Hutton - Fundamentals of Finite Element Analysis (523155), страница 61

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Note that, in this case, if you do not obtain a uniformflow field, you have made errors in either your formulation or your calculations.The horizontal boundaries are to be taken as fixed surfaces. The coordinates ofnode 3 are (1.5, 1).4 (0, 2)5(3, 2)3U1342Y2(3, 0)1 (0, 0)XFigure P8.118.12Now repeat Problem 8.11 with the inlet flow shown in Figure P8.12. Does thebasic finite element formulation change? Do you have to redefine geometry orelements? Your answer to this question will give you insight as to how to usefinite element software.

Once the geometry and elements have been defined,various problems can be solved by simply changing boundary conditions orforcing functions.4531342U12Figure P8.128.13Consider the flow situation depicted in Figure P8.13. Upstream, the flow isuniform. At a known point between the two solid walls, a source of constantstrength Q (volume per unit time) exists (via the action of a pump for example).How would the source be accounted for in a finite element formulation?(Examine the heat transfer analogy.)325Hutton: Fundamentals ofFinite Element Analysis3268.

Applications in FluidMechanicsCHAPTER 8Text© The McGraw−HillCompanies, 2004Applications in Fluid MechanicsQUFigure P8.138.14Reconsider Example 8.1 and assume the cylinder is a heating rod held at constantsurface temperature T0. The uniform inlet stream is at known temperatureTi < T0 . The horizontal boundaries are perfectly insulated and steady-stateconditions are assumed. In the context of finite element analysis, can the flowproblem and the heat transfer problem be solved independently?Hutton: Fundamentals ofFinite Element Analysis9.

Applications in SolidMechanicsText© The McGraw−HillCompanies, 2004C H A P T E R9Applicationsin Solid Mechanics9.1 INTRODUCTIONThe bar and beam elements discussed in Chapters 2–4 are line elements, as onlya single coordinate axis is required to define the element reference frame, hence,the stiffness matrices. As shown, these elements can be successfully used tomodel truss and frame structures in two and three dimensions.

For applicationof the finite element method to more general solid structures, the line elementsare of little use, however. Instead, elements are needed that can be used tomodel complex geometries subjected to various types of loading and constraintconditions.In this chapter, we develop the finite element equations for both two- andthree-dimensional elements for use in stress analysis of linearly elastic solids.The principle of minimum potential energy is used for the developments, asthat principle is somewhat easier to apply to solid mechanics problems thanGalerkin’s method.

It must be emphasized, however, that Galerkin’s method isthe more general procedure and applicable to a wider range of problems.The constant strain triangle for plane stress is considered first, as the CST isthe simplest element to develop mathematically. The procedure is shown to becommon to other elements as well; a rectangular element formulated for planestrain is used to illustrate this commonality.

Plane quadrilateral, axisymmetric,and general three-dimensional elements are also examined. An approach forapplication of the finite element method to solving torsion problems of noncircular sections is also presented.327Hutton: Fundamentals ofFinite Element Analysis3289. Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid Mechanics9.2 PLANE STRESSA commonly occurring situation in solid mechanics, known as plane stress, isdefined by the following assumptions in conjunction with Figure 9.1:1.

The body is small in one coordinate direction (the z direction byconvention) in comparison to the other dimensions; the dimension in thez direction (hereafter, the thickness) is either uniform or symmetric aboutthe xy plane; thickness t, if in general, is less than one-tenth of the smallestdimension in the xy plane, would qualify for “small.”2. The body is subjected to loading only in the xy plane.3.

The material of the body is linearly elastic, isotropic, and homogeneous.The last assumption is not required for plane stress but is utilized in this text aswe consider only elastic deformations.Given a situation that satisfies the plane stress assumptions, the only nonzerostress components are ␴x , ␴y , and ␶x y . Note that the nominal stresses perpendicular to the xy plane (␴z , ␶x z , ␶ yz ) are zero as a result of the plane stress assumptions. Therefore, the equilibrium equations (Appendix B) for plane stress are∂ ␴x∂ ␶x y+=0∂x∂x∂ ␴y∂ ␶x y+=0∂y∂y(9.1)where we implicitly assume that ␶x y = ␶ yx .

Utilizing the elastic stress-strainrelations from Appendix B, Equation B.12 with ␴z = ␶x z = ␶ yz = 0, the nonzerostress components can be expressed as (Problem 9.1)␴x =E(ε x + ␯ε y )1 − ␯2␴y =E(ε y + ␯ε x )1 − ␯2␶x y =E␥x y = G ␥x y2(1 + ␯)(9.2)where E is the modulus of elasticity and ␯ is Poisson’s ratio for the material.In the shear stress-strain relation, the shear modulus G = E /2(1 + ␯) has beenintroduced.The stress-strain relations given by Equation 9.2 can be conveniently writtenin matrix form:1 ␯0␴εx  x E ␯ 10  ε␴y =(9.3)y1 − ␯  ␥  ␶  1 − ␯2 xyxy0 02Hutton: Fundamentals ofFinite Element Analysis9.

Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.2 Plane StressyyF2zxF1f (x, y)tFigure 9.1 An illustration of plane stressconditions.or{␴} = [D]{ε}where(9.4)  ␴x {␴} = ␴y␶ xyis the column matrix of stress components,1 ␯E ␯1[D] =1 − ␯2 0 0(9.5)00 1−␯ 2is the elastic material property matrix for plane stress, and εx {ε} = ε y␥ xy(9.6)(9.7)is the column matrix of strain components.For a state of plane stress, the strain energy per unit volume, Equation 2.43,becomes1u e = (␴x ε x + ␴y ε y + ␶x y ␥x y )(9.8)2or, using the matrix notation,ue =1 T1{ε} {␴} = {ε} T [D]{ε}22(9.9)Use of {ε} T allows the matrix operation to reproduce the quadratic form of thestrain energy.

Note that a quadratic relation in any variable z can be expressed329Hutton: Fundamentals ofFinite Element Analysis3309. Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid Mechanicsas {z} T [ A]{z} , where [ A] is a coefficient matrix.

This is the subject of an end-ofchapter problem.The total strain energy of a body subjected to plane stress is then1Ue ={ε} T [D]{ε} dV(9.10)2Vwhere V is total volume of the body and dV = t dx dy . The form of Equation 9.10will in fact be found to apply in general and is not restricted to the case of planestress.

In other situations, the strain components and material property matrixmay be defined differently, but the form of the strain energy expression does notchange. We use this result extensively in applying the principle of minimumpotential energy in following developments.9.2.1 Finite Element Formulation: Constant Strain TriangleFigure 9.2a depicts a three-node triangular element assumed to represent a subdomain of a body subjected to plane stress.

Element nodes are numbered asshown, and nodal displacements in the x-coordinate direction are u 1 , u 2 , and u 3 ,while displacements in the y direction are v1, v2, and v3. (For plane stress, displacement in the z direction is neglected). As noted in the introduction, the displacement field in structural problems is a vector field and must be discretizedaccordingly. For the triangular element in plane stress, we write the discretizeddisplacement field asu(x , y) = N 1 (x , y)u 1 + N 2 (x , y)u 2 + N 3 (x , y)u 3 = [N ]{u}v(x , y) = N 1 (x , y)v1 + N 2 (x , y)v2 + N 3 (x , y)v3 = [N ]{v}(9.11)where N 1 , N 2 , and N 3 are the interpolation functions as defined in Equation 6.37.Using the discretized representation of the displacement field, the element strainv33f3yu3f3xv2f2yv11f1y2u1u2f2xf1x(a)(b)Figure 9.2(a) Nodal displacement notation for a planestress element.

(b) Nodal forces.Hutton: Fundamentals ofFinite Element Analysis9. Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.2 Plane Stresscomponents are thenεx =∂u∂ N1∂ N2∂ N3=u1 +u2 +u3∂x∂x∂x∂xεy =∂v∂ N1∂ N2∂ N3=v1 +v2 +v3∂y∂y∂y∂y␥x y =(9.12)∂u∂v∂ N1∂ N2∂ N3∂ N1∂ N2∂ N3+=u1 +u2 +u3 +v1 +v2 +v3∂y∂x∂y∂y∂y∂x∂x∂xDefining the element displacement column matrix (vector) as u1  u2   (e)  u 3 ␦=v1  v2 v3the element strain matrix can be expressed as∂ N1 ∂ N2 ∂ N3000 ∂x∂x∂x∂ N1 ∂ N2 ∂ N300{ε} =  0∂y∂y∂y ∂ N1 ∂ N2 ∂ N3 ∂ N1 ∂ N2 ∂ N3∂y∂y∂y∂x∂x∂x(9.13) u 1u  2  u3= [B] ␦(e)v1  v2 v3(9.14)where [B] is the 3 × 6 matrix of partial derivatives of the interpolation functionsas indicated, also known as the strain-displacement matrix.

Referring to Equation 6.37, we observe that the partial derivatives appearing in Equation 9.14 areconstants, since the interpolation functions are linear in the spatial variables.Hence, the strain components are constant throughout the volume of the element.Consequently, the three-node, triangular element for plane stress is known as aconstant strain triangle.By direct analogy with Equation 9.10, the elastic strain energy of the element is (e) T 11(e)T(e){ε} [D]{ε} dV =U e =␦[B] T [D][ B] ␦(e) dV (e)22(9.15)V (e)V (e)As shall be seen in subsequent examples, Equation 9.15 is a generally applicablerelation for the elastic strain energy of structural elements.

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