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This is in accord with the assumption that loads are applied only atthe element nodes, as indicated by the bending moment diagram of a loadedbeam element shown in Figure 4.5. If a distributed load were applied to the element across its length, the bending moment would vary at least quadratically.Application of the boundary conditions 4.13–4.16 in succession yieldsv(x = 0) = v1 = a0(4.18)v(x = L ) = v2 = a0 + a1 L + a2 L + a3 Ldv = 1 = a1dx x=0dv = 2 = a1 + 2a2 L + 3a3 L 2dx 23x=LF2F1M11Mz2M2F1L M2 M1F1L M1xM1Figure 4.5 Bending moment diagram fora flexure element. Sign convention per thestrength of materials theory.(4.19)(4.20)(4.21)Hutton: Fundamentals ofFinite Element Analysis4. Flexure ElementsText© The McGraw−HillCompanies, 20044.3 Flexure ElementEquations 4.18–4.21 are solved simultaneously to obtain the coefficients in termsof the nodal variables asa0 = v1(4.22)a1 = 1(4.23)a2 =31(v2 − v1 ) − (21 + 2 )L2L(4.24)a3 =21(v1 − v2 ) + 2 (1 + 2 )3LL(4.25)Substituting Equations 4.22–4.25 into Equation 4.17 and collecting the coefficients of the nodal variables results in the expression3x 22x 32x 2x3v(x ) = 1 − 2 + 3 v1 + x −+ 2 1LLLL 23x2x 3x3x2+−v+−2(4.26)2L2L3L2Lwhich is of the formv(x ) = N 1 (x )v1 + N 2 (x )1 + N 3 (x )v2 + N 4 (x )2(4.27a)or, in matrix notation,v(x) = [N1N2N3 v 11N4 ]= [N ] {} v2 2(4.27b)where N 1 , N 2 , N 3 , and N 4 are the interpolation functions that describe the distribution of displacement in terms of nodal values in the nodal displacementvector {}.For the flexure element, it is convenient to introduce the dimensionlesslength coordinatex =(4.28)Lso that Equation 4.26 becomesv(x ) = (1 − 3 2 + 2 3 )v1 + L ( − 2 2 + 3 )1 + (3 2 − 2 3 )v2+ L 2 ( − 1)2(4.29)where 0 ≤ ≤ 1.

This form proves more amenable to the integrations requiredto complete development of the element equations in the next section.As discussed in Chapter 3, displacements are important, but the engineer ismost often interested in examining the stresses associated with given loadingconditions. Using Equation 4.11 in conjunction with Equation 4.27b, the normal97Hutton: Fundamentals ofFinite Element Analysis984. Flexure ElementsCHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure Elementsstress distribution on a cross section located at axial position x is given byx (x , y) = − y Ed2 [N ]{}dx 2(4.30)Since the normal stress varies linearly on a cross section, the maximum and minimum values on any cross section occur at the outer surfaces of the element,where distance y from the neutral surface is largest. As is customary, we take themaximum stress to be the largest tensile (positive) value and the minimum to bethe largest compressive (negative) value.

Hence, we rewrite Equation 4.30 asx (x ) = ymax Ed2 [N ]{}dx 2(4.31)and it is to be understood that Equation 4.31 represents the maximum and minimum normal stress values at any cross section defined by axial coordinate x. Alsoymax represents the largest distances (one positive, one negative) from the neutralsurface to the outside surfaces of the element. Substituting for the interpolationfunctions and carrying out the differentiations indicated, we obtain12x66x4612xx (x ) = ymax E− 2 v1 +−1 +− 3 v2L3LL2LL2L 6x2+−2(4.32)L2LObserving that Equation 4.32 indicates a linear variation of normal stress alongthe length of the element and since, once the displacement solution is obtained,the nodal values are known constants, we need calculate only the stress valuesat the cross sections corresponding to the nodes; that is, at x = 0 and x = L .

Thestress values at the nodal sections are given by62x (x = 0) = ymax E(v−v)−(2+)(4.33)2112L2L62x (x = L ) = ymax E(v−v)+(2+)(4.34)1221L2LThe stress computations are illustrated in following examples.4.4 FLEXURE ELEMENT STIFFNESS MATRIXWe may now utilize the discretized approximation of the flexure element displacement to examine stress, strain, and strain energy exhibited by the elementunder load.

The total strain energy is expressed as1Ue =x ε x dV(4.35)2VHutton: Fundamentals ofFinite Element Analysis4. Flexure ElementsText© The McGraw−HillCompanies, 20044.4Flexure Element Stiffness Matrixwhere V is total volume of the element. Substituting for the stress and strain perEquations 4.5 and 4.6, 2 2E2 d vUe =ydV(4.36)2dx 2Vwhich can be written asEUe =2L d2 vdx 22 y 2 dA dx(4.37)A0Again recognizing the area integral as the moment of inertia I z about the centroidal axis perpendicular to the plane of bending, we haveEI zUe =2L d2 vdx 22(4.38)dx0Equation 4.38 represents the strain energy of bending for any constant crosssection beam that obeys the assumptions of elementary beam theory.

For thestrain energy of the finite element being developed, we substitute the discretizeddisplacement relation of Equation 4.27 to obtainEI zUe =2L d2 N 2d2 N 3d2 N 4d2 N 1v++v+2112dx 2dx 2dx 2dx 22dx(4.39)0as the approximation to the strain energy. We emphasize that Equation 4.39 is anapproximation because the discretized displacement function is not in general anexact solution for the beam flexure problem.Applying the first theorem of Castigliano to the strain energy function withrespect to nodal displacement v1 gives the transverse force at node 1 as∂U e= F1 = EI z∂ v1L 0 2d2 N 1d N1d2 N 2d2 N 3d2 N 4v1 +1 +v2 +2dx2222dxdxdxdxdx 2(4.40)while application of the theorem with respect to the rotational displacementgives the moment as∂U e= M 1 = EI z∂ 1L 0 2d2 N 1d N2d2 N 2d2 N 3d2 N 4v1 +1 +v2 +2dx2222dxdxdxdxdx 2(4.41)99Hutton: Fundamentals ofFinite Element Analysis1004.

