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However, the following approach is more direct and algebraically simpler. Substitute Equation 6.14 into Equation 6.7 and equate to Equation 6.6 to obtain1000  0100 v 13231 ␪1v(x) = [1 x x 2 x 3 ] −−− L2LL2L v2  2␪2211− 3L3L2LL2 v 1␪1= [N1 N2 N3 N4 ](6.15) v2 ␪2The interpolation functions are[N1N2N3N4 ] = [ 1xx21 0 33 −x ] L2 2L3012−L1L2003L22− 3L001−L1L2(6.16)and note that the results of Equation 6.16 are identical to those shown inEquation 4.26.The reader may wonder why we repeat the development of the beam elementinterpolation functions.

The purpose is twofold: (1) to establish a general procedure for use with polynomial representations of the field variable and (2) to revisit the beam element formulation in terms of compatibility and completenessrequirements. The general procedure begins with expressing the field variable asa polynomial of order one fewer than the number of degrees of freedom exhibited by the element.

Using the examples of the truss and beam elements, it hasbeen shown that a two-node element may have 2 degrees of freedom, as in thetruss element where only displacement continuity is required, or 4 degrees offreedom, as in the beam element where slope continuity is required. Next thenodal (boundary) conditions are applied and the coefficients of the polynomialare computed accordingly. Finally, the polynomial coefficients are substitutedinto the field variable representation in terms of nodal variables to obtain theexplicit form of the interpolation functions.Examination of the completeness condition for the beam element requires amore-detailed thought process.

The polynomial representation of the displacement field is such that only the third derivative is guaranteed to have a constant169Hutton: Fundamentals ofFinite Element Analysis1706. Interpolation Functionsfor General ElementFormulationCHAPTER 6© The McGraw−HillCompanies, 2004TextInterpolation Functions for General Element Formulationvalue, since any lower-order derivative involves the spatial variable. However, ifwe examine the conditions under which the element undergoes rigid body translation, for example, we find that the nodal forces must be of equal magnitude andthe same sense and the applied nodal moments must be zero. Also, for rigid bodytranslation, the slopes at the nodes of the element are zero. In such case, the second derivative of deflection, directly proportional to bending moment, is zeroand the shear force, directly related to the third derivative of deflection, is constant.

(Simply recall the shear force and bending moment relations from themechanics of materials theory.) Therefore, the field variable representation as acubic polynomial allows rigid body translation. In the case of the beam element,we must also verify the possibility of rigid body rotation. This consideration, aswell as those of constant bending moment and shear force, is left for end-ofchapter problems.6.3.1 Higher-Order One-Dimensional ElementsIn formulating the truss element and the one-dimensional heat conduction element (Chapter 5), only line elements having a single degree of freedom at eachof two nodes are considered.

While quite appropriate for the problems considered, the linear element is by no means the only one-dimensional element thatcan be formulated for a given problem type. Figure 6.2 depicts a three-node lineelement in which node 2 is an interior node. As mentioned briefly in Chapter 1,an interior node is not connected to any other node in any other element in themodel. Inclusion of the interior node is a mathematical tool to increase the orderof approximation of the field variable. Assuming that we deal with only 1 degreeof freedom at each node, the appropriate polynomial representation of the fieldvariable is␾(x ) = a0 + a1 x + a2 x 2(6.17)and the nodal conditions are␾(x = 0) = ␾1L␾ x== ␾22(6.18)␾(x = L) = ␾3L21L223xFigure 6.2 A three-nodeline element.

Node 2 is aninterior node.Hutton: Fundamentals ofFinite Element Analysis6. Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004Text6.3 Polynomial Forms: One-Dimensional ElementsApplying the general procedure outlined previously in the context of the beamelement, we apply the nodal (boundary) conditions to obtain1 00 ␾1a0L L2  a11␾2 = (6.19)24  a␾321 L L2from which the interpolation functions are obtained via the following sequence100  341a0 ␾−  1−a1 =  L(6.20a)LL  ␾2a2 242  ␾3− 2L2LL2100 3 41 ␾−  12 −␾(x) = [1 x x ]  LLL  ␾2 242  ␾3−L2L2L2 ␾1= [N 1 N 2 N 3 ] ␾2(6.20b)␾332N 1 (x ) = 1 − x + 2 x 2LL4xxN 2 (x ) =1−(6.20c)LLx 2xN 3 (x ) =−1L LNote that each interpolation function varies quadratically in x and has valueof unity at its associated node and value zero at the other two nodes, as illustratedin Figure 6.3.

These observations lead to a shortcut method of concocting theinterpolation functions for a C 0 line element as products of monomials as follows. Let s = x /L such that s1 = 0, s2 = 1/2, s3 = 1 are the nondimensionalcoordinates of nodes 1, 2, and 3, respectively. Instead of following the formalprocedure used previously, we hypothesize, for example,N 1 (s) = C 1 (s − s2 )(s − s3 )(6.21)where C1 is a constant.

