Hutton - Fundamentals of Finite Element Analysis (523155), страница 67

Просмтор этого файла доступен только зарегистрированным пользователям. Но у нас супер быстрая регистрация: достаточно только электронной почты!

Текст из файла (страница 67)

Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.6 General Three-Dimensional Stress ElementsIn Equation 9.114, the strain-displacement matrix is given by∂00 ∂x∂ 00 ∂y∂  0 [N ] [0] [0]0∂z [B] = [L][N3 ] =  ∂ [0] [N ] [0]∂ [0] [0] [N ]0 ∂y ∂x ∂∂ 0 ∂z∂z ∂∂ 0∂z ∂ y(9.115)and is observed to be a 6 × 3M matrix composed of the first partial derivativesof the interpolation functions.Application of the principle of minimum potential energy to Equation 9.114yields, in analogy with Equation 9.22,[B]T [D][B] dV {␦} = { f }(9.116)Vas the system of nodal equilibrium equation for a general three-dimensional stresselement.

From Equation 9.116, we identify the element stiffness matrix as[k] =[B] T [D][ B] dV(9.117)Vand the element stiffness matrix so defined is a 3M × 3M symmetric matrix,as expected for a linear elastic element. The integrations indicated in Equation 9.117 depend on the specific element type in question. For a four-node,linear tetrahedral element (Section 6.7), all the partial derivatives of the volumecoordinates are constants, so the strains are constant—this is the 3-D analogy toa constant strain triangle in two dimensions.

In the linear tetrahedral element, theterms of the [B] matrix are constant and the integrations reduce to a constantmultiple of element volume.If the element to be developed is an eight-node brick element, the interpolationfunctions, Equation 6.69, are such that strains vary linearly and the integrandsin Equation 9.117 are not constant. The integrands are polynomials in the spatialvariables, however, and therefore amenable to exact integration by Gaussian quadrature in three dimensions.

Similarly, for higher-order elements, the integrationsrequired to formulate the stiffness matrix are performed numerically.The eight-node brick element can be transformed into a generally shapedparallelopiped element using the isoparametric procedure discussed in Section 6.8. If the eight-node element is used as the parent element, the resulting367Hutton: Fundamentals ofFinite Element Analysis3689. Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid Mechanicsisoparametric element has planar faces and is analogous to the two-dimensionalquadrilateral element.

If the parent element is of higher-order interpolation functions, an element with general (curved) surfaces results.Regardless of the specific element type or types used in a three-dimensionalfinite element analysis, the procedure for assembling the global equilibrium equations is the same as discussed several times, so we do not belabor the point here.As in previous developments, the assembled global equations are of the form[K ]{} = {F }(9.118)with [K] representing the assembled global stiffness matrix, {} representing thecolumn matrix of global displacements, and {F } representing the column matrixof applied nodal forces.

The nodal forces may include directly applied externalforces at nodes, the work-equivalent nodal forces corresponding to body forcesand forces arising from applied pressure on element faces.9.7 STRAIN AND STRESS COMPUTATIONUsing the stiffness method espoused in this text, the solution phase of a finiteelement analysis results in the computation of unknown nodal displacements aswell as reaction forces at constrained nodes.

Computation of strain components,then stress components, is a secondary (postprocessing) phase of the analysis.Once the displacements are known, the strain components (at each node in themodel) are readily computed using Equation 9.104, which, given the discretization in the finite element context, becomes u{ε} = [L]vw= [L][N3 ]{␦} = [B]{␦}(9.119)It must be emphasized that Equation 9.119 represents the calculation of straincomponents for an individual element and must be carried out for every elementin the finite element model. However, the computation is straightforward, sincethe [B] matrix has been computed for each element to determine the elementstiffness matrix, hence the element contributions to the global stiffness matrix.Similarly, element stress components are computed as{␴} = [D][ B]{␦}(9.120)and the material property matrix [D] depends on the state of stress, as previouslydiscussed.

Equations 9.119 and 9.120 are general in the sense that the equationsare valid for any state of stress if the strain-displacement matrix [B] and thematerial property matrix [D] are properly defined for a particular state of stress. (Inthis context, recall that we consider only linearly elastic deformation in this text.)The element strain and stress components, as computed, are expressed inthe element coordinate system.

In general, for the elements commonly used inHutton: Fundamentals ofFinite Element Analysis9. Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.7 Strain and Stress Computationstress analysis, the coordinate system for each element is the same as the globalcoordinate system.

It is a fact of human nature, especially of engineers, that weselect the simplest frame in which to describe a particular occurrence or event.This is a way of saying that we tend to choose a coordinate system for convenience and that convenience is most often related to the geometry of the problemat hand. The selected coordinate system seldom, if ever, corresponds to maximum loading conditions. Specifically, if we consider the element stress calculation represented by Equation 9.120, the stress components are referred, andcalculated with reference, to a specified Cartesian coordinate system. To determine the critical loading on any model, we must apply one of the so-called failure theories.

