When a matrixis banded and symmetric, we need to store only entries in the upper or lower band ofthe matrix. Equation solvers written for the solution of banded symmetric equations areavailable for use in such cases (see Chapter 7 for additional discussion). The symmetry ofthe coefficient matrix depends on the type of the differential equation, its variational form,and the numbering of the finite element equations. The sparseness of the matrix is a result ofthe finite element interpolation functions, which have nonzero values only over an elementof the domain (i.e., so-called “compactness” property of the approximation functions).

Remark 10. The balance (or “equilibrium”) of the secondary variables Q at the interelement boundaries is expressed by (3.2.43). This amounts to imposing the condition that thesecondary variable adu/dx at the node, where a is the actual solution, be continuous. However, this does not imply continuity of adu/dx, where u is the finite element solution.Thus, we have

or Qo (3.3.7)but

(3.3.8)

In most books on the finite element method, this point is not made clear to the reader.These books consider the quadratic form or weak form of the total problem and omit thesum of the interelement contributions (for linear elements),

(3.3.9)

in the quadratic form (or functional) of the problem. However, this amounts to imposingequilibrium conditions of the form (3.3.8). When the secondary variable is specified to benonzero (say, Qo) at an interelement boundary (say, at global node 2), we have

In other books, Q is included in the functional as Q0U2,where U2 is the value of u atglobal node 2.

To fully understand the difference between the direct use of the global statement versus using the assembly of element equations, consider the model problem described byEqs. (3.2.1) and (3.2.2) with c = 0. The variational form of these equations over the entiredomain is given by

(3.3.10)

When u is approximated by functions that are defined only on a local interval (which is thecase in the finite element method), use of the above variational form implies the omissionof the sum of the interelement contributions of (3.3.9).

Consider a mesh of three linear elements. Since (e = 1, 2, 3) is zero in any elementfore f, the (global) finite element solution for the entire domain is given by

uh(x) = U1,(x) (3.3.11)e=I i=l 1=1

where c11(x) (I = 1,2,3,4) are the piecewise-continuous global interpolation functions(see Fig. 3.2.9),

4’’(x) forxj.1 <x <x1

(1) —

— (3.3.12)(x) forxj<x<xj+i

146 A) EODUCON TO TEE EINrI’E ELEMENENEOIOD CHAPTER 3: SECONDORDER DIFFERENTIAL EQUATiONS IN ONE DIMENSION: TEE EMEN’r MODELS 147

Substituting (3.311) for u and v = 4)j into (3.3.10), we obtain

01L

[a! (UJ!) _rq] dx—41(L)Q0

Since I is nonzero only between Xj_j and XJ+I, the integral becomes

(3.3.13)

d41 dcP11 1,i+i r d41 d1

x1 L dx ( dx dx dx j j0=! Ia— Uj1___+U1_+Ui+j I—PjqI dx_411(L)Qd

and we havef I d / d1 d42\

1=1:0= I la—i Ui—+U2— dx—j(L)QoJx=oL dx \ dx dx)

JX3I d42 / d1 d42 d43

1=2:0=1=oL dx \ dx dx dx

—2(L)Q0

j’x4=L r d43 / d2 d3 d441=3:0=1 Ia_(U2_+Us_+U4_)_3q] dx

J2 L dx dx dx dx

—3(L)Q0

I d / d3 d4)- 4q] dx —1=4:0= I la—lU3—+U4—

J3 L dx \ dx dx

(3.3.14)

(3.3.15)

These equations, upon performing the integration, yield (3.2.50), with the last column(containing Qs) in the latter replaced by

0Qo00

(3.3.16)

Although this procedure, known as the direct snfftess method in structural mechanics,gives the assembled equations directly, it is algebraically complicated (especially for two-dimensional problems) and not amenable to simple computer implementation.

3.4 AXISYMMETRIC PROBLEMS

3.4.1 Model Equation

The equations governing physical processes in a cylindrical geometry are described analytically in terms of cylindrical coordinates (see Fig. 3.4.1; also see Fig. 1.4.5). When thegeometry, loading, and boundary conditions are independent of the circumferential direction(i.e., 0-coordinate in Fig. 3.4.1), the problem is said to be axisymmetric and the governingequations become two-dimensional in terms of r and z. The equations are functions of only

Figure 3.4.1 Volume element and computational domain of an axisymmetric problem.

the radial coordinate r if the problem geometry and data are independent of z. Here, weconsider a model second-order equation in a single variable and formulate its finite elementmodel.

Consider the differential equation [an analogue of (3.2.1)]

ldI dul———la(r)—I=f(r) for R<r<R0

rdrL drj(3.4.1)

where r is the radial coordinate, a and f are known functions of r, and u is the dependentvariable. Such equations arise, for example, in connection with radial heat flow in a longcircular cylinder of inner radius R and outer radius R0. The radially symmetric conditionsrequire that both a = kr (k is the conductivity) andf (internal heat generation) be functionsof only r. Since the cylinder is long, the temperature distribution at any section alongits length (except perhaps at the ends) is the same, and it is sufficient to consider any crosssection away from the ends, i.e., the problem is reduced from a three-dimensional problemto a two-dimensional one, Since a and f are independent of the circumferential direction 0,the temperature distribution along any radial line is the same, reducing the t,’o-dimensionalproblem to a one-dimensional one, as described by (3.4.1).

3.4.2 Weak Form

In developing the weak form of (3.4.1), we multiply it with a weight function w(r) andintegrate over the volume of the cylinder of unit length (see Fig. 3.4.1)

c I 1d/du 10= I w{--— la—I -fldvJv L rdr \ drj j

JIf27rb

[_ (a) _f]rdrdodzo o r, rdr dr

ç r ld / du’\=2 / WI--—Ia—l-f rdr

Jr ,, rdr dr;

R0

R

— r

(3.4.2)

148 AL UbTRODUCOON TO THE PIMPlE ELEMENT METHOD CHAPTER): SECOND-ORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: FINITE ELEMENT MODELS 149

where (r0, Tb) is the domain of a typical element along the radial direction, Next, we carryout the remaining two steps of the weak formulation:

O=21rf(a_rwf)dr_2Jr[wa]

r, drdr dr

o = 22r f (a— rwf) dr

— w(ra)Q — w(rb)Q (3.4.3a)

where/ du / du

Q_2Jr(a_) QER2Yr(a_) (3.4.3b)dr dr

3,4,3 Finite Element Model

The finite element model is obtained by substituting the approximation

u(r) ui(r) (3,4.4)j=l

and w = ,, into (3.4.3a). The finite-element model is given by

[K9{u9 = {fe} + {Q} (34.5a)

wherer

Kfj=2Jrf a--L-i-dr, fe=21rf frdr (3.4.5b)

and are the interpolation functions expressed in terms of the radial coordinate r, Forexample, the linear interpolation functions are of the form (h = Tb — ra) [see Eq. (3.2.17)]

(r)=—, (r)=--- (346)

The explicit forms of the coefficients K1 and J for a = ar and f f are given below(r, denotes the global coordinate of node 1 of the element).

Linear Element

[K9 = Ef.(r + he) [1 1] = 2irfh {3ra+he ) (3.4.7)

Quadratic Element

2 3he+l4Ta (4he+l6Ta) he+2Ta—(4he+l6ra) l6he+32ra —(l2h+l6r5)

6h h+2r. (l2he+l6ra) llhe+l4Ta

22Tfh4ra + The (3.4.8)

Ta + h

•1-

IElemet1 (rat

‘ ‘-

[K21=[3 _3’j{f2) 4xlO(ool2s)b {JElement 3 ‘a =2h and PE=3h)

[K]$[5 j 4106°’25f71Element 4 (e. =-3h andr =4k)

The assemi

2 2O

2O8

-:. 1&.

-. i-240 O—149

- :‘

150 AN NraooucrloNTo mE ELEMENT MEOD CRAI’fER I: SECONDOP.DERUIEFERENTtAL EQUAI1ONS IN ONE DIMENSION: F1NrrE ELEMENT MODELS 151

3.5 SUMMARY

A systematic study of the steps involved in the finite element formulation of a model second-order differential equation in single variable is presented. The basic steps of the formulationand analysis of a typical equation are outlined in Table 3.1.1. The model equation is representative of the equations arising in various fields of engineering (see Table 3.2.1). Coupleof numerical examples are presented to illustrate the steps in the finite element analysis ofsecond-order differential equations. Additional examples, especially those arising in heattransfer, fluid mechanics, and solid mechanics will be presented in Chapter 4.

The finite element model is developed following three steps:

1. Weak (or variational) formulation of the differential equation over an element.

2. Finite element interpolation of the primary variables of the weak formulation.

3. Finite element model (i.e., a set of algebraic equations relating the “primary” and “secondaxy” variables) development over a typical element.

The weak formulation itself involves a three-step procedure, which enables identificationof primary and secondary variables. The primary variables are required to be continuousthroughout the domain, including the nodes at which elements are connected. The secondaryvariables are included in the weak form. The finite element interpolation functions have beendeveloped systematically on the basis of continuity, completeness, and linear independence.The finite element model has been developed by substituting appropriate interpolation ofthe primary variable into the weak form of the differential equation.

PROBLEMSFor Problems 3.1—3.4, carry out the following tasks:

(a) Develop the weakforms of the given differential equation(s) over atypical finite element, whichis a geometric subdomain located between x x and x = Xb. Note that there are no “specified”boundary conditions at the element level. Therefore, in going from Step 2 to Step 3 of the weak-form development, one must identify the secondary variable(s) at the two ends of the domainby some symbols (like Q and Q for the first problem) and complete the weak form.

(b) Assume an approximation(s) of the form

u(x)=ui/4(x) (1)

where u is a primary variable of the formulation, fr(x) are the interpolation functions, andare the values of the primary variable(s) at the jth node of the element. Substitute the expressionin (1) for the primary variable and 1t? for the weight function into the weak form(s) and derivethe finite element model. Be sure to define all coefficients of the model in terms of the problemdata and

3.1 Develop the weak form and the finite element model of the following differential equation overan element:

d I du’ d2 Id2u\———la—l+—lb—i+cu=f for x0<x<xb

dx \ dx) dx2 \ dx2)

where a, b, c, and f are known functions of position x. Ensure that the element coefficientmatrix [K’] is symmetric. What is the nature of the interpolation functions for the problem?

3.2 Construct the weak form and the finite element model of the differential equation

dfdu’\ du———(a——l—b-—=f for O<x<Ldx\dx) dx

over a typical element t2, = (xa, xa). Here a, b, and! are known functions of x, and u is thedependent variable. The natural boundary condition should not involve the function b(x). Whattype of interpolation functions may be used for u?

3.3 Develop the weak forms of the following pair of coupled second-order differential equationsover a typical element (x8, xb):

-- [ax (a

+= f(x) (Ia)

__ (ix) + a(x) (+

= q(x) (lb)

where u and v are the dependent variables, and a, b, f, and q are known functions of x. Alsoidentify the primary and secondary variables of the formulation.

3.4 Consider the following weak forms of a pair of coupled differential equations:

o=f(1_wIf)dx_Pawl(xa)_Pbwl(xb) (la), dxdx

ofXb

± c w2v - w2q dx — Qaw2xa) —

di di(lb)

152 AN INTRODUCTION TO ThE FNTTE ELEMENT METHOD CHAF1R 3 5ECOND-ORDERDIFFERENTIALEQUATIONS IN ONE DIMENSION FINITE ELEMENT MODELS 153

wher c(x) is a known function, w1 and w2 are weight functions, u and v are dependent variables For the minimum number of linear elements, give (a) the boundary conditions on the nodal

(prunasy variables), and P, Pb, Q, and Qb are the secondary variables of the formulation. - variables (primary as well as secondary variables) and (b) the final condensed finite element

Use the finite element approximations of the form equations for the unknowns.

u(x)=uii(x), v(x)=vq(x) (2)j=I 1=1 d=4in. d=2.5in. d=2in.

and w1 = i and w2 , and derive the finite element equations from the weak forms. Thefinite element equations should be in the form 500 kips

o = K1u +K2v —F (3a) lUiii—

m SteeLE=30xl06psi(3b) Aumnuni,E=l0xi06psi

Define the coefficients K’, K,, K, K, F, and 2 in terms of the interpotion functions,Figure P3.9

known data, and secondary variables.

