Hutton - Fundamentals of Finite Element Analysis (523155), страница 25
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The supports at A and B arecompletely fixed. Other connections are welded. Assuming the structure to bemodeled using the minimum number of beam-axial elements:WABFigure P4.23a.b.c.d.How many elements are needed?What is the size of the assembled global stiffness matrix?What are the constraint (boundary) conditions?What is the size of the reduced global stiffness matrix after application of theconstraint conditions?e. Assuming a finite element solution is obtained for this problem, what stepscould be taken to judge the accuracy of the solution?129Hutton: Fundamentals ofFinite Element Analysis1304.
Flexure ElementsCHAPTER 44.24Text© The McGraw−HillCompanies, 2004Flexure ElementsRepeat Problem 4.23 for the frame structure shown in Figure P4.24.WB(pin joint)AFigure P4.244.254.26Verify Equation 4.69 by direct calculation.The cantilevered beam depicted in Figure P4.26 is subjected to two-planebending. The loads are applied such that the planes of bending correspond to theprincipal moments of inertia.
Noting that no axial or torsional loadings arepresent, model the beam as a single element (that is, construct the 8 × 8 stiffnessmatrix containing bending terms only) and compute the deflections of the freeend, node 2. Determine the exact location and magnitude of the maximumbending stress. (Use E = 207 GPa .)y1.5 m500 Nyzx300 Nz6 cm3 cmFigure P4.264.27Repeat Problem 4.26 for the case in which the concentrated loads are replacedby uniform distributed loads q y = 6 N/cm and q z = 4 N/cm acting in the positivecoordinate directions, respectively.Hutton: Fundamentals ofFinite Element Analysis5. Method of WeightedResidualsText© The McGraw−HillCompanies, 2004C H A P T E R5Method of WeightedResiduals5.1 INTRODUCTIONChapters 2, 3, and 4 introduced some of the basic concepts of the finite elementmethod in terms of the so-called line elements. The linear elastic spring, the barelement and the flexure element are line elements because structural propertiescan be described in terms of a single spatial variable that identifies position alongthe longitudinal axis of the element.
The displacement-force relations for the lineelements are straightforward, as these relations are readily described using onlythe concepts of elementary strength of materials. To extend the method of finiteelement analysis to more general situations, particularly nonstructural applications, additional mathematical techniques are required. In this chapter, themethod of weighted residuals is described in general and Galerkin’s method ofweighted residuals [1] is emphasized as a tool for finite element formulation foressentially any field problem governed by a differential equation.5.2 METHOD OF WEIGHTED RESIDUALSIt is a basic fact that most practical problems in engineering are governed bydifferential equations.
Owing to complexities of geometry and loading, rarelyare exact solutions to the governing equations possible. Therefore, approximatetechniques for solving differential equations are indispensable in engineeringanalysis. Indeed, the finite element method is such a technique. However, thefinite element method is based on several other, more-fundamental, approximatetechniques, one of which is discussed in detail in this section and subsequentlyapplied to finite element formulation.The method of weighted residuals (MWR) is an approximate technique forsolving boundary value problems that utilizes trial functions satisfying the131Hutton: Fundamentals ofFinite Element Analysis1325.
Method of WeightedResidualsCHAPTER 5Text© The McGraw−HillCompanies, 2004Method of Weighted Residualsprescribed boundary conditions and an integral formulation to minimize error, inan average sense, over the problem domain. The general concept is describedhere in terms of the one-dimensional case but, as is shown in later chapters,extension to two and three dimensions is relatively straightforward. Given adifferential equation of the general formD[y(x ), x ] = 0a<x <b(5.1)subject to homogeneous boundary conditionsy(a) = y(b) = 0(5.2)the method of weighted residuals seeks an approximate solution in the formny*(x ) =ci N i (x )(5.3)i=1where y* is the approximate solution expressed as the product of ci unknown,constant parameters to be determined and N i (x ) trial functions. The majorrequirement placed on the trial functions is that they be admissible functions;that is, the trial functions are continuous over the domain of interest and satisfythe specified boundary conditions exactly.
In addition, the trial functions shouldbe selected to satisfy the “physics” of the problem in a general sense. Given thesesomewhat lax conditions, it is highly unlikely that the solution represented byEquation 5.3 is exact. Instead, on substitution of the assumed solution into thedifferential Equation 5.1, a residual error (hereafter simply called residual)results such thatR(x ) = D[y*(x ), x ] = 0(5.4)where R(x) is the residual. Note that the residual is also a function of theunknown parameters ci. The method of weighted residuals requires that theunknown parameters ci be evaluated such thatbwi (x ) R(x ) dx = 0i = 1, n(5.5)awhere wi (x ) represents n arbitrary weighting functions.
