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Taking this observa-tion into account, Equation 5.14 can be expressed as 2x j+1dn j (x )( y j n j (x ) + y j+1 n j+1 (x )) + f (x ) dx = 0dx 2j = 1, M + 1xj(5.15)Integration of Equation 5.15 yields M + 1 algebraic equations in the M + 1 unknown nodal solution values yj, and these equations can be written in the matrixform[K ]{y} = {F }(5.16)where [K ] is the system “stiffness” matrix, {y} is the vector of nodal “displacements” and {F } is the vector of nodal “forces.” Equation 5.14 is the formalstatement of the Galerkin finite element method and includes both element formation and system assembly steps.

Written in terms of integration over the fullproblem domain, this formulation clearly shows the mathematical basis in themethod of weighted residuals. However, Equations 5.15 show that integrationover only each element is required for each of the equations. We now proceed toexamine separate element formulation based on Galerkin’s method.5.3.1 Element FormulationIf the exact solution for Equation 5.8 is obtained, then that solution satisfies theequation in any subdomain in (a, b) as well. Consider the problemd2 y+ f (x ) = 0dx 2x j ≤ x ≤ x j+1(5.17)where xj and x j+1 are contained in (a, b) and define the nodes of a finite element.The appropriate boundary conditions applicable to Equation 5.17 arey(x j ) = y jy(x j+1 ) = y j+1(5.18)and these are the unknown values of the solution at the end points of the subdomain.

Next we propose an approximate solution of the formy (e) (x ) = y j N 1 (x ) + y j+1 N 2 (x )(5.19)where superscript (e) indicates that the solution is for the finite element and theinterpolation functions are now defined asN 1 (x ) =x j+1 − xx j+1 − x jx j ≤ x ≤ x j+1(5.20a)N 2 (x ) =x − xjx j+1 − x jx j ≤ x ≤ x j+1(5.20b)Hutton: Fundamentals ofFinite Element Analysis5.

Method of WeightedResidualsText© The McGraw−HillCompanies, 20045.3 The Galerkin Finite Element MethodNote the relation between the interpolation functions defined in Equation 5.20and the trial functions in Equation 5.11. The interpolation functions correspondto the overlapping portions of the trial functions applicable in a single elementdomain. Also note that the interpolation functions satisfy the conditionsN 1 (x = x j ) = 1N 2 (x = x j ) = 0N 1 (x = x j+1 ) = 0N 2 (x = x j+1 ) = 1(5.21)such that the element boundary (nodal) conditions, Equation 5.18, are identicallysatisfied.

Substitution of the assumed solution into Equation 5.19 gives the residual asd2 y (e)d2R (e) (x ; y j , y j+1 ) =+ f (x ) =[y j N 1 (x ) + y j+1 N 2 (x )] + f (x ) = 02dxdx 2(5.22)where the superscript is again used to indicate that the residual is for the element.Applying the Galerkin weighted residual criterion results inx j+1x j+1N i (x ) R (x ; y j , y j+1 ) dx =(e)xjd2 y (e)N i (x )+ f (x ) dx = 0dx 2i = 1, 2xj(5.23)orx j+1d2 y (e)N i (x )dx +dx 2xjx j+1N i (x ) f (x ) dx = 0i = 1, 2(5.24)xjas the element residual equations.Applying integration by parts to the first integral results inxx j+1x j+1dy (e) j+1dN i dy (e)N i (x )−N i (x ) f (x ) dx = 0dx +dx x jdx dxxjxji = 1, 2(5.25)which, after evaluation of the nonintegral term and rearranging is equivalent tothe two equations, isx j+1dN 1 dy (e)dx =dx dxxjx j+1xjN 1 (x ) f (x ) dx +dy (e) dx x j(5.26a)N 2 (x ) f (x ) dx −dy (e) dx x j+1(5.26b)x j+1xjdN 2 dy (e)dx =dx dxx j+1xjNote that, in arriving at the form of Equation 5.26, explicit use has been made ofEquation 5.21 in evaluation of the interpolation functions at the element nodes.143Hutton: Fundamentals ofFinite Element Analysis1445.

Method of WeightedResidualsCHAPTER 5Text© The McGraw−HillCompanies, 2004Method of Weighted ResidualsIntegration of Equation 5.24 by parts results in three benefits [2]:1. The highest order of the derivatives appearing in the element equations hasbeen reduced by one.2. As will be observed explicitly, the stiffness matrix was made symmetric.If we did not integrate by parts, one of the trial functions in each equationwould be differentiated twice and the other trial function not differentiatedat all.3. Integration by parts introduces the gradient boundary conditions at theelement nodes. The physical significance of the gradient boundaryconditions becomes apparent in subsequent physical applications.Setting j = 1 for notational simplicity and substituting Equation 5.19 intoEquation 5.26 yieldsx2x2dy (e) dN 1dN 1dN 2N 1 (x ) f (x ) dx +y1+ y2dx =(5.27a)dxdxdxdx x1x1x2x1x1x2dy (e) dN 2dN 1dN 2N 2 (x ) f (x ) dx −y1+ y2 2 dx =dxdxdxdx x2which are of the form(5.27b)x1k 11k 21k 12k 22y1y2=F1F2The terms of the coefficient (element stiffness) matrix are defined byx2dN i dN jki j =dxi, j = 1, 2dx dx(5.28)(5.29)x1and the element nodal forces are given by the right-hand sides of Equation 5.27.If the described Galerkin procedure for element formulation is followed andthe system equations are assembled in the usual manner of the direct stiffnessmethod, the resulting system equations are identical in every respect to thoseobtained by the procedure represented by Equation 5.13.