3.5 Derive the Lagrange cubic interpolation functions for a four-node (one-dimensional) element(with equally spaced nodes) using the alternative procedure based on interpolation properties 3.10 Resolve the problem in Example 3.2.1 using the uniform mesh of three linear finite elements.

(3.2.18a) and (3.2J8b). Use the local coordinate f for simplicity. Answer: U2 = —0.02999, U3 = —0.04257, (Q)def =0.08998, (Q)def =0,12771.

3.6 Evaluate the element matrices [K11), [Ku], and [K’] for the linear interpolation of u(x) and 3.11 Solve the differential equation in Example 3.2.1 for the mixed boundary conditions

v(x) in Problem 3.4./du\

3.7 Evaluate the following coefficient matrices and source vector using the linear Lagrange inter- u(0) = 0,_)

=polation functions:

X xI

fXbe

dd1fr—

Usetheuniformmeshofthreelinearelements,Theexactsolutionis

i (a0+a1x)——dx, M0_ , (c0+c1x)ifri/rdx 2cos(l—x)—sinxdx dx u(x) = + x2 —2cos(1)

(f+fx)dx Answer: U=0.4l34, U3=0.7958, (14=1.1420, (Q=—l.2402.

3.12 Solve the differential equation in Example 3.2.1 for the natural (or Neumann) boundary con-where a0, a, c0, c, f, and f1 are constants. ditions

3.8 (Hear transfer in a rod) The governing differential equation and convection boundary conditionareoftheform: () =1

d20\dx ‘ dx)

— + c8 = 0, 0 <X (l) Use the uniform mesh of three linear finite elements to solve the problem. Venfy your solution

r d9 with the analytical solution9(0)=T0—T, k—+8 =0 (2)

L dx s.L , cos(1—x)+2cosx 2+x —2where 8 = T — T, c = $P/(Ak), is the heat transfer coefficient, P is the perimeter, A is sin(1)

the areaof crosssection, andk is theconductivity. Forameshof twolinearelements (of equal Answer: U1 = 1.0280, (12=1.3002, U4 = 1.4447, Li5 = 1,4821.length), give (a) the boundaiy conditions on the nodal variables (primary as well as secondary 3.13 Solve the problem described by the following equationsvariables) and (b) the final condensed finite element equations for the unknowns (both primaryand secondary nodal variables). Use the following data: T0 = 120°C, T = 20°C, L = 0.25 m, d2uc=256,8=64,andk=50(withproperunits).

_1jcosrx, 0<x <1; u(0)=0, u(l)=0

3.9 (Axial deformation of a bar) The governing differential equation is of the form (E and A are Use the uniform mesh of three linear elements to solve the problem and compare against theconstant): exact solution

_- {ilA]=o 0<x.<L (1) u(x)=—(cosrx+2x—1)

1

154 AN !WrRODUCOON TO THE FINITH ELEMENT METHOD

3.14 Solve the differential equation in Problem 3.13 using the mixed boundary conditions

u(O)=O, () =0dx

Use the uniform mesh of three linear elements to solve the problem and compare against the

exact solution

u(x)=!(cosrrx1)

3.15 Solve the differential equation in Problem 3.13 using the Neumann boundary conditions

(du’—0

(du\11—0l)

Use the uniform mesh of three linear elements to solve the problem and compare against the

exact solutioncos irx

u(x)=—

Note: For Neumann boundary conditions, none of the primary dependent variables are speci

fied, and therefore the solution can be determined within an arbitrary constant for this equation

(i.e., when the cu term is not present, the coefficient matrix is singular and cannot be inverted).

In such cases, one of the U should be set equal to a constant to remove the “rigid-body” mode

(i.e., to determine the arbitrary constant in the solution).

REFERENCES FOR ADDITIONAL READING1. Crandall, S. H., EngineeringAnalysis, McGraw-Hill, New York, 1956.

2. Holman, J. R, Heat Transfer, 7th at., McGraw-Hill, New York, 1990.

3. Kreith, F. and Bobs, M. S., Principles of Neo.e Transfer, 5th ed., West Publishing Company, SL Paul, MN,1993.

4. Mikhlin, S. C., Variational Methods in Mathematical Physics, Pergamon Press, New York, 1964.

5. Mikhlin, S. G., The Numerical Performance of Variational Methods, Wolters-Noordhoff, Groningen, 1971.

6. Odes, J. t and Carey, G. F., Finite Elements. Mathematical Aspects, Volume Pv Prentice-Hall, Englewood

Cliffs, NJ, 1983.

7. Odes, J. T. and Reddy, J. N., 1&uiational Methods in Theoretical Mechanics. Springer-Verlag, New York,

1976; 2d ed., 1983.

8. Ozisik, M. N., Heat Conduction, 2nd ed., John Wiley, New York, 1993.

9. Reddy, J. N., Energy Principles and Variational Methods in Applied Mechanics, 2nd ed. John Wiley, NewYork, 2002.

10. Reddy, 1. N., .4pplied FunctionalAnalysis and Variational Methods in Engineering, McGraw-Hill, New York.

1986; Keieger, Melbourne, FL, 1991.

11. Reddy, 3. N. and Rasmussen, M. L,Advanced Engineering Analysis, John Wiley, New York, 1982; Krieger,

Melbourne, Fl, 1990.

12. Rrddy, J. N., An introduction to Nonlinear Finite Element Analysis, Oxford University Press, Oxford, UK,

2004.

13. Rektoryu, K., Variational Methods in Mathematics, Science and Engineering, Rridel, Boston, 1977.

Chapter 4

SECOND-ORDERDIFFERENTIAL EQUATIONS

IN ONE DIMENSION:APPLICATIONS

4.1 PRELIMINARY COMMENTS

In Chapter 3 we developed weak forms and finite element models of continuum problemsdescribed by a fairly general second-order differential equation. For discrete systems, suchas a network of springs or electrical circuits, no differential equations exist and the weakform concept is not applicable. Therefore, an alternate approach based on the laws ofphysics must be used to develop finite element models (i.e., relations between the causeand effect) of such systems. Physical principles can also be used to develop finite elementmodels of continuum problems (as discussed in Remark 7 of Chapter 3) but the approachcannot be used to derive finite element models with higher-order approximation of the fieldvariable.

The objective of this chapter is two-fold. First, we derive finite element models of sometypical discrete systems. Fmiui element models of discrete systems are developed usingphysical laws familiar to most engineering and applied science majors, and the approachrequires no concept of weak form. Second, we present numerical examples of application offinite element models developed for both discrete systems and continuum systems. We willconsider several examples to illustrate the steps involved in the finite element analysis of one-dimensional second-order differential equations arising in heat transfer, fluid mechanics, andsolid mechanics. The examples presented here make use of the element equations alreadydeveloped in Chapter 3. While the notation used for the dependent variables, independentcoordinates, and data of problems from field to field is different, the reader should keep thecommon mathematical structure in mind and not get confused with the change of notationfrom problem to problem and field to field.

156 AN INTRODUCTION TO THE PINTTE ELEMENT METHOD CHAPTER 4: SECOND-ORDER DIFFERENTIAL EQUATIONS iN ONE DIMENSION; APPLICATIONS 157

4.2 DISCRETE SYSTEMS

4.2.1 Linear Elastic Spring

A linear elastic spring is a discrete element (i.e., not a continuum) whose load-displacementrelationship can be expressed as

F = k8 (4.2.1)

where F is the force (N) at the right end, 5 is the displacement (m) of the right end of thespring relative to the left end in the direction of the force, and k is the constant known as thespring constant (N/m), The spring constant depends on the elastic modulus, area of crosssection, and number of turns in the coil of the spring. Often a spring is used to characterizethe elastic behavior of complex physical systems.

A relationship between the end forces (Fr, F) and end displacements (&, ) of atypical spring element [see Fig. 4.2.1(a)] can be developed as discussed in Remark 7 for abar element. The force F at node 1 is equal to the spring constant multiplied by the relativedisplacement of node 1 with respect to node 2, l —

Fk(l kel

Similarly, the force at node 2 is equal to

Fk4(& 5eJ3e

Note that the force equilibrium, F + F = 0, is automatically satisfied. The above equationscan be written in matrix form as

1 —1 l

—l 1 = F(4.2.2)

Equation (4.2.2) is applicable to any spring element whose force-displacement relation islinear. Thus a typical spring in a network of springs of different spring constants obeyEq. (4.2.2).

The quthbnum condition (42 3b) suggests that we must add the second equation of element Ithe first equation of element 2, and the first equation of element 3 together to replace the sum -

fth_ree unknowns forces F’} .-F21 .1- F3t with the knowii.force F2. Thus, we have fdur-- enhlationc She firct ernl.hI-TAE nf element 1 the nim af the three eniitnn t,tel ,F,nve the

-

— Globali t14

J ç - -

i—’ * ‘ DIEiiISULhIUU

*.**

F —

2

-: A A A i:*;;

*

(b)

Fsgur4.2 1 (a) A spring finite element (b) Iluec spring assemblage

- -Thethreeelementsareconnectedatnode- - - -. at- andequffibriumnditionsatnode-2tequire - - -. - - - - - -

-. - -

-

= 21_ = - -- - - --

- - - + F -F3 -

158 AN INTRODUCTION TO THE FINIYE ELEMENT METHOD CHAPTER 4: SECOND-ORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: APPLICATIONS 159

4.2.2 Torsion of Circular Shafts

Another problem that can be directly formulated as a discrete element is the torsion ofcircular shafts. From a course on mechanics of deformable solids, the angle of twist 9 of aconstant cross-section circular cylindrical member is related to the torque T (about the axisof the member) by [see Fig. 4.2.2(a)]

TO (4,2.8)

where J is the polar moment of area, L is the length, and G is the shear modulus of thematerial of the shaft. The above equation can be used to write a relationship between theend torques (T, T) and the end twists (9, 9) of a circular cylindrical member of lengthhe [see Fig. 4.2.2(b)]:

4.2.3 Electrical Resistor Circuits

There is a direct analogy between a network of mechanical springs and a direct currentelectric resistor network. Ohm’s law provides the relationship between flow of electriccurrent I (amperes) through an ideal resistor and voltage drop V (volts) between the endsof the resistor

V=IR (4.2.10)

where R denotes the electric resistance (ohms) of the wire.Kirchhoff’s voltage rule states tha(the algebraic sum of the voltage changes in any ioop

must be equal to zero. Applied to a single resistor, the rule gives [see Fig. 4.2.3(a)]

IR+V—V=0, Re+VV20

Re

(a)Global node numbers

R=lO2 /R=sV=200V= 200 V

Element numbersR = 5 Element node

numbers

27 0 VV6 = DVR=152 R=20t2

(b) (c)

Figure 4.2.3 (a) Direct current electhc element (current flows from high to low voltage). (b) A resistorcircuit. (c) Finite element mesh of the resistor cIrcuit.

(a) (b)

or

Figure 4.2.2 Torsion of a circular shaft.

GeJeE 1 _l11911TIei(4.2.9)

[-l l]l9JlTf

R=512’

A 5 TVV

R=152 R=20G

160 AN Uefl6ODUCI1ON TO THE FINITE ELEMENT METHOD CHAPTER 4: SECOND-ORDER DiFFERB4flAL EQUAtiONS IN ONE DIMENSION: APPUCATIONS 161

Example 4.2.2 -Consider the resistor circuit shown in Fig 42 3(b) We wish to determine the currents in theIoo and voltage-at th node& The element numbering,element node numbering, aiid globalnode numbering are indicated in Fig. 4.2.3(c) The element node humbering is importaiitin -assembling, the element.èquations From the element notation indicated in Fig. 4.2.3(a), it is

• clear that current flows fton node I’ to node 2 of the element. The element ndde numbering in -Fig. 4.2.3(c) indicates the assumed direction of the currents. - - -‘ - - . - -

The assembled coefficient matrix is given by -1 :,-2. 3 4 5 6

0 0 0 0

K +K — K K 0 0 2

-. .- - ‘• - - - K1 +4 - - ... - 0 - . 0 3

- .. - ‘- - 4tK1. 4 0 4

syrnm. - -. K, + K1 + K1 4- . .- 6

32

0.2 - —0.2 - - - - 0 , - -. 0 - - 0 -- - - O.2+O.l+O.1 - .-0J - 0. - -0.1 .. 0 -

01-02 —02 0 0- - . . - - 0.2+0.0667 —0.0667 - .0

symm. -- 0.0667+0.1±0.05 —0.05--. : . - :...- -

(4.2.12)

The boundar conditions are V1 200 V and V6 = 0 V. The condensed equations for thenodalvoltages are obtained by omitting the first row andlastrow of the system, and then moving -

the-terms involving and V6 to theright.Weobtain.

