Hutton - Fundamentals of Finite Element Analysis, страница 84

The concept of matrix partitioning is most useful in reducing thesize of a system of equations and accounting for specified values of a subset ofthe dependent variables. Consider a system of n linear algebraic equations governing n unknowns x i expressed in matrix form as[ A]{x } = { f }(A.41)in which we want to eliminate the first p unknowns. The matrix equation can bewritten in partitioned form as [ A 11 ] [ A 12 ]{X 1 }{F1 }=(A.42)[ A 21 ] [ A 22 ]{X 2 }{F2 }where the orders of the submatrices are as follows[ A 11 ][ A 12 ][ A 21 ][ A 22 ]{X 1 }, {F1 }{X 2 }, {F12 }⇒⇒⇒⇒⇒⇒p× pp × (n − p)(n − p) × p(n − p) × (n − p)p×1(n − p) × 1(A.43)The complete set of equations can now be written in terms of the matrix partitions as[ A 11 ]{X 1 } + [ A 12 ]{X 2 } = {F1 }[ A 21 ]{X 1 } + [ A 22 ]{X 2 } = {F2 }(A.44)The first p equations (the upper partition) are solved as{X 1 } = [ A 11 ]−1 ({F1 } − [ A 12 ]{X 2 })(A.45)(implicitly assuming that the inverse of A 11 exists).

Substitution of Equation A.45into the remaining n − p equations (the lower partition) yields−1−1[ A 22 ] − [ A 21 ] A 11[ A 12 ] {X 2 } = {F2 } − [ A 21 ] A 11]{F1 } (A.46)Equation A.46 is the reduced set of n − p algebraic equations representing theoriginal system and containing all the effects of the first p equations. In the context of finite element analysis, this procedure is referred to as static condensation.As another application (commonly encountered in finite element analysis),we consider the case in which the partitioned values {X 1 } are known but the corresponding right-hand side partition {F1 ] is unknown.

In this occurrence, thelower partitioned equations are solved directly for {X 2 ] to obtain{X 2 } = [ A 22 ]−1 ({F2 } − [ A 21 ]{X 1 })(A.47)The unknown values of {F1 ] can then be calculated directly using the equationsof the upper partition.Hutton: Fundamentals ofFinite Element AnalysisBack MatterAppendix B: Equations ofElasticity© The McGraw−HillCompanies, 2004A P P E N D I XBEquations of ElasticityB.1 STRAIN-DISPLACEMENT RELATIONSIn general, the concept of normal strain is introduced and defined in the contextof a uniaxial tension test. The elongated length L of a portion of the test specimenhaving original length L 0 (the gauge length) is measured and the correspondingnormal strain defined asε=L − L0L=L0L0(B.1)which is simply interpreted as “change in length per unit original length” and isobserved to be a dimensionless quantity.

Similarly, the idea of shear strain is oftenintroduced in terms of a simple torsion test of a bar having a circular cross section. In each case, the test geometry and applied loads are designed to produce asimple, uniform state of strain dominated by one major component.In real structures subjected to routine operating loads, strain is not generallyuniform nor limited to a single component. Instead, strain varies throughout thegeometry and can be composed of up to six independent components, includingboth normal and shearing strains.

Therefore, we are led to examine the appropriate definitions of strain at a point. For the general case, we denote u = u(x , y, z),v = v(x , y, z) , and w = w(x , y, z) as the displacements in the x, y, and z coordinate directions, respectively. (The displacements may also vary with time; fornow, we consider only the static case.) Figure B.1(a) depicts an infinitesimal element having undeformed edge lengths dx , dy , dz located at an arbitrary point(x, y, z) in a solid body.

For simplicity, we first assume that this element is loadedin tension in the x direction only and examine the resulting deformation as shown(greatly exaggerated) in Figure B.1(b). Displacement of point P is u while that ofpoint Q is u + (∂ u/∂ x ) dx such that the deformed length in the x direction isgiven by∂u∂udx = dx + u Q − u P = dx + u +dx − u = dx +dx(B.2)∂x∂x455Hutton: Fundamentals ofFinite Element Analysis456Back MatterAPPENDIX BAppendix B: Equations ofElasticity© The McGraw−HillCompanies, 2004Equations of Elasticityyxzdx⬘ydxdyxP␴xP⬘QQ⬘␴xdzuu⫹dx⭸udx⭸x(b)(a)␤⫽⭸u⭸yC⬘CA⬘BBA␣⫽⭸v⭸x(d)(c)Figure B.1(a) A differential element in uniaxial stress; (b) resulting axialdeformation; (c) differential element subjected to shear;(d) angular changes used to define shear strain.The normal strain in the x direction at the point depicted is thenεx =dx − dx∂u=dx∂x(B.3)Similar consideration of changes of length in the y and z directions yields thegeneral definitions of the associated normal strain components asεy =∂v∂yandεz =∂w∂z(B.4)To examine shearing of the infinitesimal solid, we next consider the situationshown in Figure B.1(c), in which applied surface tractions result in shear of theHutton: Fundamentals ofFinite Element AnalysisBack MatterAppendix B: Equations ofElasticity© The McGraw−HillCompanies, 2004B.1 Strain-Displacement Relationselement, as depicted in Figure B.1(d).