It is important toobserve that, during the assembly process, when two elements are joined at acommon node as in Figure 5.5, for example, the assembled system equation forthe node contains a term on the right-hand side of the formdy (3) dy (4) −+(5.30)dx x4dx x4If the finite element solution were the exact solution, the first derivatives for eachelement indicated in expression 5.30 would be equal and the value of the expression would be zero. However, finite element solutions are seldom exact, so theseHutton: Fundamentals ofFinite Element Analysis5. Method of WeightedResidualsText© The McGraw−HillCompanies, 20045.3 The Galerkin Finite Element Method3x31454x4x5y(3)(x4) y(4)(x4)dy(4)dy(3)dx x4dx x4Figure 5.5 Two elementsjoined at a node.terms are not, in general, zero.

Nevertheless, in the assembly procedure, it isassumed that, at all interior nodes, the gradient terms appear as equal and opposite from the adjacent elements and thus cancel unless an external influence actsat the node. At global boundary nodes however, the gradient terms may be specified boundary conditions or represent “reactions” obtained via the solutionphase. In fact, a very powerful technique for assessing accuracy of finite elementsolutions is to examine the magnitude of gradient discontinuities at nodes or,more generally, interelement boundaries.EXAMPLE 5.5Use Galerkin’s method to formulate a linear finite element for solving the differentialequationxd2 ydy+− 4x = 0dx 2dx1≤x ≤2subject to y(1) = y(2) = 0 .■ SolutionFirst, note that the differential equation is equivalent toddyx− 4x = 0dxdxwhich, after two direct integrations and application of boundary conditions, has the exactsolutiony(x ) = x 2 −3ln x − 1ln 2For the finite element solution, the simplest approach is to use a two-node element forwhich the element solution is assumed asy(x ) = N 1 (x ) y1 + N 2 (x ) y2 =x2 − xx − x1y1 +y2x2 − x1x2 − x1where y1 and y2 are the nodal values.

The residual equation for the element isx 2Nix1ddyx− 4x dx = 0dxdxi = 1, 2Hutton: Fundamentals ofFinite Element Analysis1465. Method of WeightedResidualsCHAPTER 5Text© The McGraw−HillCompanies, 2004Method of Weighted Residualswhich becomes, after integration of the first term by parts,xx 2x 2dy 2dN i dyNi x−xdx − 4x N i dx = 0dx x 1dx dxx1i = 1, 2x1Substituting the element solution form and rearranging, we havex 2dN ixdxx1xx 2dN 1dN 2dy 2− 4x N i dxy1 +y2 dx = N i xdxdxdx x 1i = 1, 2x1Expanding the two equations represented by the last result after substitution for the interpolation functions and first derivatives yieldsx 21(x 2 − x 1 ) 2x ( y1 − y2 ) dx = −x 1x1x1x 21(x 2 − x 1 ) 2x1x 2dy x2 − x−4xdxdx x 1x2 − x1x 2dy x − x1x ( y2 − y1 ) dx = x 2−4xdxdx x 2x2 − x1x1Integration of the terms on the left reveals the element stiffness matrix ask (e) =x 22 − x 2112−12(x 2 − x 1 )−11while the gradient boundary conditions and nodal forces are evident on the right-handside of the equations.To illustrate, a two-element solution is formulated by taking equally spaced nodes atx = 1, 1.5, 2 as follows.Element 1x1 = 1x 2 = 1.51.5(1)F1= −4k = 2.5x1.5 − xdx = −1.166666 .

. .1.5 − 1xx−1dx = −1.33333 . . .1.5 − 111.5F(1)2= −41Element 2x 1 = 1.5x2 = 22(2)F1= −4k = 3.5x2−xdx = −1.66666 . . .2 − 1.5xx − 1.5dx = −1.83333 . . .2 − 1.51.52(2)F2= −41.5Hutton: Fundamentals ofFinite Element Analysis5. Method of WeightedResidualsText© The McGraw−HillCompanies, 20045.3 The Galerkin Finite Element MethodThe element equations are thendy −1.1667−dx x 1 2.5 −2.5y (1)1=(1)dy−2.5 2.5 y2 −1.3333 + 1.5 dx x2  −1.6667 − 1.5 dy  (2)  dxx23.5 −3.5y1=dy −3.5 3.5y (2)2 −1.8333 + 2 dx x3Denoting the system nodal values as Y1 , Y2 , Y3 at x = 1, 1.5, 2, respectively, the assembled system equations are  −1.1667 − dy   dx2.5 −2.50x1 Y1   −2.5−36−3.5  Y2 =  Y30−3.5 3.5 −1.8333 + 2 dy dx x3Applying the global boundary conditions Y1 = Y3 = 0, the second of the indicated equations gives Y2 = −0.5 and substitution of this value into the other two equations yieldsthe values of the gradients at the boundaries asdy = −2.4167dx x 1For comparison, the exact solution givesy(x = 1.5) = Y2 = −0.5049dy dx = 1.7917x3dy = −2.3281dx x 1dy = 1.8360dx x 3While the details will be left as an end-of-chapter problem, a four-element solutionfor this example (again, using equally spaced nodes x i ⇒ (1, 1.25, 1.5, 1.75, 2)) resultsin the global equations dy   −0.5417 −4.5 −4.5000Y1 dxx1 −4.510−5.500   Y2  −1.25−5.512−6.50  Y3 =−1.5 0  0Y4 0−6.514−7.5  −1.75  dy Y5000−7.5 7.5−0.9583+2dx x5Applying the boundary conditions Y1 = Y5 = 0 and solving the remaining 3 × 3 systemgives the resultsY2 = −0.4026Y3 = −0.5047Y4 = −0.3603dy = −2.350dx x 1dy = 1.831dx x5147Hutton: Fundamentals ofFinite Element Analysis1485.

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