0401 0.*l•-‘.“•

O.2V1.,.

- —0.1 . .0.3 —0.2 0.0 V3 — 0 .0.0. —0.2 0.2667 —0.0667 . 0 -

—01 00 —00667 02167 V5 005l’

The solution of these equations is given by (obtained with the aid of aconsputer)

2=16923V V3=15385V V4=146l5V V5=12308V

The condensed equations for the unknown currents at nodes I and 6 can be calculated from -equations I and 6 of the system. We have

- ,j) =O.2V —O.2V2=40—33.846=6.154A . .—005V5+005V6=—6 154A

The negative sign on j6) indIcates that the current i flowing out of global node- 6.’The currents through each element can be calculated using the element eqations (42.11)- For example, the nodal currents in resistor 5 are given by: - . .- . .

-. : . -_[0.1 - -0.1 123.08 - -4.615 -

i5) — [-..o.t 0.1 169.23 —. 4.615

.which indicates that the net current flow in resistorS is from its node 2 to.node I(or globalnode 2 to global node 5), and its value is 4.615 amps (which is the same as 6.154—1.539).The finite element solutions for the voltages and curreotsare shown in Figure 4.2.4.

- V3= 153.85 V2= 169.23

R=lO3 R’-5f2

1:::...:R=15t2 R=20fl

V4146.l5 Vl23.O8 . . ..-Figure 4.2.4 The solution for currents and voltages obtained with the finite element method.

4.2.4 Fluid Flow through Pipes

Another example of a discrete element is provided by steady, fully deeloped, flows ofviscous incompressible fluids through circular pipes. The velocity offully developed laminar

or in matrix form

i i 1 VC je

Re —l 1 V =(4.2.11)

Thins, once again we have the same form of relationship between voltages and currents asin the case of springs. The quantity i/Re is known as the electrical conductance. -

The assembly of resistor equations is based on the following rules:

1. Voltage is single-valued.

2. Kirchhoff current rule: The sum of all currents entering a node is equal to zero.

162 1N110DUC11ON 10 ThE FIrEEISMENTMEnI0D CHAFfER 4: sECOND-ORDER DtFFERENITAL EQUAI1ONS IN ONE DIMENSION: APPUCATIONS 163

iBf BT\ IT——lkr— l+g=pc—rlr\ Br) It

_____________

vr) R =zde

pe

(a) (b)

Figure 4.2.5 Flow of viscous fluids through pipes.

flow of viscous fluids through circular pipes is given by

v=_±()2

(4.2.13a)4pdx d

where dP/dx is the pressure gradient, d is the diameter of the pipe, and jis the viscosity of

the fluid [see Fig. 4.2.5(a)]. The volume rate of flow, Q, is obtained by integrating v5 over

the pipe cross section. Thus, the relationship between Q and the pressure gradient dP/dx

is given by the equation

rd4 dP(4.2.13b)

128dx

The negative sign indicates that the flow is in the direction of negative pressure gradient.Equation (4.2.13b) can be used to develop a relationship between the nodal values of

the volume rate of flow, (Q, Q) and the pressure, (Pr, Pt), of a pipe element of length

he and diameter de. The volume rate of flow entering node 1 is given by [see Fig. 4.2.5(b)]

ird4Q1128l.the(P2

— P)

Similarly, the volume rate of flow entering node 2 is

Q2l2Sh(PiP2)

Thus, we have

yrd4 1 —I = (4.2.14)l28h —1 1 P Q

The constant, R = 128,ih/rd is called the pipe resistance, in analogy with the electricalresistance [see Eq. (4.2.11)].

4.3 HEAT TRANSFER

4.3.1 Governing Equations

The equations governing conduction heat transfer were discussed in Example 1.2.2. Herewe briefly-review the pertinent equations for our use.

The Fourier heat conduction law for one-dimensional systems states that the heat flowq(x) is relate the temperature gradient IT/Ix by the relation (with heat flow in the positivedirection of x)

q = _kAL (4.3.1)Ix

where k is the thermal conductivity of the material, A the cross-sectional area, and Tthe temperature. The negative sign in (4.3.1) indicates that heat flows downhill on thetemperature scale. The balance of energy requires that

_ (--) + Ag = pcA (4.3.2)

where g is the heat energy generated per unit volume, p is the density, c is the specificheat of the material, and t is time, Equation (4.3.2) governs the transient heat conductionin a slab or fin (i.e., a one-dimensional system) when the heat flow in the normal to thex-direction is zero. For a plane wall, we take A = 1.

In the case of radially symmetric problems with cylindrical geometries, (4.3.2) takes adifferent form. Consider a long cylinder of inner radius R1, outer radius R0, and length L.When L is very large compared with the diameter, it is assumed that heat flows in the radialdirection r. The transient radially symmetric heat flow in a cylinder is governed by

(4.3.3)

A cylindrical fuel element of a nuclear reactor, a current-carrying electrical wire, and athick-walled circular tube provide examples of one-dimensional radial systems.

The boundary conditions for heat conduction involve specifying either the temperatureT or the heat flow Q at a point:

T = T0 or Q —kA- = Qo (4.3.4)Ix

It is kisown that when a heated surface is exposed to a cooling medium, such as air or liquid, the surface will cool faster. We say that the heat is convected away. The convection heattransfer between the surface and the medium in contact is given by Newton ‘slow ofcooling:

Q=BA(T3—T) (4.3.5)

where T5 is the surface temperature, T is the temperature of the surrounding medium,called the ambient temperature; and ,6 is the convection heat transfer coefficient or filmconductance (or film coefficient). The heat flow due to conduction and convection at aboundary point must be in balance with the applied flow Qo:

±kA+A(T-T)+Q0=O (4.3.6)Ix

The sign of the first term in (4.3.6) is negative when the heat flow is from the fluid at T tothe surface at the left end of the element, and it is positive when the heat flow is from thefluid at T to the surface at the right end.

164 AN D,mrODUCflON TO THE FINITE ELEMEtt MEtHOD CHAPTER 4: SECOND-ORDER DEFERENTIAL EQUATiONS IN ONE DIMENSION: ucnoo 165

Convection of heat from a surface to the surrounding fluid can be increased by attaching I

heat flow along its length, heat can convect across the lateral surface of the fin [see Fig. I ..•-

- eapesul to ambient temperature, 7’,,

4.3.1(a)1. To account for the convection of heat throuch the surface, we must add the rate ./ -

of heat loss by convection to the right-hand side of (4.3.2): - - -

thin strips of conducting metal to the surface. The metal strips are calledfins. For a fin with I Lateral surface and right end are

where P is the peeter and is the film coefficient. Equation (4.3.7a) can be expressed I. tunilars

1/ dT a—

(Ak— ) + Aq =pcA + P(T - Tm) (4.3.7a)IX\ IXJ

in the alternative form Body from which heat is to be extracted

pcA’ — -- (_‘_‘) + P,3T = Ag + PfiTm (4.3.7b) I (a)

It dx\ 8x1

The units of various quantities (in metric system) are as follows: 4:T °C (celsius) k W/(m . I of the wall

g W/m3 p kglm3

[“/([/.$ Crosssection - -- -. -

equal to zero. The steady-state equations for various one-dimensional systems are summa

c J/(kg°C) $ W/(m2•°C)

For a steady state, we set the time derivatives in (4.3.2), (4.3.3), (4.3.7a), and (4.3.7b) IFurnace

(b)rized below [see Fig. 4.3.1(b) and (c); see Eqs. (1.2.14) and (1.2.17)].

Plane Wall [Q=k(dT/dx)1

d / dT\(4.3.8)

Fin [Q=kA(dT/dx)J

__(kAi)+cT=Ag+cTm c=P (4.3.9) (c)dxk dx

Figure 4.3.1 Heat transfer in (a) fins, (b) plane wall, and (c) radially symmetric system.Cylindrical System [Q = k(dT/dr)1

I d / dTlkr—

r dr dr)= g(r) (4.3.10) Here

Xb / dre di/i )The essential and natural boundary conditions associated with these equations are of= j I kALL + P,8tfri/i. dx, J =

i/t°(Ag + PflTm) dx

the form X, \. dx dx Jx,

=1 —kA— , = I kA— II (4.3.llb)TT0, Q±$A(TT)+Q00/ dT) Qe / dT

\ dx \ dx,Ij4

Equations (4.3.8)—(4.3.10) are a special case of the model equation (3.2.1) discussed in

Section 3.2, with a = kA, c = P, and f -÷ Ag + P$Tm. We immediately have the finite where Q and Q denote heat flow into the element at the nodes.

element model of Eqs. (4.3.8) and (4.3.9) from (3.2.3la) and (3.2.31b): Equation (4.3.10) is also a special case of the model boundary value problem. However,in developing the weak forms of (4.3.10), integration must be carried over a typical volume

[r]{Tt}= (fC) + (Q9 (4.3.1 la) element of each system, as discussed in Section 3.4 [see Eq. (3.4.2)].

166 AN INTRODUCCION To THE FINTrE ELEMENT METHOD CHAPTER 4: SECOND-ORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: APPUCATTONS 167

Material 2, k2

Material I, k1 Material 3, k3

= --- R, = ---- R3 —- qk3A

T1 T, T5 1’4

Figure 4.3.2 One-dimensional heat transfer through composite walls and their thermal circuits.

4.3.2 Finite Element Models

It is interesting to note the analogy between Eq. (4.2.11) of an electric resistor and Eq.

(3.3.5b) of one-dimensional heat transfer (see Remark 7 of Section 3,3):

Aeke[1 ]{:;}={g (4.3.12)

If we identify thermal resistance R by

(4.3.13)

Equations (4.2.11) and (4.3.12) are the same with the following correspondence:

R R, 1=‘ Q, (4.3.14)

This allows us to model complicated problems involving both series and parallel thermalresistances. Typical problems and their electrical analogies are shown in Figure 4,3.2.

4.3.3 Numerical Examples

TI

A compoawal1on

1O:-

--. - a

168 AN INTRODUCTION TO THE FINrrE ELEMENT METHOD CHAPTER 4: SECOND-ORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: APPLICATIONS 169

A;’A -

A’A

“h’

A

The- V

A ‘ ‘ ‘itcanbeownthand beats oincidesith the* ‘ -

-

1()= B1tBx.c1+c, 4Pi+hE-<

-

V ‘V “V A A wwhere ‘ - ,

: V

- -• T—To”

V

BAl

‘A;A;’ .A: “‘A;’ “‘ViA

Rectangular fins are used to remove heat from a heated surface ee 4 (d)j the fisare exposed ambient air at T. Thehet sfer doeffic SS9 wfin aanctTtheairisflAAumingthatJIeatiscondUctedalOI1gtheieItgth otthefiwifithand theknefronsevsitand heat loss per fiuifor twp differeutset

The goveniin Itt lqat on ausUxVAU ‘ ;“f ,A’

V -

‘AA’A’V Vi*

- SetJ Tçoj; - -

- 4 1‘A -

Set2 T(Ql Pi1

V - - -‘

V

Examp 432

- 1111- I Io

-V ‘L:-;:

‘A -

A A A

‘VA;“ -V

-‘ ‘ ‘A. V ‘ VV AA

Se I Boundary ConditionV

-, Theb idar andbahmus ond ant ar

V ‘‘U7 Q-+O,Q QUV A

, -, ‘‘VAV , ‘

_______

A

V: ‘

V

V A AV A’- atP4t’ AA

A’’ AA

VV ‘ ‘

V A’

Fir4’34 ieelnentriIesho(atectanguldrfi’- ‘-A V

• - - - —

170 AN NTRODUCflON TO ThE FNrrE ELEMENT MEtHOD CHAPTER 4: SECOND-ORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: APPUCATIONS 171

172 ANINrRODUCTION TO THE F ifE ELEMENT METHODCHAPTER 4: SECOND-ORDER DIFFERENTIAL EQUAtIONS IN ONE DIMENSION: APPLICATIONS 173

Thus, the heat loss from the end of the fin is o th — Lat the ambient temperature. The total heat loss

The exact solution of s, 3.1 (4.3.

