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Chapter 12VECTOR FIELDS.12.1 IntroductionHere we will extend the concepts from the previous chapter on scalar fields to thosethat involve vector unknowns. The most common (and historically the first) application isthat is displacement based stress analysis. Other electrical field applications will only benoted through recent publications on those techniques which often require different ’edgebased’ interpolation methods not covered here. The main new considerations here are toselect ordering for the vector unknowns and related vector or tensor quantities so thatthey can be cast in an expanded matrix notation.12.2 Displacement Based Stress Analysis SummaryLet uu denote a displacement vector at a point xx in a solid. The finite elementdisplacement formulation for stress analysis is based on the energy concept of findinguu(xx) that both satisfies the essential boundary conditions on uu and minimizes the TotalPotential Energy:Π(uu) = U − Wwhere W is the external mechanical work due to supplied loading data with body forcesper unit volume, XX, surface tractions per unit area, TT, and point loads, PPj, at point xxjsoW(u) =∫ΩuuTXX dΩ+∫ΓuuTTT dΓ+jΣuuTjPPjand U(uu) consists of the strain energy due to the deformation, uu, of the material. It isdefined asU =12∫ΩεεTσσdΩwhereεεare the strain components resulting from the displacements, uu, andσσare thestress components that correspond to the strain components.The strains,εε, are defined by a "Strain-Displacement Relation" for each class ofstress analysis. This relation can be represented as a differential operator matrix, say LL,acting on the displacementsεε= LL uu. There are four commonly used classes that we willconsider here. They are4.3 Draft − 5/28/04© 2004 J.E. Akin 382Finite Elements, Vector Fields 3831. Axial stress: uu = u, LL =∂/∂x,εε=εx=∂u/∂x, Bj=∂Hj/∂x2. Plane Stress and Plane Strain: uuT=uvLL=∂∂x0∂∂y0∂∂y∂∂x, Bj=∂Hj∂x0∂Hj∂y0∂Hj∂y∂Hj∂xεεT=[εxεyγ]=∂u∂x∂v∂y∂u∂y+∂v∂x3. Axisymmetric solid with radius r and axial position z : uuT=uvLL=∂∂r0∂∂z1/r0∂∂z∂∂r0, Bj=∂Hj∂r0∂Hj∂z1/r0∂Hj∂z∂Hj∂r0,εεT=εrεzγεθ=∂u∂r∂u∂z(∂u∂z+∂v∂r)u/r4. The full three-dimensional solid: uuT= [uv w]LL=∂∂x00∂∂y∂∂z00∂∂y0∂∂x0∂∂z00∂∂z0∂∂x∂∂y,Bj=∂Hj∂x00∂Hj∂y∂Hj∂z00∂Hj∂y0∂Hj∂x0∂Hj∂z00∂Hj∂z0∂Hj∂x∂Hj∂yεεT=εxεyεzγxyγxzγyz=∂u∂x∂v∂y∂w∂z∂u∂y+∂v∂x∂u∂z+∂w∂y∂v∂z+∂w∂y.4.3 Draft − 5/28/04 © 2004 J.E. Akin. All rights reserved.384 J. E. AkinThe above notations for strains is called the engineering definition. There is a moregeneral definition, that we will not use, called the strain tensor which has the formεεjk= (uj,k+ uk, j) /2. The classic elasticity definitions are expressed in a matrix form anddo not yet include any finite element assumptions. Within any element we interpolate thedisplacement vector in terms of nodal displacement vector components,δδe:u(x) ≡ N(x)δδex∈Ωewhere N denotes that spatial interpolation for vector components. Then, in all cases, wecan define a common notation for the element strains in terms of the elementdisplacement:εε= L(x)Ne(x)δδe= Be(x)δδewhere the differential operator action on the displacement vector interpolations definesthe "B-matrix":Be(x) = L(x)Ne(x)Usually we use the same interpolation for each of the scalar components of thedisplacement vector, u. For exampleu(x) = He(x)ue,v(x)=He(x)vve,w(x)=He(x)wweand we order the unknown element displacements asδδeT= [u1v1w1u2v2...vmwm]for an element with "m" nodes. Thus the vector interpolation matrix, N, simply containsthe H scalar interpolation functions and usually some zeros;N(x) = N(H(x)) = [N1N2...Nm],where Njis the matrix partition asociated with the j − th node. It follows that the Bematrix can also be partitioned into a set of nodal matricesB(x) = B(H(x)) = [B1B2...Bm],where Bj= L(x) I Hsubj(x) = [L(bdx) Hj(x)] . With the above choices for the order ofthe unknowns inδδewe note that for our four analysis classes the vector interpolation sub-matrices at node j are:1. Axial stress: Nj(x) = Hj(x)2. - 3. Two-dimensional and axisymmetric casesNj(x) =Hj(x)00Hj(x)4. Three-Dimensional CaseNj(x) =Hj(x)000Hj(x)000Hj(x)In general we can use an identity matrix, II, the size of the displacement vector, to defineNj(x) = II Hj(x). Returning to the elasticity notations, the corresponding stresscomponents for our four cases are1. Axial Stress:σσ=σx4.3 Draft − 5/28/04 © 2004 J.E. Akin. All rights reserved.Finite Elements, Vector Fields 3852. Plane Stress (σz= 0), Plane Strain (εz= 0):σσT=σxσyτ3. Axisymmetric Solid:σσT=σrσzτσθ4. Three-dimensional Solid:σσT=σxσyσxτxyτxzτyzwhere theτterms denote shear stresses and the other terms are normal stresses. For thestate of plane strainσzis not zero, but is recovered in a post-processing operation. Themechanical stresses are related to the mechanical strains (ignoring initial effects) by theminimal form of a material constitutive law usually known as the basic Hooke’s Law:σσ=ΕΕεε, where ΕΕ = ΕΕTis a symmetric material properties matrix that relates the stressand strain components for each case. Actually, it is defined for the three-dimensionalcase and the others are special forms of it. For an isotropic material they are:1. Axial stress: ΕΕ = Ε, Youn g′s Modulus2. Plane Stress (σz≡ 0)ΕΕ =Ε1−ν21ν0ν1000(1 −ν)/2ν, Poisso n′ s ratio,0≤v<0.53. Plane Strain (εz≡ 0)ΕΕ =Ε(1 +ν)(1 − 2ν)(1 −ν)ν00(1 −ν)000(1 − 2ν)/24. Axisymmetric SolidΕΕ =Ε(1 +ν)(1 − 2ν)(1 −ν)ν0νν(1 −ν)0ν00(1 − 2ν) /20νν0(1 −ν)5. General


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Rice MECH 517 - Vector Fileds

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