As we limit the discussion to linearly elastic behavior, the “failure”in our context is yielding of the material. There are several commonly acceptedfailure theories for yielding in a general state of stress. The two most commonlyapplied are the maximum shear stress theory and the distortion energy theory. Wediscuss each of these briefly. In a general, three-dimensional state of stress, theprincipal stresses ␴1 , ␴2 , and ␴3 are given by the roots of the cubic equation represented by the determinant [2] ␴x − ␴␶x y␶x z ␶x y(9.121)␴y − ␴␶ yz = 0 ␶x z␶ yz␴z − ␴ Customarily, the principal stresses are ordered so that ␴1 > ␴2 > ␴3 . Via theusual convention, a positive normal stress corresponds to tension, while a negative normal stress is compressive.

So, while ␴3 is algebraically the smallest of thethree principal stresses, it may represent a compressive stress having significantly large magnitude. Also recall that the principal stresses occur on mutuallyorthogonal planes (the principal planes) and the shear stress components onthose planes are zero.Having computed the principal stress components, the maximum shearstress is|␴1 − ␴2 | |␴1 − ␴3 | |␴2 − ␴3 |␶max = largest of,,(9.122)222The three shear stress components in Equation 9.122 are known to occur onplanes oriented 45◦ from the principal planes.The maximum shear stress theory (MSST) holds that failure (yielding) in ageneral state of stress occurs when the maximum shear stress as given by Equation 9.122 equals or exceeds the maximum shear stress occurring in a uniaxialtension test at yielding.

It is quite easy to show that the maximum shear stressin a tensile test at yielding has value equal to one-half the tensile yield strengthof the material. Hence, the failure value in the MSST is ␶max = S y /2 = S ys . Inthis notation, Sy is tensile yield strength and S ys represents yield strength in shear.The distortion energy theory (DET) is based on the strain energy stored in amaterial under a given state of stress.

The theory holds that a uniform tensile or369Hutton: Fundamentals ofFinite Element Analysis3709. Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid Mechanicscompressive state of stress (also known as hydrostatic stress) does not cause distortion and, hence, does not contribute to yielding. If the principal stresses havebeen computed, total elastic strain energy is given by1Ue =(␴1 ε1 + ␴2 ε2 + ␴3 ε3 ) dV2V=1 2␴1 + ␴22 + ␴32 − 2␯(␴1 ␴2 + ␴1 ␴3 + ␴2 ␴3 ) V2E(9.123)To arrive at distortion energy, the average (hydrostatic) stress is defined as␴av =␴1 + ␴2 + ␴33(9.124)and the corresponding strain energy isU hyd =23␴av(1 − 2␯)V2E(9.125)The distortion energy is then defined asU d = U e − U hyd(9.126)After a considerable amount of algebraic manipulation, the distortion energy interms of the principal stress components is found to be given by!"1/21 + ␯ (␴1 − ␴2 ) 2 + (␴1 − ␴3 ) 2 + (␴2 − ␴3 ) 2Ud =V(9.127)3E2The DET states that failure (yielding) occurs in a general state of stress when thedistortion energy per unit volume equals or exceeds the distortion energy per unitvolume occurring in a uniaxial tension test at yielding.

It is relatively easy toshow (see Problem 9.20) that, at yielding in a tensile test, the distortion energy isgiven byUd =1+␯ 2S V3E y(9.128)and, as before, we use Sy to denote the tensile yield strength. Hence, Equations 9.127 and 9.128 give the failure (yielding) criterion for the DET as!"1/2(␴1 − ␴2 ) 2 + (␴1 − ␴3 ) 2 + (␴2 − ␴3 ) 2≥ Sy(9.129)2The DET as described in Equation 9.129 leads to the concept of an equivalentstress (known historically as the Von Mises stress) defined as!"1/2(␴1 − ␴2 ) 2 + (␴1 − ␴3 ) 2 + (␴2 − ␴3 ) 2␴e =(9.130)2Hutton: Fundamentals ofFinite Element Analysis9.

Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.7 Strain and Stress Computationand failure (yielding) can then be equivalently defined as␴e ≥ S y(9.131)Even though we do not present the algebraic details here, the DET can be shownto be equivalent to another elastic failure theory, known as the octahedral shearstress theory (OSST). For all practical purposes, the OSST holds that yieldingoccurs when the maximum shear stress exceeds 0.577Sy. In comparison to theMSST, the OSST gives the material more “credit” for strength in shear.Why do we go into detail on these failure theories in the context of finite element analysis? As noted previously, strain and stress components are calculatedin the specified coordinate system.

The coordinate system seldom is such thatmaximum stress conditions are automatically obtained. Here is the point: Essentially every finite element software package not only computes strain and stresscomponents in the global and element coordinate systems but also principalstresses and the equivalent (Von Mises) stress for every element. In decidingwhether a design is acceptable (and this is why we use FEA, isn’t it?), we mustexamine the propensity to failure. The examination of stress data is the responsibility of the user of FEA software. The software does not produce results thatindicate failure unless the analyst carefully considers the data in terms of specificfailure criteria.Among the stress- and strain-related items generally available as a result ofsolution are the computed stresses (in the specified coordinate system), the principal stresses, the equivalent stress, the principal strains, and strain energy.

Характеристики

Список файлов книги