-

- Q(O) = —kA =M

Evaluating the exact solution at the nodes, we o

f(O.025) = 74C, T(O.05) = 53.

àndQ=2.268W. -

The next example is concerned with heat transfer in a rod and comparison of finitedifference and finite clement analysis steps.

174 INTRODUCTiON TOTHE FINITE ELEMENT METhOD

*!I! : . • .

\ - I ‘

;prit c’4:, .

,

, .

Wt H

: :• •

bf 0 frrb‘ ‘

,

.

4m2*i,

... •

,

. .

7 . . ; . :QUq1 V

$:p bV*

.V

VV

VVVIb,i:TV

•V

t;4wV •

V3

V•

VV

;I:fia%rrV ‘ V

g • V V

V

r Pii ‘

V •

1i1p: (i? Vi

V

V

V%

V

VVV VThe tätte is i&d Qii dhht$ a(429b we set&

V & V V hio

CIIAFItE 4: SECOND-ORDER DhlWRENflALEQUATJONS N ONE DIMENSION: APPLICATIONS 175

Un

-

±±

f--

—

x-

I

______

H

II

Q;r.I-

:P

-ç

r

Un

L—4—

-

-.

4_

4)

4)

4)

4-)-

—p

- _

I•

Un

-C

-)

I

-—

-

,

II

-H

S

4)--

4)

=-

Un

-•

-

rj

-

JL

--

-----

-

4)

--

0ii- -

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4)

4)4

4

=4)

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0044

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0-Sii

-

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-

178 AN INTRODUCTION TO TUE FINITE ELEMENT METhOD CHAPTER 4: SECOND-ORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: APPLICATIONS 179

The finite element solution becomes -

7,(r)=

[fa+U2(+Th)(l._1)+To(2-_l)

— j (5—3_)+m. tbr0<r<R0

7goR r_T_(1)+TU. forR0<r<R0 --

The exact solution of the problem is

Tlr)= 4k

— ()2

+ To(C) - 4.3.35i

q(r) = !gxr W/m2, Q(R0.= 27rkr) —Irg0R (W f4.3.362 dr

IR -

The temperature at the center of the cylinder according to the exact solution is T(0) =

gaRg/4k + T0, whereas it is goR/3k ± T0 and 5grR/l8k + 7’4 according to the one- andtwo-element models, respectively.

The finite element solutions obtained using one-, two-, foui, and eight-element meshesof linear elements are compared with the exact solution in Table 4 3.2. Convergence of thefinite element solutions, = (T — Ts)k/go R, to the exact solution with an increasing numberof elements is clear (see Fig. 4.3.6). Figure 4.3.7 shows plots of Q1ñ = Qñ/2orR go andQ (r) = 2rrkrdT/dr, versus F = r/Ro, as computed in the finite element analysis and the exactsolution.

Table 4.32 Comparison of the finite element and. exact solutions for heat transferin a radially symmetric cylinder R0=0.0I m, g=2 x 108 W/m3, k=20 WI(m.°C), T0 100°C.

- Temperature T(r)t

One To Four Eightr/ R0 element elements elements e1emeels Exact

0.000 433.33 377.78 358.73 352.63 350.00-0.125 391.67 356.23 348.31 347.42 346.090.250 35000 335.11 337.90 335.27 334380.375 308.33 315.28 313.59 315.48 334.840.500 266.67 29444 289.29 287,95 287.51)0.625 225.00 245.83 24970 252.65 252.340.750 183.33 197.22 210.12 209.56 209.380.875 - 141.67 148.61 155.06 158.68 158.591.000 100.00 100.00 100.00 100.00 100.00

I The wide lined teresa are nodal a1ueo and others are interpolated ‘-alum

1.0 uooicno TOThNLEMENTMEflIOD CHAPTER 4: SECOND-ORDER DIEFERENTIAL EQUATIONS IN ONE DIMENSION: APPLICATIONS ill

aP—=0ay

-

5...

— 0.09 jIiIIIIIiI1ILliIIIIIjIliIIiIIIIIIIII —

0.0- 0.2 0.4 0.6 0.5 -

-

Coordinate, F -_ -.-.-

Figure 4.3.6 Comparison of the finite element solutions with the exact solution fcin a radially symmetric problem with cylindrical geometrY. - -.

000—urliII,IFItIImlI h1 —

--

-

- —

Analytical- - - 0 telements

.— - 4elements—0.40— a 2elenients -

- 1element

— —0.50— 1111111 III I 11111 I ii

0.0 0.2 0.4 0.6 0.1-

- Coordinate, r

Figure 4.3.7 Comparison.of the finite element solution with the exact solutiature gradient in a radially symmetric problem with cylindrical i

4.4 FLUID MECHAMCS

4.4.1 Governing Equations

All built matter in nature exists in one of two forms: solid or fluid. A solid body is characterized by the relative immobility of its molecules whereas a fluid state is characterizedby relative mobility of its molecules. Fluids can exist either as gases or liquids. The fieldof fluid mechanics is concerned with the motion of fluids and the conditions affecting themotion (see Reddy and Gartling, 2001).

The basic equations of fluid mechanics are derived from the global laws of conservationof mass, momentum, and energy. Conservation of mass gives the continuity equation, whilethe conservation of momentum results in the equations of motion. The conservation ofenergy, considered in the last section, is the first law of thermodynamics, and it results inEqs. (4,3.8)—(4.3.l0) for one-dimensional systems when thermal-fluid coupling is omitted.For additional details, see Schlichting (1979), Bird et al. (1960), and Reddy and Gartling(2001). More details are provided in Chapter 10, which is dedicated to finite element modelsof two-dimensional flows of viscous incompressible fluids.

Here, we consider so-called parallel flow, where only one velocity component is differentfrom zero resulting in all the fluid particles moving in one direction, i.e., u = u(x, y, z),where u is the velocity component along the x coordinate. We assume that there are no bodyforces. The z-momentum equation requires that u = u (x, y). The conservation of mass inthis case reduces to

which implies that u = u(y). They-momentum equation simplifies to

whichimplies that P = P (x), where P is the pressure. The x-momentumequation simplifiesto

d2u dP(4.4.1)

dy2 dx

The energy equation for this problem reduces to

IT (82T 02T\ Idu\2pcu — = k I — +

— I + i I — I (4.4.2)Ix \1x2 \dy)

Here we are primarily interested in the finite element analysis of Eq. (4.4.1).

4.4.2 Finite Element Model

Equation (4.4.1) is a special case of the model equation (3.2.1) with the following correspondence:

dPf=——, a=u=constant, c=0, x=y

dx(4.4.3)

182 AN INmODUCCIONTO THE FINITE ELEMENT METHOD CHAFfER 4: SECOND-ORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: APPLICATIONS 183

Therefore, the finite element model in (3.2.31a) and (3.2.31b) is valid for this problem:

where

= {f9 + {Q} or Keue = je +

‘ / dPK=j iLfLdy, f=I __)d

dy dy

du\ I/

du))I,

\ dy

Next, we consider an example.

(4.4.4i)

(4,4.4b)

• •:,

Figure 4f4 () wbetween parallel plates (b5

- - - -

and the fimte element solutions for the two sets of bouadaiy conditions

U2= for$etl-

: 2

-

U2—+—U forSet2 v

Although the nodal values predicted in the linear and quadratic elethey vary linearly and quadratically between nodes of Imear- and quadratic elementtively

•. The exact solutions for the two sets ofbonñdary àonditibiis ip

forSetl

u(y)=Uo(L+)_(l_!.) fb

Note that thefinite element solutions aitheisodes are exact, as expectesolution agrees with the exaecsotutions (44 10) for every value of

4.5 SOLID AND STRUCTURAL MECHANICS

4.5.1 Preliminary Comments

Solid mechanics is that branch of mechanics dealing with the motion and deformation ofsolids. The Lagrangian description of motion is used to express the global conservation laws.The conservation of mass for solid bodies is trivially satisfied because of the fixed materialviewpoint used in the Lagrangian description. The conservation of momentum is nothingbut Newton’s second law of motion. Under isothermal conditions, the energy equationuncouples from the momentum equations, and we need oniy consider the equations ofmotion or equilibrium (see Example 2.3.3).

dP[drconstant

____ ____

—u

EX.1

184 AN NTRODUCrION TO THE FINITE ELEMENT METHOD CHAPTER 4: SECOND-ORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: APPLICATIONS 185

Unlike in fluid mechanics, the equations governing solid bodies undergoing different and the strains and stresses take the form

forms of deformations are derived directly without specializing the three—dimensional elas-- du d

= — = — (Nv) _Beueticity equations to one dimension. Various types of load-carrying members are called bydx dx

different names, e.g., bars, beams, and plates. A bar is a structural member that is subjected= Es = EeBeUE

to only axial loads (see Examples 1.2.3 and 2.3.2), while a beam is a member that is subjected to loads that tend to bend it about an axis perpendicular to the axis of the member (see and the expression for the total potential energy becomesExample 2.4.2). The equations governing the motion of such structural elements are not

1 Xb

deduced directly from (2.3.52), but they are derived either by considering the equilibrium =— I dx

— f UTBTJ dx — uTQ

of an element of the member with all its proper forces and using Newton’s second law 2 ix,(Example 1.2.3), or by using an energy principle (Examples 2.3.2 and 2.4.2). where the element label on B and u is omitted for brevity. Then the principle of minimum

total potential energy, 1W = 0, yields the finite element model4.5.2 Finite Element Model of Bars and Cables Ifle Su

\ ‘X

The equation of motion governing axial deformation of a bar is (see Example 1.2.3) Jx, j x, ] == AeEeBTBdxIu_j BTfdx_Q

or02u I / du

It2 Ix Ix)= f(x, t) (4.5.1)pA———(EA—

whereFor static problems, Eq. (4.5.1) reduces to

d / du\ K= [ AEB”B dx, l

= [ BTf dx, B = (4.5.4b)(EA_) =f(x) (4.5.2) dx, dx,, dxdx\ dxThese are just matrix form of the equations already presented in (3.2.3la) and (3,2.31b).

It should be recalled that Eq. (4.5.2) is derived under the assumption that all material pointson the line x = constant (i.e., all points on any cross section) move by the same amount

4.53 Numerical Examplesu (x). This is equivalent to the assumption that the stress on any cross section is uniform.Equation (4.5.2) is the same as the model equation (3.2.1), with a = EA and c = 0. Hence, In this section we consider a number of examples of finite element analysis of bars,the finite element model in (3.2.31a) and (3.2.31b) is valid for bars.

The average transverse deflection u (x) of a cable made of elastic material is also governed by an equation of the form:

d(T)=f(x) (4.5.3)

dx \ dx‘4eig4t$th4Iwhere T is the uniform tension in the cable and! is the distributed transverse force. Again,

bPEq. (4.5.3) is a special case of Eq. (3.2.1), with a = T and c = 0.In structural mechanics problems, the quadratic functional of (3.2.10) takes the special 1ea nd (on4

1 px fx& ni -=— j

5T0, dx— J uf dx — u Q’

(3.2.24) of u can be written as

u 1 Wioi4ôdjul

= ... J=Neue

Lu

186 AN INTRODUCOON TO THE FIN1TE ELEMENT METHOD

4 U:Thi cmplts th# load tepresentation Of the problem.

Thtggverning e qpado4nfrthe rbl

& nt-s çtioi aree*x}

1.

Thos, the conctet pier prb1ñ isJdalizéd asa diutthplacement tis goendb> thecqnation

CHAFfER 4: SECOND-ORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: APPUCATIONS 187

i 4tho1 4: -4. 0.000 —0,625 0625

‘

The otttensetteqtiations for the unknown dispfaaemeptt

r 0.37& o.375 u) -

UU [-0751 .oooJ ffq 1250G- - 4 4

Thesolutiouisgivaabypositivedpiacementsbecàucthutthe pIer iS4Izcompression - -. - -

- -UI ZIli 10m =2I

4’.-‘ 4

Hence, the sress a’tthe fixd ends(compmssive

-; -‘s -The fo -ele’mnttho&l gi’es Z.00 c ‘l0 untL

The exact olution fEqL4.5and

4 44 .[625...625E :U

-- -. - -,-‘ -‘5- - 4 44 -

-Example42 : - JConsider the composie bar ronsisting a tapered steel bar fasteed aatñ zrt Iunithrm crosslsection and subjected to loads-as showuinFigpre4.5ewldstodetennhiiethe displacement4field in the bar using the finite element methorL ‘‘

Uecxictvniuesofuatnodes 2am.