Unlike normal strain, the effects of shearing are seen to be distortions of the original rectangular shape of the solid. Suchdistortion is quantified by angular changes, and we consequently define shearstrain as a “change in the angle of an angle that was originally a right angle.” Onfirst reading, this may sound redundant but it is not.

Consider the definition in thecontext of Figure B.1(c) and B.1(d); angle A BC was a right angle in the undeformed state but has been distorted to A BC by shearing. The change of theangle is composed of two parts, denoted ␣ and ␤, given by the slopes of B A andBC , respectively as ∂ v/∂ x and ∂ u/∂ y. Thus, the shear strain is␥x y =∂u∂v+∂y∂x(B.5)where the double subscript is used to indicate the plane in which the angularchange occurs. Similar consideration of distortion in x z and yz planes results in␥x z =∂u∂w+∂z∂xand␥yz =∂v∂w+∂z∂y(B.6)as the shear strain components, respectively.Equations B.3–B.6 provide the basic definitions of the six possible independent strain components in three-dimensional deformation.

It must be emphasizedthat these strain-displacement relations are valid only for small deformations.Additional terms must be included if large deformations occur as a result ofgeometry or material characteristics. As continually is the case as we proceed, itis convenient to express the strain-displacement relations in matrix form. Toaccomplish this task, we define the displacement vector as u(x, y, z) {␦} = v(x, y, z)(B.7)w(x, y, z)(noting that this vector describes a continuous displacement field) and the strainvector asεx  εy  ε z{ε} =(B.8)␥x y ␥x z ␥yzThe strain-displacement relations are then expressed in the compact form{ε} = [L ]{␦}(B.9)457Hutton: Fundamentals ofFinite Element Analysis458Back MatterAPPENDIX BAppendix B: Equations ofElasticity© The McGraw−HillCompanies, 2004Equations of Elasticitywhere [L] is the derivative operator matrix given by∂00  ∂x∂ 00∂y∂  00∂z [L] =  ∂∂0 ∂y ∂x ∂∂ 0 ∂z∂x ∂∂ 0∂z ∂ y(B.10)B.2 STRESS-STRAIN RELATIONSThe equations between stress and strain applicable to a particular material areknown as the constitutive equations for that material.

In the most general type ofmaterial possible, it is shown in advanced work in continuum mechanics that theconstitutive equations can contain up to 81 independent material constants. However, for a homogeneous, isotropic, linearly elastic material, it is readily shownthat only two independent material constants are required to completely specifythe relations. These two constants should be quite familiar from elementarystrength of materials theory as the modulus of elasticity (Young’s modulus) andPoisson’s ratio.

Again referring to the simple uniaxial tension test, the modulus ofelasticity is defined as the slope of the stress-strain curve in the elastic region orE=␴xεx(B.10)where it is assumed that the axis of loading corresponds to the x axis. As strain isdimensionless, the modulus of elasticity has the units of stress usually expressedin lb/in.2 or megapascal (MPa).Poisson’s ratio is a measure of the well-known phenomenon that an elasticbody strained in one direction also experiences strain in mutually perpendiculardirections.

In the uniaxial tension test, elongation of the test specimen in the loading direction is accompanied by contraction in the plane perpendicular to the loading direction. If the loading axis is x, this means that the specimen changes dimensions and thus experiences strain in the y and z directions as well, even though noexternal loading exists in those directions. Formally, Poisson’s ratio is defined asunit lateral contraction␯=−(B.11)unit axial elongationand we note that Poisson’s ratio is algebraically positive and the negative sign assures this, since numerator and denominator always have opposite signs. Thus, inHutton: Fundamentals ofFinite Element AnalysisBack MatterAppendix B: Equations ofElasticity© The McGraw−HillCompanies, 2004B.2 Stress-Strain Relationsthe tension test, if ε x represents the strain resulting from applied load, the inducedstrain components are given by ε y = ε z = −␯ε x .The general stress-strain relations for a homogeneous, isotropic, linearly elastic material subjected to a general three-dimensional deformation are as follows:E␴x =[(1 − ␯)ε x + ␯(ε y + ε z )](B.12a)(1 + ␯)(1 − 2␯)E␴y =[(1 − ␯)ε y + ␯(ε x + ε z )](B.12b)(1 + ␯)(1 − 2␯)E␴z =[(1 − ␯)ε z + ␯(ε x + ε y )](B.12c)(1 + ␯)(1 − 2␯)␶x y =E␥x y = G ␥x y2(1 + ␯)(B.12d)␶x z =E␥x z = G ␥x z2(1 + ␯)(B.12e)␶ yz =E␥yz = G ␥yz2(1 + ␯)(B.12f)where we introduce the shear modulus or modulus of rigidity, defined byEG=(B.13)2(1 + ␯)We may observe from the general relations that the normal components of stressand strain are interrelated in a rather complicated fashion through the Poisson effect but are independent of shear strains.