Hutton - Fundamentals of Finite Element Analysis (523155), страница 28
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Method of WeightedResidualsCHAPTER 5Text© The McGraw−HillCompanies, 2004Method of Weighted Residuals1.2501.51.752.00.10.20.30.4ExactTwo elementsFour elements0.50.6Figure 5.6 Two-element, four-element, and exactsolutions to Example 5.5.For comparison, the exact, two-element, and four-element solutions are shown in Figure 5.6.
The two-element solution is seen to be a crude approximation except at theelement nodes and derivative discontinuity is significant. The four-element solution hasthe computed values of y(x) at the nodes being nearly identical to the exact solution. Withfour elements, the magnitudes of the discontinuities of first derivatives at the nodes arereduced but still readily apparent.5.4 APPLICATION OF GALERKIN’S METHODTO STRUCTURAL ELEMENTS5.4.1 Spar ElementReconsidering the elastic bar or spar element of Chapter 2 and recalling that thebar is a constant strain (therefore, constant stress) element, the applicable equilibrium equation is obtained using Equations 2.29 and 2.30 asdxdd2 u(x )=( E εx ) = E=0dxdxdx 2(5.31)where we assume constant elastic modulus. Denoting element length by L, thedisplacement field is discretized by Equation 2.17:xxu(x ) = u 1 N 1 (x ) + u 2 N 2 (x ) = u 1 1 −+ u2(5.32)LLHutton: Fundamentals ofFinite Element Analysis5.
Method of WeightedResidualsText© The McGraw−HillCompanies, 20045.4 Application of Galerkin’s Method to Structural ElementsAnd, since the domain of interest is the volume of the element, the Galerkinresidual equations becomeLd2 ud2 udV =A dx = 0i = 1, 2N i (x ) ENi E(5.33)dx 2dx 2V0where dV = A dx and A is the constant cross-sectional area of the element.Integrating by parts and rearranging, we obtainLLdu dN i duAEdx = N i AE (5.34)dx dxdx 00which, utilizing Equation 5.32, becomesLAE0LAEdu dN 1 d= − AE ε| x=0 = − A| x=0(u 1 N 1 + u 2 N 2 ) dx = − AEdx dxdx x=0(5.35a)dN2 ddu (u 1 N1 + u 2 N2 ) dx = AE dx dxdx 0= AEε|x = L = Ax = Lx=L(5.35b)From the right sides of Equation 5.35, we observe that, for the bar element, thegradient boundary condition simply represents the applied nodal force since A = F.Equation 5.35 is readily combined into matrix form asdN1 dN1 dN1 dN2L dx dx uF1 dx dxAE (5.36)dx 1 =u2F2 dN1 dN2 dN2 dN2 0dx dxdx dxwhere the individual terms of the matrix are integrated independently.Carrying out the indicated differentiations and integrations, we obtain A E 1 −1u1F1=(5.37)−11uF2L2which is the same result as obtained in Chapter 2 for the bar element.
This simply illustrates the equivalence of Galerkin’s method and the methods of equilibrium and energy (Castigliano) used earlier for the bar element.5.4.2 Beam ElementApplication of the Galerkin method to the beam element begins with consideration of the equilibrium conditions of a differential section taken along the149Hutton: Fundamentals ofFinite Element Analysis1505. Method of WeightedResidualsCHAPTER 5Text© The McGraw−HillCompanies, 2004Method of Weighted Residualsq(x)yMMxVdxVdMdxdxdVdxdxFigure 5.7 Differential section of a loaded beam.longitudinal axis of a loaded beam as depicted in Figure 5.7 where q(x) represents a distributed load expressed as force per unit length.
Whereas q may varyarbitrarily, it is assumed to be constant over a differential length dx . The condition of force equilibrium in the y direction isdV−V + V +dx + q (x ) dx = 0(5.38)dxfrom whichdV= −q (x )dxMoment equilibrium about a point on the left face is expressed asdMdVdxM+dx − M + V +dx dx + [q (x ) dx ]=0dxdx2(5.39)(5.40)which (neglecting second-order differentials) givesdM= −Vdx(5.41)Combining Equations 5.39 and 5.41, we obtaind2 M= q (x )dx 2(5.42)Recalling, from the elementary strength of materials theory, the flexure formulacorresponding to the sign conventions of Figure 5.7 isM = E Izd2 vdx 2(5.43)(where in keeping with the notation of Chapter 4, v represents displacement inthe y direction), which in combination with Equation 5.42 provides the governing equation for beam flexure asd2d2 vE I z 2 = q (x )(5.44)dx 2dxHutton: Fundamentals ofFinite Element Analysis5. Method of WeightedResidualsText© The McGraw−HillCompanies, 20045.4 Application of Galerkin’s Method to Structural ElementsGalerkin’s finite element method is applied by taking the displacement solution in the formv(x ) = N 1 (x )v1 + N 2 (x )1 + N 3 (x )v2 + N 4 (x )2 =4N i (x )i(5.45)i=1as in Chapter 4, using the interpolation functions of Equation 4.26.