-. =2.08lGm, 4H(U—-‘5 - -

_.‘5J ‘544

:‘Th& finite elementsolutioliat the nodes sndt exactbeauseproblem. •- -“ - -

-

.xp(;1+)

jf=4k

‘pp.p-xj’!‘-1J

n,JQA!tfl

XqA1J7

;v

ItX96O+LEc€QO—4fl±•1•

6O9OO(+qJ

xwo4•.

.;.

6R1SNOLLVDrMdVNOJSN3I4iG3N0NISNOLtVflI31VLLN3a4UU3IflIOcINOD3SJ4VJ4GOHJSV1I1J]NUII1OINOUDflGOIIININV

—i

--

.

-?

g

-0

—-,

0

2II

..-..-..

-L

.o-

.—

..

--

2:0

o-.---.-

‘.

-o----

.--

--1’:--..

---

.Ii.

I-•-

g-.c

-g•---

-._

I-..o

.-

-c-_-

-.-

_

--a’

.-

.---I

-

—.

..

-S

—..

.-.-

----

----

.—-----

--:.

-

IIII

-.

..q--.

0—

H

Hp

=—

IAIA

___

‘<

IAIA

0IA

IAIA

A°

°—

L-

‘0

I-

co

’-

__

__

-S

ca

II-

.“.—

‘--.S

..--

.-

..

..

.-.:;.-------.

.

-1

‘a-

—

-—

—

5-

--‘.

-0

-r’--

-.

—.

—-

.-.

5—

’--

r-) 0 0 0 C 0 C z 0 2 C C 2 ‘0

192 AN INTRODUCTION TO THE FINTFE ELEMENT METHOD CHAPTER 4: SECOND-ORDER DWFERENT[AL EQUATIONS [NONE DIMENSION: APPLICATIONS 193

8 --- 4 8

wrCG is sitearoidinates t ui

athflefd theabsne(

yielth the equatiOn :

-:

The boundary conditions on Uare . -

u(b)=O - U(b)=04 —2rL(r)l1,€P —t!4:

This completes the theliretical nu1atior of the probusing the fi iteelinent method.

The finite etenient model orEq. (4114) fcf E(3.4(14,5b) with n(r)=rG and f=O. One linear elementin thedomaihba) -;

- 2irUf- t 1i -1 111 [Q

-ijcrf,

andwsthUQandQPJL4weobtauv 8

IT—8

The çxact solution of Eqs8(l 5 l4and(4S i

U(r—iog(r/b U(a

Theone elereiit soludon in Eq. (4i16) correspUe.,lo&(a/bfl.iJeofonequadratic elementgives (fr8 —-a)1-7

3h+1 .-(4h+6a)

—(4h44$a) 1jt4-32a -(ih6a

h -j2a —(121i +-16a) I1h4I4a-

78 78%

If

8

c444- 418.4T - -

-d

W44

—.1Ø4

11

-, 444 -t:; , : ; :

- :

eIah

___________________

48

iiL4 8414

48

44

8

:‘4-• - -.

,__•_ .___

theana1yt4ica1\iiftualiy ives1hOx

4 -

44-i.d444 4--

4 - --

-4:14 : 4 1

_-_$_•-_ 13Q-

- -

4 - -

a 8-d-- -4- -

4 - -

- - a74.. - - - 4

4 -

84f4 41.;

44

8,- -

4 a1gkleoreattboini-

7- - - a. - -

- 4 . 4 _4.S - - - - - --

- -

-- 8 -

5;

194 AN wrRODucrION TO ThE FINflt ELEMENT MEOIOD CNAPTEE 4: SECOND’-ORDER DIFFEEEWIIAL EQUATIONS IN ONE DIMENSION: APPLICATIONS 195

4.6 PLANE TRUSSES

4.6.1 Introduction

Consider a structure consisting of several bar elements connected to each other by pins,as shown in Figure 4.6.1. The members may rotate freely about the axis of the pin. Consequently, each member carries only axial forces. The planar structure with pin-connectedmembers (i.e., all members lie in the same plane) is called a plane truss. Since each memberis oriented differently with respect to a global coordinate system (x, y), it is necessaryto transform the force displacement relations that were derived in element coordinate system (Y, ) to the global coordinate system (x, y) so that the structure stiffness can beassembled from the element stiffness referred to the same global coordinate system.

4,6,2 Basic Truss Element

First, we consider a uniform bar element with constant EA and oriented at an angle 8e

measured counterclockwise, from the positive x-axis. If the member coordinate system(Ye, is taken as shown in Figure 4.6.2(a), and (u, i)) and (E, 0) denote the displacements and forces at node i with respect to the member coordinate system (Ye,

respectively, the element equations (4.5.1 la) can be expressed as (we now use the notation

1 0 —1 0 ü F

EeAe 00 00 D 0 - - -

—1 0 1 =or [K9{z9 = {P) (4.6.1)

0 0 0 0 i) 0

We wish to write the force-deflection relations (4.6.1) in terms of the correspondingglobal displacements and forces, Toward this end, we first write the transformation relationsbetween the two sets of coordinate systems (x, y) and (Ye, 5) (see Fig. 4.6,2)

Xe = x cos 9e + Sill 8e Ye —X 5in 8e + y cos 9

X = Xe cos 9e — Ye sin 8e Y .e sin 0e + Ye cos 0e

(a) (b)

Figure 4.6.2 A bar element oriented at an angle with respect to the global coordinate system (x, y).(a) Forces and displacements in the element coordinates. (b) Forces and displacementsin the global coordinates.

or, in matrix form, we have

Xe C05 9e sin 0e X X C05 9e — 5fl 8e Xe- . , = - (4.62)Ye — Sifl 8e cos 0e ) 37 Sifi 0e C05 8e Ye

where 8, is the angle between the positive x-axis and positive Ye-axis, measured in thecounterclockwise direction. Note that all quantities with a bar over them refer to the member(or local) coordinate system (Ye, Ye), while the quantities without a bar refer to the globalcoordinate system (x, y).

The above relationship also holds for the displacements and forces of the two coordinatesystems. We have

cos8 sin8e

— 5in8e cosOe

— 0 0

.0 0

F7 F2

0 0 u

0 0 v

cosde sinde U2

5inOe cosOe 1)2

(4.6.3a)

(4.6.3b)

or

{e}= [T9{A9

where (} and {} denote the nodal displacement vectors in the member and structurecoordinate systems, respectively. Similarly, we have

[T]{F} (4.6.4)

Here {Ee} and {F9 denote the nodal force vectors in the member and structure coordinatesystems, respectively (see Figures 4.6.2).

4.6.3 General Truss Element

Next, we derive the relationship between the global displacements and global forces. UsingEqs. (4.6.3b) and (4.6.4) in Eq. (4.6.1), we obtain

Figure 4.6,1 A plane truss structure. = Te)(Fe) (4.65)

• IF,

(b)

lent representation of a plane truss. (a) Geometry andapplied loads. (b) Element numbering and reaction forces.

vertical displacements, at each node of the element. The element stiffness matrix in the localcoordinate system is given by Eq. (4.6.1) while the stiffness matrix and force vector (with

= 0) in the global coordinate system is given by Eqs. (4.6.8) and (4.6.9), respectively. Theelement data and connectivity is given in the following table.

Element Global - Geometric Materialnumber nodes - properties property Orientation

1 2 A. h1=L E 01=0

:2 23 A,h2=1.-. E 89O

3 1 3 - A, h3=./L E 0=45°

Element Matrices -

The element stiffness matrices are given by [1/(2) =0.3536]

10—1-0 0 00 0

• [Kl)=0 0 0 0

, [K21=0 1 0 1

L—j0 10 LO 00.0

0.000 0—10 1

0.3536 0.3536 —0.3536 —0.3536

EA 0.3536 0.3536 —0.3536 —0.3536 -

[K3]—

(4.6.11)• L —0,3536 —0.3536 0.3536 0.3536

—0.3536 —0.3536 0.3536 0.3536

CHAJ’TER4 SECOND-ORDER DIFFERENTIAL EQUATIONS IN OlE DIMENSIOFL AFFUCA11ONS 197196 ie INTRODIJCOON TO THE FINITh ELEI4EIff METHOD

Premultiplying both sides of the above equation with [T9T and noting that [Tel_i = [Td]T,

we obtain

[Te1T[Rn1[Te1(M) = (Fe) or [K9(d) = (F9 (4.6.6)

where

[K9 = [T9T[!(9[T9, {Fd} = [T9T{Fe} (4.6.7)

Carrying out the indicated matrix multiplications, we obtain

COS2Oe Sin28e —cos28 S1fl2Ge

[Kd] = EeAe Sin 29e Sin2 0e —Sin 2O 5111 O

(4.6.8)h — cos2G — sin 2G cos G sin 2G

4sin2Oe 5in28e SIfl26e Sifl2Ge

F F cos 8 cos 6

F FSrn8 JSin9eV} = -e + -e (4.6.9)

F F2 COS G f2 C05 0ee d . —e

F4 F2sinG f2sinG

where J are computed using Eq. (3.2.31b) [also see Eq. (3.2.34)]

jh.

f()ifr) d (4.6.10)

Equations (4.6.8) and (4.6.9) provide the means to compute the element stiffness matrix[Kd] and force vector (F9, respectively, both referred to the global coordinate system, ofa bar element oriented at an angle O. The assembly of elements with their stiffness matrix

and force vector in the global coordinates follows the same ideas as discussed before exceptthat we must note that each node now has two displacement degrees of freedom. Theseideas are illustrated in the following example.

::- ,

Exampk46l

Considera three memberlrussohowninFigure4 63’-

CHAnER 4. SECOND-ORDER DIFFERENTIAL EQUAl IONS NONE DIMNStON. APPS ICATIONS 199

where (U5, V1) and (F5,F,) denote the and ompo thforces, respectively, at the global e

Boundary Conditions

The specified displacement and fo s of

- lJ=1’= 2V20, — —

The first two boundary conditions con-espon onode 1, the next two correspond to the horizonta’ and v plast correspond to the force boundary conditions at node 3 a en(U3, ‘/) of node 3 and forces F1) at node 1 and fo es (Fr,,

Condensed Equations

The con nuations ‘ r the unknown displacements (U3, oas indicated by the dotted linef in (4.6.

r03536 0.35361 [U3 1 P= (4.6.18)

—2P

(from t four equations of

-0.3536

—ft3536 (13 -

(4.6.19)V3

1

clcnbdtovnq

4,5,

‘S

C;

4S

45,4s

I-

5’rcz)-zr9pll,

S_i 5

S_SI

ii_Si’

4

_4,4

5_i5

C45S__iS

S ‘w,,:-

S_S

gUO1Sq)9’irr‘osuOpopoiwaqq5jiqui‘)sio;q;

,S_44C,,?

ntdpojopnw

zquaSWaTpUta1t1rnXS

qi3o£aiuoapos(E4pmrn’spd‘paiaai(4‘fl)sidspjo’asuasaippuaarnpiu3pJouo9utuXauV

nepuo!Wp1dJuI4--•

S

‘5

S,;,

II’I4.-CI4

qwau 5u

‘I’

S5,:aioj4maqjnt, ,5

I’S

4S_S

‘55-C

‘,:555

S

S

5544S4C‘

5555

S‘5ajduiaxaR7JS1Qiaqaiaw

iCjUOTaxa

CS

Si

OE9t55-C

,CC

4IIC4S

‘-CI“

SC•S

0 ‘5

4445S_5

45

=II)suopajaiaq

4q14$-C‘554,

sI

54‘54S5rC5’

45

C

5’LI1 55

,,55,S_C ‘IllssC_SSl

‘:C’S

LI5-CC 4S,:

“E’ 55

S_I

555

—55

C’5

‘SI

‘1

JOZSN0aVDrJddvNOISNaNIG3N0NJSN0JJYflOa1VUNaJBthflGJ13Q0-GNODSPI!RJ4VHOQOHi3NNN31331I4IIJfl5L01NOLI3000VINJvOOZ

202 AN INTRODUCnON TO THE FINITE ELEMENT METHOD

Qo

= u cosO+ v sina= -it sina + v cosa

Figure 4,6.5 Transformation of specified boundary conditions from a local coordinate system to the

global coordinate system (for an inclined support).

