Hutton - Fundamentals of Finite Element Analysis (523155), страница 85

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Similarly, the shear stress components*are unaffected by normal strains.The stress-strain relations can easily be expressed in matrix form by definingthe material property matrix [D] as1−␯␯␯000 ␯1−␯␯000  ␯␯1−␯000 1 − 2␯E 00000 [D] =2(1 + ␯)(1 − 2␯) 1 − 2␯ 00000 21 − 2␯ 000002(B.14)*The double subscript notation used for shearing stresses is explained as follows: The first subscriptdefines the axial direction perpendicular to the surface on which the shearing stress acts, while thesecond subscript denotes the axis parallel to the shearing stress. Thus, ␶x y denotes a shearing stressacting in the direction of the x axis on a surface perpendicular to the y axis.

Via moment equilibrium, itis readily shown that ␶x y = ␶ yx , ␶x z = ␶zx, and ␶ yz = ␶zy.459Hutton: Fundamentals ofFinite Element Analysis460Back MatterAppendix B: Equations ofElasticityAPPENDIX B© The McGraw−HillCompanies, 2004Equations of Elasticityand writing{␴} = ␴x  ␴y ␴ z␶x y ␶x z  ␶ yz= [D]{ε} = [D][L]{␦}(B.15)Here {␴} denotes the 6 × 1 matrix of stress components.

We do not use the termstress vector, since, as we subsequently observe, that term has a generally accepted meaning quite different from the matrix defined here.B.3 EQUILIBRIUM EQUATIONSTo obtain the equations of equilibrium for a deformed solid body, we examinethe general state of stress at an arbitrary point in the body via an infinitesimal differential element, as shown in Figure B.2.

All stress components are assumed to␴y ⫹␶xy ⫹⭸␶xy⭸y⭸␴y⭸ydy␶yz ⫹⭸␶yz⭸y␴zdydy␶xz␶yz␶xz ⫹dx⭸x⭸␴xdx␴x ⫹⭸x␶xz␴xdy␶xydz␶xy ⫹dx␶yz␶xz ⫹⭸␶xz⭸z␴z ⫹␶xydz⭸␴z⭸zdz␶yz ⫹⭸␶yz⭸zdz⭸␶xz␴yFigure B.2 A three-dimensional element in a general state of stress.⭸␶xy⭸xdxHutton: Fundamentals ofFinite Element AnalysisBack MatterAppendix B: Equations ofElasticityB.4© The McGraw−HillCompanies, 2004Compatibility Equationsvary spatially, and these variations are expressed in terms of first-order Taylorseries expansions, as indicated. In addition to the stress components shown, itis assumed that the element is subjected to a body force having axial componentsB x , B y , B z .

The body force is expressed as force per unit volume and representsthe action of an external influence that affects the body as a whole. The mostcommon body force is that of gravitational attraction while magnetic and centrifugal forces are also examples.Applying the condition of force equilibrium in the direction of the x axis forthe element of Figure B.2 results in∂ ␴x∂ ␶x y␴x +dx dy dz − ␴x dy dz + ␶x y +dy dx dz − ␶x y dx dz∂x∂y∂ ␶x z(B.16)+ ␶x z +dz dx dy − ␶x z dx dy + B x dx dy dz = 0∂zExpanding and simplifying Equation B.16 yields∂ ␴x∂ ␶x y∂ ␶x z+++ Bx = 0∂x∂y∂z(B.17)Similarly, applying the force equilibrium conditions in the y and z coordinatedirections yields∂ ␶x y∂ ␴y∂ ␶ yz+++ By = 0∂x∂y∂z(B.18)∂ ␶x z∂ ␶ yz∂ ␴z+++ Bz = 0∂x∂y∂z(B.19)respectively.B.4 COMPATIBILITY EQUATIONSEquations B.3–B.6 define six strain components in terms of three displacementcomponents. A fundamental premise of the theory of continuum mechanics isthat a continuous body remains continuous during and after deformation.

Therefore, the displacement and strain functions must be continuous and single valued.Given a continuous displacement field u, v, w, it is straightforward to computecontinuous, single-valued strain components via the strain-displacement relations. However, the inverse case is a bit more complicated.

That is, given a fieldof six continuous, single-valued strain components, we have six partial differential equations to solve to obtain the displacement components. In this case, thereis no assurance that the resulting displacements will meet the requirements ofcontinuity and single-valuedness. To ensure that displacements are continuouswhen computed in this manner, additional relations among the strain components461Hutton: Fundamentals ofFinite Element Analysis462Back MatterAPPENDIX BAppendix B: Equations ofElasticity© The McGraw−HillCompanies, 2004Equations of Elasticityhave been derived, and these are known as the compatibility equations. There aresix independent compatibility equations, one of which is∂ 2ε y∂ 2 ␥x y∂ 2εx+=∂ y2∂x2∂x∂y(B.20)The other five equations are similarly second-order relations.