Therefore, theelement residual equations arex2d2 vd2N i (x )EI z 2 − q (x ) dx = 0dx 2dxi = 1, 4(5.46)x1Integrating the derivative term by parts and assuming a constant EIz, we obtainxx2x2d3 v 2dN i d3 vN i (x ) EI z 3 − EI zdx −N i q (x ) dx = 0i = 1, 4 (5.47)dx x1dx dx 3x1x1and sinceV =−dMd=−dxdxEI zd2 vdx 2= − EI zd3 vdx 3(5.48)we observe that the first term of Equation 5.47 represents the shear force conditions at the element nodes. Integrating again by parts and rearranging givesx2EI zd2 N i d2 vdx =dx 2 dx 2x1x2x1xd3 v 2N i q (x ) dx − N i EI z 3 dx x1xdN id2 v 2+EI z 2 dxdx x1i = 1, 4(5.49)and, per Equation 5.43, the last term on the right introduces the moment conditions at the element boundaries. Integration by parts was performed twice in thepreceding development for reasons similar to those mentioned in the context ofthe bar element.
By so doing, the order of the two derivative terms appearing inthe first integral in Equation 5.49 are the same, and the resulting stiffness matrixis thus symmetric, and the shear forces and bending moments at element nodesnow explicitly appear in the element equations.Equation 5.49 can be written in the matrix form [k]{} = {F } where theterms of the stiffness matrix are defined byx2k i j = EI zx1d2 N i d2 N jdxdx 2 dx 2i, j = 1, 4(5.50)151Hutton: Fundamentals ofFinite Element Analysis1525.
Method of WeightedResidualsCHAPTER 5Text© The McGraw−HillCompanies, 2004Method of Weighted Residualswhich is identical to results previously obtained by other methods. The terms ofthe element force vector are defined byxxx2d3 v 2 dN id2 v 2Fi =N i q (x ) dx − N i EI z 3 +EI z 2 i = 1, 4(5.51a)dx x1dxdx x1x1or, using Equations 5.43 and 5.48,x2dN iFi =N i q (x ) dx + N i V (x )| xx21 +M (x )| xx21dxi = 1, 4(5.51b)x1where the integral term represents the equivalent nodal forces and moments produced by the distributed load.
If q (x ) = q = constant (positive upward), substitution of the interpolation functions into Equation 5.51 gives the element nodalforce vector asqL−V1 22qL−M1 12{F} =(5.52) qL+V222qL+M−2 12Where two beam elements share a common node, one of two possibilitiesoccurs regarding the shear and moment conditions:1. If no external force or moment is applied at the node, the shear andmoment values of Equation 5.52 for the adjacent elements are equal andopposite, cancelling in the assembly step.2. If a concentrated force is applied at the node, the sum of the boundaryshear forces for the adjacent elements must equal the applied force.Similarly, if a concentrated moment is applied, the sum of the boundary bendingmoments must equal the applied moment.
Equation 5.52 shows that the effects ofa distributed load are allocated to the element nodes. Finite element softwarepackages most often allow the user to specify a “pressure” on the transverse faceof the beam. The specified pressure actually represents a distributed load and isconverted to the nodal equivalent loads in the software.5.5 ONE-DIMENSIONAL HEAT CONDUCTIONApplication of the Galerkin finite method to the problem of one-dimensional,steady-state heat conduction is developed with reference to Figure 5.8a, whichdepicts a solid body undergoing heat conduction in the direction of the x axisHutton: Fundamentals ofFinite Element Analysis5.
Method of WeightedResidualsText© The McGraw−HillCompanies, 20045.5 One-Dimensional Heat Conductionqx(xa)Insulatedaqx(xb)bx(a)Aqx qxdqxdxdxdx(b)Figure 5.8 Insulated body in one-dimensional heatconduction.only. Surfaces of the body normal to the x axis are assumed to be perfectlyinsulated, so that no heat loss occurs through these surfaces. Figure 5.8b showsthe control volume of differential length dx of the body, which is assumed to beof constant cross-sectional area and uniform material properties. The principle ofconservation of energy is applied to obtain the governing equation as follows:E in + E generated = E increase + E out(5.53)Equation 5.53 states that the energy entering the control volume plus energy generated internally by any heat source present must equal the increase in internalenergy plus the energy leaving the control volume.
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