4.6.4 Constraint Equations: Penalty Approach

It is not uncommon in structural systems to find that the displacement components at a

point are related. For example, when the plane of a roller support is at an angle to the

global coordinate system (see Fig. 4.6.5), the boundary conditions on displacements and

forces at the roller are known only in terms of the normal (to the support) component of the

displacement and the tangential component of the force

u=O, Ø=Qo (4.6.31)

where u is the normal component of displacement and Q is the tangential component of

the force at node I of the element 2f; Qo is any specified tangential force. These conditions,

when expressed in terms of the global components of displacements and forces by means

of the transformation of the form (4.6.3b) and (4.6.4), become

u = —usincr +ucosa=O

= Qcos+Qsina=Qo

(4.6.32a)

(4,6.32b)

where (u, t4) and (Q Q) are the x andy components of the displacements and forces,

respectively, at the support. Equations (4.6.32a) can be viewed as constraint equations

among the global displacements, which have a companion relation among the associated

forces, namely Eq. (4.6.32b). Here, we present the penalty function method through which

constraint equations of the type in (4.6.32a) and (4.6.32b) can be included in the finite

element equations.The penalty function method allows us to reformulate a problem with constraints as

one without constraints. The basic idea of the method can be described by considering an

algebraic constrained problem:

minimize the fitnctionf (x, y) subject to the constraint G(x, y) = 0

In the Lagrange multiplier method the problem is reformulated as one of determining

the stationary (or critical) points of the modified function FL(X, y),

CHAPTER 4: SECOND-OP,DER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: APPLICATIONS 203

subject to no constraints. Here A denotes the Lagrange multiplier. The solution to the problem- is obtained by setting partial derivatives of Fc with respect to x, y and A to zero:

=o, fL=0,ax ay

(4.6.34)

which gives three equations in the three unknowns (x, y, A).In the penalty function method, the problem is reformulated as one of finding the

minimum of the modified function F,

Fp(x, y) = f(x, y) + -[G(x, y)]2 (4.6.35)

where y is a preassigned weight parameter, called the penalty parameter. The factor inEq. (4.6.35) is used for convenience: When F is differentiated with respect to its arguments,the factor will be cancelled by the power on G(x, y). The solution to the modified problemis given by the following two equations:

=0 -0 (4.6.36a)

The solution of Eqs.(4.6.36a) will be a function of the penalty parameter, (xv, yy). Thelarger the value of y, the more exactly the constraint is satisfied (in a least-squares sense),and (xv, yp) approaches the actual solution (x, y) as y —* oc. An approximation to theLagrange multiplier is computed from the equation,

A=yG(x,y)

We consider a specific example to illustrate the ideas presented above.

(4.6.36b)

FL(x. y) = f(x, y) + AG(x, y) (4,6.33)

204 rRoDucnoN TO THE FINrrE ELEMENT METHOD CHAPTER 4: SECOND.ORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: APPLICATIONS 205

• where is the Lagrange multiplier to be determined. We have

= 8x + 2y + 6 + 2X ‘ -

ay

• Solving the three algebraic equations, we obtain

x=3,y=.

Penalty Function Method. The mbdified fuctio

Fpç,y)f(X.y)+

and we have -

f=8x+2y+6+2y(2x+3Y)0.

f=-6y+2x—3+3y(2x+3y)0

The solution of these equations is -

l5—36y — l8+24

—26± l2y ‘— —26+ 12y

The Lagrange multiplier is given by

84i .--

= yG(x,-26+ 12y

Clarly, in the limit y -+ no, the penalty function solution approaches the exact solution:

lire x = —3, lim v = 2, urn A = 71’-4CC )‘

An approximate solution to the problem can be obtained, within a desired accuracyr byselecting a finite value of the penalty parameter (see Table 4.6.1). This completes the example.

Table 4.6.1 Convergence of penalty function solution with increasing penalty parataieter.

0 -0.5769 -0.6923 0.0000 -3.2308

1.5000 -3.0000 -6.0000 -6.0000

10 —3.6702 2.7447 8.9362 0.8936ItO —3.0537 2.0596 7.1550 0.0716

1,000 —3.0053 2.0058 7.0152 0.0068

10,000 -3.0005 2.0006 7.0015 0.0008,

ro -3.0000 2.0000 7.0000 0.0000

+ (mUm + — m,i) (4.6.39)

The functional l1 attains a minimum oniy when rnUm + — mn is very small, i.e.,approximately satisfying the constraint (4.6.37). Setting 3fl,, 0 yields

(K+Kp)u=f+Q+Q (4.6.40a)

Now we turn our attention to constraint equations of the form used with bar elements

mUm + $nUn = Imn (46.37)

where ,,, I3 and are known constants, and u and u are the mth and nth displacement degrees of freedom in the mesh, respectively. The functional that must be minimized subject to the constraint in (4.6.37) in this case is the total potential energy of thesystem (see Section 4.5.2)

Ii = f AEuBrBta dx— j flTflTf dx — UTQ (4.6.38)

The penalty functional is given by

= j Jo TB.r —f UTBTf dx _uTQ

where [see Eq. (4.5.4a)j

K=JAEBTBdx, K=yrnfl

yn.$n

f= f Bf dx, Q = ... (4.6.40b)

Yt,nnI3n

Thus, a modification of the stiffness and force coefficients associated with the constraineddegrees of freedom will provide the desired solution to the constrained problem. As illustrated in Example 4.6,2, the value of the penalty parameter ‘ dictates the degree to whichthe constraint condition (4.6.37) is met. An analysis of the discrete problem shows that thefollowing value of y may be used:

y=maxIK1xiO, li,j<N (4.6.41)

where N is the order of the global coefficient matrix. The reaction forces associated withthe constrained displacement degrees of freedom are obtained from

Fmp = “Y/3m (finium + nUn — mn), = Yn (8mUm + nUn — 8rnn) (4.6.42)

Because of the large magnitudes of the penalty terms, it is necessary to carry out computations in double precision (hand calculations do not give accurate results).

II

Iis

/:

re-!

j”t

20S AN INTRODUCHON TO THE PINfTE ELPJ4ENTMEIHOD CHAPTER 4: SECOND-ORDER DJFPERENTL4L EQUATIONS SHONE DIMENSION: APPLICATIONS 209

These constraints bring in an additional degtee of freedom, namely U5, into the equations.Hence, the assenibled equations before including the constraint conditions are -

116.67 0 —116.67 0 0 U1 Q. -

:0 300 —300 0

iO —116.67 0 116.67 0 0 U3 = Q (4.6.45)

0 —300 0 300 0 U4 Q

0 00 00 U5 F2

The 1as row and column of the above equation are added to facilitate the addition of penaltyterms Without the addition of the penalty contributions the last equation is nonsensical.

Using the procedure developed in this section, we can include the constraints, 3U1 — U5 = 0

and 1.8(32 — U5 = 0 into assembled equations (4.6.43). The value of the penalty parameter is selected to be y = (300 x l0) l0. The stiffness additions due to the two constraints

are(i =3,=.1.8,5=—1,and15=/325=0)

1 5

1 [ (3)y 3(—l)yl0

2700.00 —900.00

5 [(—l)3y (—l)2y — —900.00 300.00

2 5 (4.6.46)

2 (1.8)2? (—l)18y 972.00 —540.00

5 —l)l.8,’ (_fl2y — —540.00 300.00

The force additions are zero on account of s = = 0. Hence, the modified finite element

equations become

27,000,117 0 —117 0 —9,000,000 U1 Q0 9,720,300 0 —300 —5,400,000 U2 Q

106 —116.67 0 117 0 0 U =

0 —300 0 300 0 (34 Q

—9,000,000 —5,400.000 0 0 6.000.000 115 F2

46.47)

- The condensed equations are

27,000,117 0 —9,000,000 U 0

106 Q 9,720.300 —5,400,000 U2 = 0 (4.6.48)

—9,000,000 —5,400,000 6,000,000 U5 30 x i03

whose solutios (using an equation solver) is (defiections are in the direction of the force)

U1 = 0.09474 (mm), U2 = 0.15789 (mm), U = 0.28422 (mm) . (4.6.49a)

Next, we consider a plane truss with an inclined support. The penalty approach is usedto include the constraint condition among the displacement components at the support.

210 INTRODUCTION TO THE FINITE ELEMENT MEINOD CHAPTER 4: SECOND-ORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: APPUCATIONS 211

4.6.5 Constraint Equations; A Direct Approach

Here, we present an exact method by which constraint equations of the type in (4.6.32a)and (4.6.32b) can be included in the assembled equations for the unknowns. The methodinvolves expressing the global displacement degrees of freedom at the node with a constraintin terms of the local displacement degrees of freedom so that the boundary conditions canbe readily imposed.

Recall from Eqs. (4,6,2) and (4.6.3a) that the displacements (i2, ), referred to the localcoordinate system (, 9) at a point, are related to the displacements (u, v), referred to theglobal coordinates system (x, y), by

cos sin u— fl=Au (4.6.51a)

v —smfl cos v

and the inverse relation is given by

cos sin= AtI, where A = , (4.6.5lb)

—sin$ cos

where the subscript “c” refers to the constrained degrees of freedom. Since we wish toexpress the global displacements at a given node in terms of the local displacements at aspecific node, we construct the transformation of the whole (global) system as

[I] [0] [0]

[TI = [0] [A] [01 (4.6.52)

[0] [0] [I]

Thus, all displacement degrees of freedom that are not constrained are unaffected andonly global displacements that are constrained are transformed to the local displacements.Further, note that the transformation matrix [A] is placed in [Ti in such a way that only theconstrained degrees of freedom are transformed. We have

& 190 &= 0 AT 0 ii or i. = TTA (46.53)

2 0 0 I 2 -.

212 AN INTRODUCOON SO THE FINITE ELEMENT METhOD CHAPTER 4: SECOND-ORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: APPLICATIONS 213

where t and 2 denote vectors of the global displacement components ahead and behind

(in terms of numbering) the constrained displacement degrees of freedom u in the mesh.

We also note that the transformed displacement vector contains the global displacement

vectors & and & and the local (constrained) displacement vector ü,

The remaining steps of the procedure are the same as that described in Sections 4.6.2

and 4.6,3. Thus, we obtain

K =F (4.654)

where the transformed global stiffness matrix and global force vector are known in

terms of the assembled global stiffness matrix K and force vector F as [see Eq. (4.6.7)1

= TTKT, E = TTF (4.6.55)

Since the constrained displacements are a part of the global system of equations, we may

impose the boundary conditions on them directly (such as ü = 0 at an inclined roller

support, where n denotes the coordinate normal to the roller).

In summary, we may introduce a transformation of the displacements that facilitates the

imposition of boundary conditions or inclusion of constraints on the displacements. Once

the transformation T is identified, we may use Eq. (4.6.7) or (4.6.55) to obatin the modified

equations that have the desired effect, We revisit the problems of Examples 4.6.3 and 4.6.4

to illustrate the ideas described here.

214 AN INTR005CI1ON TO ThE F1540t aEMENr METhOD CHAPTER 4: SECOND-ORDER DIFFERENUAL EQUPflONS IN ONE DIMENSION: APPUCATIONS 215

4.7 SUMMARY

In this chapter, finite element models of discrete systems have been developed and applica

tions of finite element models to the solution of problems of heat transfer, fluid mechanics,

and solid mechanics have been presented. To aid the reader, a brief review of the basic

terminology and governing equations of each of the three fields has also been given. Anal

ysis of plane trusses and inclusion of constraint relations between primary variables by two

different methods is also discussed.It has been shown that the secondary variables of a problem can be computed using ei

ther the global algebraic equations of the finite element mesh (i.e., condensed equations for

the secondary variables) or by their original definition through finite element interpolation.