While not usedexplicitly in this text, the compatibility equations are absolutely essential inadvanced methods in continuum mechanics and the theory of elasticity.Hutton: Fundamentals ofFinite Element AnalysisBack MatterAppendix C: SolutionTechniques for LinearAlgebraic Equations© The McGraw−HillCompanies, 2004A P P E N D I XCSolution Techniquesfor Linear AlgebraicEquationsC.1 CRAMER’S METHODCramer’s method, also known as Cramer’s rule, provides a systematic means ofsolving linear equations.

In practicality, the method is best applied to systems ofno more than two or three equations. Nevertheless, the method provides insightinto certain conditions regarding the existence of solutions and is included herefor that reason.Consider the system of equationsa11 x 1 + a12 x 2 = f 1a21 x 1 + a22 x 2 = f 2(C.1)[ A]{x } = { f }(C.2)or in matrix formMultiplying the first equation by a22 , the second by a12 , and subtracting the second from the first gives(a11 a22 − a12 a21 )x 1 = f 1 a22 − f 2 a12(C.3)Therefore, if (a11 a22 − a12 a21 ) = 0, we solve for x 1 asx1 =f 1 a22 − f 2 a12a11 a22 − a12 a21(C.4)x2 =f 2 a11 − f 1 a21a11 a22 − a12 a21(C.5)Via a similar procedure,463Hutton: Fundamentals ofFinite Element Analysis464Back MatterAPPENDIX CAppendix C: SolutionTechniques for LinearAlgebraic Equations© The McGraw−HillCompanies, 2004Solution Techniques for Linear Algebraic EquationsNote that the denominator of each solution is the same and equal to the determinant of the coefficient matrix a11 a12 = a11 a22 − a12 a21| A| = (C.6)a21 a22 and again, it is assumed that the determinant is nonzero.Now, consider the numerator of Equation C.4, as follows.

Replace the firstcolumn of the coefficient matrix [ A] with the right-hand side column matrix { f }and calculate the determinant of the resulting matrix (denoted [ A 1 ]) to obtain f a12 = f 1 a22 − f 2 a12| A 1 | = 1(C.7)f 2 a22 The determinant so obtained is exactly the numerator of Equation C.4. If we similarly replace the second column of [ A] with the right-hand side column matrixand calculate the determinant, we have a11 f 1 = f 2 a11 − f 1 a21| A2| = (C.8)a21 f 2 and the result of Equation C.8 is identical to the numerator of Equation C.5.Although presented for a system of only two equations, the results are applicableto any number of linear algebraic equations as follows:Cramer’s rule: Given a system of n linear algebraic equations in n unknowns x i ,i = 1 , n , expressed in matrix form as[ A]{x } = { f }(C.9)where { f } is known, solutions are given by the ratio of determinantsxi =| Ai || A|i = 1, n(C.10)provided | A| = 0.Matrices [ A i ] are formed by replacing the ith column of the coefficientmatrix [ A] with the right-hand side column matrix.Note that, if the right-hand side { f } = {0} , Cramer’s rule gives the trivialresult {x } = {0} .Now consider the case in which the determinant of the coefficient matrix is0.

In this event, the solutions for the system represented by Equation C.1 are,formally,0x 1 = f 1 a22 − f 2 a120x 2 = f 2 a11 − f 1 a21(C.11)Equations (C.11) must be considered under two cases:1. If the right-hand sides are nonzero, no solutions exist, since we cannotmultiply any number by 0 and obtain a nonzero result.Hutton: Fundamentals ofFinite Element AnalysisBack MatterAppendix C: SolutionTechniques for LinearAlgebraic Equations© The McGraw−HillCompanies, 2004C.2 Gauss Elimination2. If the right-hand sides are 0, the equations indicate that any values of x 1and x 2 are solutions; this case corresponds to the homogeneous equationsthat occur if { f } = {0} . Thus, a system of linear homogeneous algebraicequations can have nontrivial solutions if and only if the determinant of thecoefficient matrix is 0.

The fact is, however, that the solutions are not justany values of x 1 and x 2 , and we see this by examining the determinant| A| = a11 a22 − a12 a21 = 0(C.12)a11a12=a21a22(C.13)orEquation C.13 states that the coefficients of x 1 and x 2 in the two equations are inconstant ratio. Thus, the equations are not independent and, in fact, represent astraight line in the x 1 x 2 plane. There do, then, exist an infinite number of solutions (x 1 , x 2 ) , but there also exists a relation between the coordinates x 1 and x 2 .The argument just presented for two equations is also general for any number ofequations. If the system is homogeneous, nontrivial solutions exist only if thedeterminant of the coefficient matrix is 0.C.2 GAUSS ELIMINATIONIn Appendix A, dealing with matrix mathematics, the concept of inverting the coefficient matrix to obtain the solution for a system of linear algebraic equations isdiscussed.

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