The first monomial term ensures that N1 has a value of zeroat node 2 and the second monomial term ensures the same at node 3. Therefore,we need to determine only the value of C1 to provide unity value at node 1.171Hutton: Fundamentals ofFinite Element Analysis1726. Interpolation Functionsfor General ElementFormulationCHAPTER 6© The McGraw−HillCompanies, 2004TextInterpolation Functions for General Element FormulationN21N30.80.60.4N10.20x⫺0.2Figure 6.3 Spatial variation of interpolationfunctions for a three-node line element.Substituting s = 0, we obtain1N 1 (s = 0) = 1 = C 1 0 −(0 − 1)2yielding C 1 = 2 and1N 1 (s) = 2 s −(s − 1)2(6.22)(6.23)Following similar logic and procedure shows thatN 2 (s) = −4s(s − 1)1N 3 (s) = 2s s −2(6.24)(6.25)Substituting s = x /L in Equations 6.23–6.25 and expanding shows that theresults are identical to those given in Equation 6.20. The monomial-based procedure can be extended to line elements of any order as illustrated by the followingexample.EXAMPLE 6.1Use the monomial method to obtain the interpolation functions for the four-node lineelement shown in Figure 6.4.L31L32L33Figure 6.4 Four-node lineelement of Example 6.1.4xHutton: Fundamentals ofFinite Element Analysis6.

Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004Text6.3 Polynomial Forms: One-Dimensional Elements■ SolutionUsing s = x /L , we have s1 = 0, s2 = 1/3, s3 = 2/3 , and s4 = 1 . The monomial terms ofinterest are s, s − 1/3, s − 2/3 , and s − 1 . The monomial products12N 1 (s) = C 1 s −s−(s − 1)332(s − 1)N 2 (s) = C 2 s s −31(s − 1)N 3 (s) = C 3 s s −321s−N 4 (s) = C 4 s s −33automatically satisfy the required zero-value conditions for each interpolation function.Hence, we need evaluate only the constants C i such that N i (s = si ) = 1, i = 1, 4 .

Applying each of the four unity-value conditions, we obtain12N 1 (0) = 1 = C 1 −−(−1)33 1112= 1 = C2−−N23333 1122= 1 = C3−N33333 12N 4 (1) = 1 = C 4 (1)3392from which C 1 = − , C 2 =27279, C3 = − , C4 = .222The interpolation functions are then given as9N 1 (s) = −21s−32s−3272s s−(s − 1)N 2 (s) =23271(s − 1)N 3 (s) = − s s −23291s−N 4 (s) = s s −233(s − 1)173Hutton: Fundamentals ofFinite Element Analysis1746. Interpolation Functionsfor General ElementFormulationCHAPTER 6© The McGraw−HillCompanies, 2004TextInterpolation Functions for General Element Formulation6.4 POLYNOMIAL FORMS:GEOMETRIC ISOTROPYThe previous discussion of one-dimensional (line) elements revealed that thepolynomial representation of the field variable must contain the same number ofterms as the number of nodal degrees of freedom.

In addition, to satisfy the completeness requirement, the polynomial representation for an M-degree of freedom element should contain all powers of the independent variable up to andincluding M − 1 . Another way of stating the latter requirement is that the polynomial is complete. In two and three dimensions, polynomial representations ofthe field variable, in general, satisfy the compatibility and completeness requirements if the polynomial exhibits the property known as geometric isotropy [1].

Amathematical function satisfies geometric isotropy if the functional form doesnot change under a translation or rotation of coordinates. In two dimensions, acomplete polynomial of order M can be expressed as(2)PM (x , y) =N ti+ j ≤Mak x i y j(6.26)k=0where N (2)t = [( M + 1)( M + 2)]/2 is the total number of terms. A completepolynomial as expressed by Equation 6.26 satisfies the condition of geometricisotropy, since the two variables, x and y, are included in each term in similarpowers.

Therefore, a translation or rotation of coordinates is not prejudicial toeither independent variable.A graphical method of depicting complete two-dimensional polynomials isthe so-called Pascal triangle shown in Figure 6.5. Each horizontal line representsa polynomial of order M. A complete polynomial of order M must contain allterms shown above the horizontal line. For example, a complete quadratic polynomial in two dimensions must contain six terms.

Hence, for use in a finite element representation of a field variable, a complete quadratic expression requiressix nodal degrees of freedom in the element. We examine this particular case inthe context of triangular elements in the next section.In addition to the complete polynomials, incomplete polynomials alsoexhibit geometric isotropy if the incomplete polynomial is symmetric. In thiscontext, symmetry implies that the independent variables appear as “equal and1xx2x3x4yx 2yx 3yLineary2xyxy 2x 2y 2Quadraticy3xy 3Cubicy4QuarticFigure 6.5 Pascal triangle for polynomialsin two dimensions.Hutton: Fundamentals ofFinite Element Analysis6. Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004Text6.4 Polynomial Forms: Geometric Isotropyopposite pairs,” ensuring that each independent variable plays an equal role inthe polynomial. For example, the four-term incomplete quadratic polynomialP (x , y) = a0 + a1 x + a2 y + a3 x 2(6.27)is not symmetric, as there is a quadratic term in x but the corresponding quadraticterm in y does not appear.

On the other hand, the incomplete quadratic polynomialP (x , y) = a0 + a1 x + a2 y + a3 x y(6.28)is symmetric, as the quadratic term gives equal “weight” to both variables.A very convenient way of visualizing some of the commonly used incomplete but symmetric polynomials of a given order is also afforded by the Pascaltriangle. Again referring to Figure 6.5, the dashed lines show the terms that mustbe included in an incomplete yet symmetric polynomial of a given order. (Theseare, of course, not the only incomplete, symmetric polynomials that can beconstructed.) All terms above the dashed lines must be included in a polynomialrepresentation if the function is to exhibit geometric isotropy.

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