The former method gives more accurate results, which will satisfy the equilibrium at inter-element nodes, whereas the latter gives less accurate results that are discontinuous at thenodes. The secondary variables computed using the lipear elements are elementwise constant, while they are elementwise linear for the Lagrange quadratic elements. The magnitudeof discontinuity can be reduced by refining the mesh (Is or p refinement). The discontinuityof the secondary variables at nodes is due to the fact that the secondary variables are notmade continuous across the elements.

In closing this chapter, the reader is reminded that the finite element method is a powerful tool for engineering analysis. The power of the method lies in transforming the traditional variational methods (e.g., Ritz, Galerkin. least-squares, and other weighted-residualmethods) into a powerful computer-based technique by developing suitable approximation functions for complex problems. Without a good understanding of the method as wellas the engineering background behind each problem, one is ill-equipped to analyze theproblem.

PROBLEMSMany of the following problems are designed for hand calculation while some are intended specificallyfor computer calculations using the program FEM1D (see Chapter 7 for details on how to use theprogram). The problem set should give the student deeper understanding of what is involvedin thesetting up of the finite element equations, imposition of boundary conditions, and identifyingthe condensed equations for the unknown primary and secondary variables of a given problem. Whenthe number of equations to be solved is greater than three, the student should opt for a computersolution of the equations. The calculations can be verified, in most cases, by solving the same problem using FEM1D.

Discrete Elements

4.1 Consider the system of linear elastic springs shown in Fig. P4.1. Assemble the element equationsto obtain the force-displacement relations for the entire system. Use the boundary conditionsto write the condensed equations for the unknown displacements and forces.

Figure P4.1

216 n orrEonucooNro me mirre i.encaurnsoo atsrrt sEcoNDORDER DreurENTIAL EQUATiONS IN ONE DIMENSION: APPLICATIONS 217

4.2 Repeat Problem 4.1 for the system of linear springs shown in Fig. P4.2. 4.5 Write the condensed equations for the unknown pressures and flows (use the minimum number- of elements) for the hydraulic pipe network shown in Fig. P4.5. Answer: P1 = Qa, P2 =

Qa,andP3=Qa.

IVAV%R2=3a

Figure P4.2

Figure P4.5

4.3 Consider the direct current electric network shown in Fig. P4.3. We wish to determine the

voltages V and currents I in the network using the finite element method. Set up the algebraic 4.6 Consider the hydraulic pipe network (the flow is assumed to be laminar) shown in Fig. P4.6.equations (i.e., condensed equations) for the unknown voltages and currents, Write the condensed equations for the unknown pressures and flows (use minimum number of

elements.)

R=3012 2 R=35flV110

. l2lthQ=5xl0na3/s PIperesistance,R= —y1

R=5f2 R=lOfl

_____

6 V6=200V \ \ 1 2

R=l5fl R5t1 L=70m heD’iOcm I

Figure P4.3

()4.4 Repeat Problem 4.3 for the direct current electric network shown in Fig. P4.4. 2• :

6

R = 20 i1 R = 50 t2Figure P4.6

V1=llOV V7=40V

4.7 Determine the maximum shear stresses in the solid steel (G = 12 Msi) and aluminum (GeFigure P4.4 4 Msi) shafts shown in Fig. P4.7.

218 AN INTRODUCI1ON TO ThE FINITE EW.4ENE METhOD C1{AFTER4: SECOND.OEDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: APPLICATIONS 219

4.8 A steel shaft and an aluminum tube are connected to a fixed support and to a rigid disk, asshown in Fig. P4.8,11 the torque applied at the end is equal toT = 6,325 Nm, determine the

shear stresses in the steel shaft and aluminum tube. Use G, = 77 GPa and G,, = 27 GPa.

Heat Transfer

4.9 Consider heat transfer in a plane wall of total thickness L. The left surface is maintained attemperature T0 and the right surface is exposed to ambient temperature T, with heat transfercoefficient . Determine the temperature distribution in the wall and heat input at the left

surface of the wall for the following daur L = 0.1 m, k = 0.01 W/(m ‘‘C), = 25 W/(m2 ‘‘C),

T0 = 50°C, and T,,, = 5°C. Solvefornodaltemperatures andtheheatatthe leftwallusing (a) two

linear finite elements and (b) one quadratic element. Answer: (a) U2 = 27.59°C, U3 = 5.179°C,

Q =4.482W/m=—Q.

4.10 An insulating wall is constnicted of three homogeneous layers with conductivities k1, k,, and

k3 in intimate contact (see Fig. P4.10). Under steady-state conditions, the temperatures of the

media in contact at the left and right surfaces of the wall are at ambient temperatures of T

and T, respectively, and film coefficients and respectively. Determine the temperatureson the left and right surfaces as well as at the interfaces. Assume that there is no internal heat

generation and that the heat flow is one-dimensional (OT/ay =0). Answer: U1 =61.582°C,

U2 = 6 1.198°C, U3 = 60.749°C, U4 = 60.612°C.

Air at temperature, T. 35°CFilm coefficient, R = 15 W/(m2’‘C)

k150W/(m’C)k2 30 W/(m ‘C)k370W/(m. ‘C)h1=50mm82-3Smm83= 25 mm

4.11 Rectangular fins are used to remove heat from the surface of a body by conduction alongthe fins and convection from the surface of the Ibis into the surroundings. The fins are 100mm long, 5mm wide, and 1 mm thick, and made of aluminum with thermal conductivity I =

170 W/(m. ‘C). The natural convection heat transfer coefficient associated with the surroundingair is = 35 W/(m2 . ‘C) and the ambient temperature is T =20°C. Assuming that the heattransfer is one dimensional along the length of the fins and that the heat transfer in each fin isindependent of the others, determine the temperature distribution along the fins and the heatremoved from each fin by convection, Use (a) four linear elements, and (b) two quadraticelements,

4.12 Find the heat transfer per unit area through the composite wall shown in Fig. P4.12. Assumeone-dimensional heat flow.

k1= 150 W/(m.°C)k 30W1(m’°C)13 = 70 WI(m. °C)14 = 50 WI(m ‘C)h125mmh2°°7Smmh3SOmm

4.13 A steel rod of diameter D = 2cm, length L = 5cm, and thermal conductivity I = 50 W/(m.°C)is exposed to ambient air at 7’,, = 20°C with a heat transfer coefficient = 100 WI(m2-°C). Ifthe left end of the rod is maintained at temperature T5 = 320°C, determine the temperaturesat distances 25mm and 50mm from the left end, awl the heat at the left end. The governingequation of the problem is

__+m29=0 for 0<x<Ldx2

where 0 = T — T,,, T is the temperature, and m2 = P/A1c. The boundary conditions are

0(0)=T(0)—T,,=300°C, =0\dX kJX.L

Use (a) two linear elements and (b) one quadratic element to solve the problem by the finiteelement method. Compare the finite element nodal temperatures against the exact values. Answer: (a) U1 =300°C, U2 =211.97°C, U3 = 179.24°C, Q =3,521.1 W/m2. (b) U1 =300°C,U2 =213.07°C, U3 = 180.77°C, Q =4,569.9W/rn2.

4.14 Find the temperature distribution in the tapered fin shown in Fig. P4.14. Assume that thetemperature at the root of the fin is 250°F, the conductivity k = 120 BtuI(h.ft.°F), and the filmcoefficient = 15 Btu/(h.ft2.°F); use three linear elements. The ambient temperature at the topand bottom of the fin is T,, = 75°F, Answer: T1(tip) = 166.23°F, T2 = 191.1°F, 1’ = 218.89°F.

d=lSin 7”°2001b.ft dlin\ K

76mm12.5ft Ift 2ft

Figure P4.7 Figure P4.8

hi h h3I I4

/k1

h2 113

IT= tOO°C

6t 10 W/(m2’°C

Figure P4.12

Figure P4.10

220 AN INTR000CrION TO THE FINflE ELEMENT METHOD dAFTER 4: SECOND-ORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: APPLICATIONS 221

T,,=75°F 250°F

in.

Figure P4.14

4.15 Consider steady heat conduction in a wire of circular cross section with an electrical heatsource. Suppose that the radius of the wire is R0, its electrical conductivity is K, (2/cm),and it is carrying an electric current density of 1 (A/cm’). During the transmission of anelectric current, some of the electrical energy is converted into thermal energy. The rate of heatproduction per unit volume is given by q, = I’/K,. Assume that the temperature rise in the wireis sufficiently small that the dependence of the thermal or electric conductivity on temperaturecan be neglected. The governing equations of the problem are

_!.._(rki)qe for O<rR0, (rk) 0, T(Ro)T0rdr dr dr rO

Determine the distribution of temperature in the wire using (a) two linear elements and(b) one quadratic element, and compare the finite element solution at eight equal intervals withthe exact solution

T(r) T+ 4k

[1

Also, determine the heat flow, Q = —2rRok(dT/dr)Js0,at the surface using (i) the temperaturefield and (ii) the balance equations.

4,16 Consider a nuclear fuel element of spherical form, consisting of a sphere of “fissionable”material surrounded by a spherical shell of aluminum “cladding” as shown in Fig. P4.16,Nuclear fission is a source of thermal energy, which varies nonuniformly from the center ofthe sphere to the interface of the fuel element and the cladding. We wish to determine thetemperature distribution in the nuclear fuel element and the aluminum cladding.

The governing equations for the two regions are the same, with the exception that there isno heat source term for the aluminum cladding. We have

Id 1, dT1\———jrki—J=q for 0rRr2dr drj

_

-- (r2ica11) = 0 for RF r Rcrdr dr

where subscripts I and 2 refer to the nuclear fuel element and cladding, respectively. The heatgeneration in the nuclear fuel element is assumed to be of the form

Use two linear elements to determine the finite element solution for the temperature distribution,and compare the nodal temperatures with the exact solution

fr’\ IIT1 ,qoR 1r / r \‘l r6kI [‘) j+c[lj jj+(l+c)(1

Fluid Mechanics

RF

R

4.17 Consider the flow of a Newtonian viscous fluid on an inclined flat surface, as shown inFig. P4.17. Examples of such flow can be found in wetted-wall towers and the applicationof coatings to wallpaper rolls. The momentum equation, for a fully developed steady laminarflow along the z coordinate, is given by

d’w—IL—=pgcosfl

dx’

where w is the z component of the velocity, j.c is the viscosity of the fluid, p is the density, g isthe acceleration due to gravity, and is the angle between the inclined surface and the vertical.

The boundary conditions associated with the problem are that the shear stress is zero atx = 0 and the velocity is zero at x =

() =0,dx

Use (a) two linear finite elements of equal length and (b) one quadratic finite element in thedomain (0, L) to solve the problem and compare the two finite element solutions at four points

T,-TO=(l+.!._!-’\3k, 5r Rd

Figure P4.16

I=0o[l+c()]

where q and c are constants depending on the nuclear material. The boundary conditions are

kr2Ji=0 at r=0

T1=T, at r=Rp, and T,=T0 at r=Rc

222 AN ttcflI0DUCnON TO ThE t’tNIE ELEMENT METhOD CHAI’IER 4: SECOND-ORDER D1EFERENTIAL EQUATIONS IN ONE DIMENSION: APPLiCATIONS 223

x =0, L, L, and of the domain with the exact solution

pgL2cos,9 r2c L’ZJ

Evaluate the shear stress (r = — dw/dx) at the wall using (i) the velocity fields and (ii)the equilibrium equations, and compare with the exact value. Answer: (a) U1 = fo. U2 =

fr Jo = (pg cos )L2/p.

4.18 Consider the steady laminar flow of a viscous fluid through a long circular cylindrical tube.The governing equation is

1 d / dw’ Po—P—-

—rc— I =

rdro drj L

where w is the axial (i.e., z) component of velocity, r is the viscosity, and f is the gradient ofpressure (which includes the combined effect of static pressure and gravitational force). Theboundary conditions are

=0, w(Ro)=0dr r=O

Using the symmetry and (a) two linear elements and (b) one quadratic element, determine thevelocity field and compare with the exact solution at the nodes:

w,(r)= f0Rg [ (r)2]

4.19 In the problem of the flow of a viscous fluid through a circular cylinder (Problem 4.18), assumethat the fluid slips at the cylinder wall; i.e., instead of assuming that w = 0 at r = R0, use theboundary condition that

at r=R0

in which k is the “coefficient of sliding friction:’ Solve the problem with two linear elements.