Flexure ElementsCHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure ElementsFor node 2, the results are∂U e= F2 = EI z∂ v2∂U e= M 2 = EI z∂ 2L 2d2 N 2d2 N 3d2 N 4d2 N 1d N3v++v+dx1122dx 2dx 2dx 2dx 2dx 20L (4.42) 2d2 N 1d N4d2 N 2d2 N 3d2 N 4v++v+dx1122dx 2dx 2dx 2dx 2dx 20(4.43)Equations 4.40–4.43 algebraically relate the four nodal displacement values tothe four applied nodal forces (here we use force in the general sense to includeapplied moments) and are of the form  F1 v1 k11 k12 k13 k14 k   21 k22 k23 k24  1 = M1(4.44) k31 k32 k33 k34   v2   F2   M22k41 k42 k43 k44where k mn , m, n = 1, 4 are the coefficients of the element stiffness matrix.

Bycomparison of Equations 4.40–4.43 with the algebraic equations represented bymatrix Equation 4.44, it is seen thatLkmn = knm = EI zd2 Nm d2 Nndxdx 2 dx 2m, n = 1, 4(4.45)0and the element stiffness matrix is symmetric, as expected for a linearly elasticelement.Prior to computing the stiffness coefficients, it is convenient to convert theintegration to the dimensionless length variable = x /L by notingL1f (x ) dx =0f ( ) L d(4.46)0d1 d=dxL d(4.47)so the integrations of Equation 4.45 becomeLk mn = k nm = EI z0d2 N m d2 N nEI zdx = 322dx dxL1d2 N m d2 N ndd 2 d 2m, n = 1, 40(4.48)Hutton: Fundamentals ofFinite Element Analysis4.

Flexure ElementsText© The McGraw−HillCompanies, 20044.4Flexure Element Stiffness MatrixThe stiffness coefficients are then evaluated as follows:k 11EI z= 3L10=36EI zL3k 12 = k 2136EI z(12 − 6) d =L31(4 2 − 4 + 1) d24−2+13EI z= 3L0=12EI zL31(12 − 6)(6 − 4) L d =6EI zL20k 13 = k 31EI z= 3L1(12 − 6)(6 − 12 ) d = −12EI zL30k 14 = k 41 =EI zL31(12 − 6)(6 − 2) L d =6EI zL20Continuing the direct integration gives the remaining stiffness coefficients ask 22 =4EI zL6EI zL22EI zk 42 =L12EI zL36EI zk 43 = − 3L4EI zLk 23 = k 32 = −k 24 =k 33 =k 34 =k 44 =The complete stiffness matrix for the flexure element is then written as126L−12 6LEI z 4L 2 −6L 2L 2 [ke ] = 3  6L(4.49)−12 −6L12 −6LL6L2L 2 −6L 4L 2Symmetry of the element stiffness matrix is apparent, as previously observed.Again, the element stiffness matrix can be shown to be singular since rigid bodymotion is possible unless the element is constrained in some manner.

The element stiffness matrix as given by Equation 4.49 is valid in any consistent systemof units provided the rotational degrees of freedom (slopes) are expressed inradians.101Hutton: Fundamentals ofFinite Element Analysis1024. Flexure ElementsCHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure Elements4.5 ELEMENT LOAD VECTORIn Equations 4.40–4.43, the element forces and moments were treated as requiredby the first theorem of Castigliano as being in the direction of the associated displacements.

These directions are in keeping with the assumed positive directionsof the nodal displacements. However, as depicted in Figures 4.6a and 4.6b, theusual convention for shear force and bending moment in a beam are such thatF−V1  1M1−M1⇒(4.50) F2  V2 M2M2In Equation 4.50, the column matrix (vector) on the left represents positivenodal forces and moments per the finite element formulations. The right-handside contains the corresponding signed shear forces and bending moments perthe beam theory sign convention.If two flexure elements are joined at a common node, the internal shearforces are equal and opposite unless an external force is applied at that node, inwhich case the sum of the internal shear forces must equal the applied load.Therefore, when we assemble the finite element model using flexure elements,the force at a node is simply equal to any external force at that node. A similarargument holds for bending moments.

At the juncture between two elements(i.e., a node), the internal bending moments are equal and opposite, thus selfequilibrating, unless a concentrated bending moment is applied at that node. Inthis event, the internal moments sum to the applied moment. These observationsare illustrated in Figure 4.6c, which shows a simply supported beam subjected toa concentrated force and concentrated moment acting at the midpoint of thePMcxVF1F2M1VM2(a)PMVMcMM(b)xx(c)Figure 4.6(a) Nodal load positive convention.

(b) Positive convention from the strength of materials theory.(c) Shear and bending moment diagrams depicting nodal load effects.Hutton: Fundamentals ofFinite Element Analysis4. Flexure ElementsText© The McGraw−HillCompanies, 20044.5 Element Load Vector103beam length. As shown by the shear force diagram, a jump discontinuity exists atthe point of application of the concentrated force, and the magnitude of the discontinuity is the magnitude of the applied force. Similarly, the bending momentdiagram shows a jump discontinuity in the bending moment equal to the magnitude of the applied bending moment.

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