We observe that, onintegration, Equation 5.5 results in n algebraic equations, which can be solved forthe n values of ci. Equation 5.5 expresses that the sum (integral) of the weightedresidual error over the domain of the problem is zero. Owing to the requirementsplaced on the trial functions, the solution is exact at the end points (the boundaryconditions must be satisfied) but, in general, at any interior point the residualerror is nonzero. As is subsequently discussed, the MWR may capture the exactsolution under certain conditions, but this occurrence is the exception rather thanthe rule.Several variations of MWR exist and the techniques vary primarily in howthe weighting factors are determined or selected.
The most common techniquesare point collocation, subdomain collocation, least squares, and Galerkin’sHutton: Fundamentals ofFinite Element Analysis5. Method of WeightedResidualsText© The McGraw−HillCompanies, 20045.2 Method of Weighted Residuals133method [1]. As it is quite simple to use and readily adaptable to the finite elementmethod, we discuss only Galerkin’s method.In Galerkin’s weighted residual method, the weighting functions are chosento be identical to the trial functions; that is,wi (x ) = N i (x )i = 1, nTherefore, the unknown parameters are determined viabbwi (x ) R(x ) dx = N i (x ) R(x ) = 0i = 1, na(5.6)(5.7)aagain resulting in n algebraic equations for evaluation of the unknown parameters.
The following examples illustrate details of the procedure.EXAMPLE 5.1Use Galerkin’s method of weighted residuals to obtain an approximate solution of thedifferential equationd2 y− 10x 2 = 5dx 20≤x ≤1with boundary conditions y(0) = y(1) = 0.■ SolutionThe presence of the quadratic term in the differential equation suggests that trial functionsin polynomial form are suitable.
For homogeneous boundary conditions at x = a andx = b, the general formN (x ) = (x − x a ) p (x − x b ) qwith p and q being positive integers greater than zero, automatically satisfies the boundary conditions and is continuous in x a ≤ x ≤ x b . Using a single trial function, the simplest such form that satisfies the stated boundary conditions isN 1 (x ) = x (x − 1)Using this trial function, the approximate solution per Equation 5.3 isy*(x ) = c1 x (x − 1)and the first and second derivatives aredy*= c1 (2x − 1)dxd2 y*= 2c1dx 2respectively. (We see, at this point, that the selected trial solution does not satisfy thephysics of the problem, since we have obtained a constant second derivative. The differential equation is such that the second derivative must be a quadratic function of x.
Nevertheless, we continue the example to illustrate the procedure.)Hutton: Fundamentals ofFinite Element Analysis1345. Method of WeightedResidualsCHAPTER 5Text© The McGraw−HillCompanies, 2004Method of Weighted ResidualsSubstitution of the second derivative of y*(x) into the differential equation yields theresidual asR(x ; c1 ) = 2c1 − 10x 2 − 5which is clearly nonzero. Substitution into Equation 5.7 gives1x (x − 1)(2c1 − 10x 2 − 5) dx = 00which after integration yields c1 = 4, so the approximate solution is obtained asy*(x ) = 4x (x − 1)For this relatively simple example, we can compare the approximate solution result withthe exact solution, obtained by integrating the differential equation twice as follows:dy=dxy(x ) =d2 ydx =dx 2dydx =dx(10x 2 + 5) dx = 10x 3+ 5x + C 135x 45x 210x 3+ 5x + C 1 dx =++ C1 x + C2362Applying the boundary condition y(0) = 0 gives C2 = 0, while the condition y(1) = 0becomes55+ + C1 = 062from which C 1 = −10/3 .
Hence, the exact solution is given byy(x ) =000.10.25 4510x + x2 −x6230.30.4x0.50.60.20.4y(x) 0.60.81.0ExactGalerkin1.2Figure 5.1 Solutions to Example 5.1.0.70.80.91.0Hutton: Fundamentals ofFinite Element Analysis5. Method of WeightedResidualsText© The McGraw−HillCompanies, 20045.2 Method of Weighted Residuals135A graphical comparison of the two solutions is depicted in Figure 5.1, which shows thatthe approximate solution is in reasonable agreement with the exact solution. However,note that the one-term approximate solution is symmetric over the interval of interest.That this is not correct can be seen by examining the differential equation.
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