4.20 Consider the steady laminar flow of a Newtonian fluid with constant density in a long annularregion between two coaxial cylinders of radii R and R0 (see Fig. P4.20). The differential

equation for this case is given by

I d / dw 1.-J2rc— I = —— =fardr\ drj L

where w is the velocity along the cylinders (i.e., the z component of velocity), c is the viscosity,L is the length of the region along the cylinders in which the flow is fully developed, and P1and P2 are the pressures at z = 0 and z = L, respectively (P1 and P2 represent the combinedeffect of static pressure and gravitational force).

The boundary conditions are

w=0 at r=R0 and R

Solve the problem using (a) two linear elements and (b) one quadratic element, and comparethe finite element solutions with the exact solution at the nodes:

fr\ 1—k2 ir1f0R 2

) +j7ln)j

Figure P4.20

Velocitydistribution

- wherek = R1/R0.Determine the shear stress r0 = dwJdr at the walls using (i) the velocityfield and (li) the equilibrium equations, and compare with the exact values. (Note that the steadylaminar flow of a viscous fluid through a long cylinder or a circular tube can be obtained as alimiting case of k — 0.)

4.21 Consider the steady laminar flow of two immiscible incompressible fluids in a region betweentwo parallel stationary plates under the influence of a pressure gradient. The fluid rates areadjusted such that the lower half of the region is filled with fluid I (the denser and more viscousfluid) and the upper half is filled with fluid II (the les-s dense and less viscous fluid), as shownin Fig. P4.21. We wish to determine the velocity distributions in each region using the finite

Velocity distribution w(x)

Figure P4.17

element method.

224 ii D-nsoOuCno04 TO THE FINErE ELEMENT METHOD CMAPTHE4: SEcOND-ORDER DIFFERENTIAL ts3UATIONS IN ONE DIMENSION: APPUCATIONS 225

The governing eqiations for the two fluids are

d2u1 d2u2—JL1--j- = fo, 27 = fo

dx dx

where f = (P0 — PL)/L is the pressure gradient. The boundary conditions are

ui(—b)wO u2(b)=O, u1(0)=u2(0)

Solve the problem using four linear elements, and compare the finite element solutions withthe exact solution at the nodes

fob2I2Ioi+p4_,i2Y(y2]

Hi = — I2 LI+2 1iI+i2bj (i=1,2)

Figure P4.21

4,22 The governing equation for an unconñned aquifer with flow in the radial direction is given bythe differential equation

1 d / du’

where k is the coefficient of permeability, f the recharge, and u the piezometric head, Pumping

is considered to be a negative recharge. Consider the following problem. A well penetratesan aquifer and pumping is performed at r =0 at a rate Q = 150 m3/h. The permeability ofthe aquifer is k = 25 m3/h. A constant head u0 = 50 m exists at a radial distance L = 200 m.Determine the piezometric head at radial distances of 0, 10, 20, 40, 80, and 140 m (see Fig.

P4.22). You are required to set up the finite element equations for the unknowns using anonuniform mesh of six linear elements.

Figure P4.22

4.23 Consider a slow, laminar flow of a viscous substance (for example, glycerin solution) througha narrow channel under controlled pressure drop of 150 Palm. The channel is 5 no long (flow

direction), 10cm high, and 50cm wide. The upper wall of the channel is maintained at 50°Cwhile the lower wall is maintained at 25°C. The viscosity and density of the substance aretemperature dependent, as given in Table P4.23. Assuming that the flow is essentially onedimensional (justified by the dimensions of the channel), determine the velocity field and massflow rate of the fluid through the channel.

y (m) Temp. (°C) Viscosity (kg/(m’ s)l Density (kg/rn3)

0.00 50 0.10 12330.02 45 0.12 12380.04 40 0.20 12430.06 35 0.28 12470.08 30 0.40 12500.10 25 0.65 1253

Solid and Structural Mechanics

4.24 The equation governing the axial deformation of an elastic bar in the presence of appliedmechanical loads f and P and a temperature change T is

for 0<x<L

where a is the thermal expansion coefficient, E the modulus of elasticity, and A the cross-sectional area. Using three linear finite elements, determine the axial displacements in anonuniform rod of length 30 in., fixed at the left end and subjected to an axial force P =400 lb and a temperature change of 60°F. Take A(x) = 6— x in.2 E = 30 x 106 lb/in.2,anda = 12 x l06/(in. -°F).

4.25 Find the stresses and compressions in each section of the composite member shown in Fig.P4,25. Use E, = 30 x 106 psi, Ed = l0 psi, Eb = 15 x 106 psi, and the minimum number oflinear elements.

4.26 Find the three-element finite element solution to the stepped-bar problem. See Fig. P4.26 forthe geometry and data. Hint: Solve the problem to see if the end displacement exceeds the gap.If it does, resolve the problem with modified boundary condition at x = 24 in.

Table P4.23: Properties of the viscous substance of Problem 4.23.

\Less dense and

____less

viscous fluid

Interface

Denser andmore viscous fluid

Pi

MI

5rLi

Steel (A38in.°)

lb

Aluminum (A 6 in.2)

Figure P4.25

226 st ornlooucrtoN n ma m4rra SLEMENT METHOD CHAPtER 4: 5EC0NDORDER DIFFERENTIAL EQUATIONS IN ONE DIMENSION: APPLICATIONS 227

fr84.2.SteeLE30Xl06pSi, Aiuminurn,Ea lOx I0psi

Figure P4.26

4.27 Analyze the stepped bar with its right end supported by a linear axial spring (see Fig. P4.27).The boundary condition at x = 24 in. is

EA+ku=0dx

SteeL E3= 30 x 106 psi, Aluminum, Ea lOx 106 j

Figure P4.27

4.28 A solid circular brass cylinder E5 = 15 x iO psi, 4 = 0.25 in.) is encased in a hollow circular

steel (E = 30 x 106 psi, d = 0.21 in). A load of P = 1330 lb compresses the assembly, as

shown in Fig. P4.28. Determine (a) the compression, and (b) compressive forces and stresses in

the steel shell and brass cylinder. Use the minimum number of linear finite elements. Assume

that the Poisson effect is negligible.

4.29 A rectangular steel bar (E = 30 x 106 psi) of length 24 in. has a slot in the middle half of its

length, as shown in Fig. 4.29. Determine the displacement of the ends due to the axial loads

P = 2000 lb. Use the minimum number of linear elements.

4.30 Repeat Problem 4.29 for the steel bar shown in Fig. P4.30.4.31 The aluminum and steel pipes shown in Fig. P4.3 1 are fastened to rigid supports at ends A and

B and to a rigid plate C at their junction. Determine the displacement of point C and stressesin the aluminum and steel pipes. Use the minimum number of linear finite elements.

4.32 A steel bar ABC is pin-supported at its upper end A to an immovable wall and loaded by aforce F1 at its lower end C, as shown in Fig. P4.32. A rigid horizontal beam BDE is pinnedto the vertical bar at B, supported at point D, and carries a load F2 at end E. Determin thedisplacements u5 and UC at points B and C.

4.33 Repeat Problem 4.32 when point C is supported vertically by a spring (k = 1000 lb/in.).

A

x Steel (E = 200 GPa,

p=so p iücm A3=60rnm2)

Aluminum (Ea = 70 GPa,

20cmAa=600fflffl2)

4.34 Consider the steel column (a typical column in a multi-storey building structure) shown in Fig.P4.34. The loads shown are due to the loads of different floors, The modulus of elasticity isE = 30 x 106 psi and cross-sectional area of the column is A = 40 in2. Determine the verticaldisplacements and axial stresses in the column at various floor-column connection points.

5in.

r6in. ‘ 6’6in.’6in. (a)Actualplate (b)Ideatizedplate

Figure P4.29 Figure P4.30

P=1330tb Brass Steel

Figure P428

Figure P4.31 Figure P4.32

228 AN INTRODUCflON 10 THE FINITE ELEMENT METHOD CHAPTER 4: SECOND-ORDER DIFTERENTIAL EQUATIONS IN ONE DIMENSION: AppucArIoNs 229

Figure P4.34

4.35 The bending moment (Al) and transverse deflection (w) in a beam according to the Euler-Bernoulli beans theory are related by

d2w—El-— =M(x)

dx2

For statically determinate beams, we can readily obtain the expression for the bending moment

in terms of the applied loads. Thus, M(x) is a known function of x. Determine the maximumdeflection of the simply supported beam under uniform load (see Fig. P4.35) using the finiteelement method.

wuu)X

I. L

Figure P4.35 Figure P4.36

4.36 Repeat Problem 4.35 for the cantilever beam shown in Fig. P4,36.

4.37 Turbine disks are often thick near their hub and taper down to a smaller thickness at theperiphery. The equation governing a variable-thickness t = r(r) disk is

where 2 is the angular speed of the disk and

fdu u\ fu du\ EI, O=C—±V—

,\dr rj \r drj i_2

(a) Construct the weak integral form of the governing equation such that the bilinear form issymmetric and the natural boundary condition involves specifying the quantity tro.

(b) Develop the finite element model associated with the weak form derived in part (a).

4.38—4.44 Forthe plane truss structures shown in Figs. P4.38—P4.44, give (a) the transformed elementmatrices, (b) the assembled element matrices, and (c) the condensed matrix equations forthe unknown displacements and forees.

8 taps—e 8 taps

All members:E=30x lft6psiA3in

__

042

_________________

fi— loft —.f+__loft.

Load

leor

25,000th

30000 lb’’fl]T2.5.

32000 lb 2.5 ft2

35,000 1b 2.5

oI

1)

E 30 msiA115in,2

A220in.1A315in,2

20 taps

15 ft

____

©

-

30ft

figure P4.38 1lgure P4.39

/80

q0 lb/ft

(t

3

lJ t

4El = constant

2P

3 —‘P

memers

L 0 2

El = constant

d

Figure P4.40 Figure P4.41

H— loft. + l0ft__-l

Iç4=l.02A3= 1.5 ilL2

0 E=29x1061b/in.2

i6kips

TtUr) — tJ0 ± Ipso2?’2= 0 Figure P4.42

230 AN ItmODUCttON TO T RE fl’E ELEMENT METHOD CHAPTER 4: SECOND-ORDER DIFFERENTIAL EQUAtIONS IN ONE DIMENSION: APPUCATIONS 231

4.45 Determine the forces and elongations of each bar in the structure shown in Fig. P4.45.

Also, determine the vertical displacements of points A and D.

REFERENCES FOR AD1MTIONAL READING1. Bird, R. B., Stewart, W. E. and Lightfoot, E. N., Transport Phenomena, John Wiley, New York, 1960.2. Hohnan, 3. P., Heat Transfer, 7th ed., McGraw-Hill, New York, 1990.3. Kreith, F. and Bohn,M. S., Principles ofHeat Transfer, 5thed.,WestPublishingCompany, SL Paul, MN, 1993,4. 0isik, M. N., Heat Conduction, 2nd Ed., John Wiley, New York, 1993.5. Reddy, 3. N., Energy Principles and Variational Methods in Applied Mechanics, 2nd ed., John Wiley,

New York, 2002.

6. Reddy, 3. N. and Gartling, D. K., The Finite Element Method in Heat Transfer and Fluid Dynamics, 2nd ed,CRC Press, Boca Raton, FL, 2001.

7. Slaughter, W. S., The Linearized Theory of Elasticity, Birkhdsiser, Boston, 2002.8. Schlichting, H., Boundary Layer Theory (translated by 3. Kestin), 7th ed., McGraw-Hill, New York (1979).9. Timeashenko, S. P. and Goodier, I. N., Theory ofElasticity, McGraw-Hill, New York, 1970.

4.46 Determine the forces and elongations of each bar in the structure shown in Fig. P4.45

when end A is pinned to a rigid wall (and P1 is removed).

P= l,000kN

Figure P4.43

For all members:E = 207 GPa,A 5cm’

For all members:E = 207 GPa,A=5x l0’4m’P=ltytkN

P

Figure P4.44

L 5ft. 50. 1J 70.Ar

Rigidbar—’i, B.:

P1=9okipslr

Steni barsE=29x10°psi

Ase= 19.5E

4

100.

F

L=soinpsjEAcF= 168 in.2

Figure P4.45