Hutton - Fundamentals of Finite Element Analysis (523155), страница 66
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Finally, note that [D] is significantly different in comparison to thecounterpart material property matrices for plane stress and plane strain. Takingthe first observation into account and recalling Equation 6.93, the stiffnessmatrix is defined by (e) k= 2[B] T [D][ B]r dr dz(9.99)A(e)and is a 6 × 6 symmetric matrix requiring, in theory, evaluation of 21 integrals.Explicit term-by-term integration is not recommended, owing to the algebraiccomplexity. When high accuracy is required, Gauss-type numerical integrationusing integration points specifically determined for triangular regions [3] is used.Another approach is to evaluate matrix [B] at the centroid of the element in an rzplane.
In this case, the matrices in the integrand become constant and the stiffness matrix is approximated by (e) k≈ 2 r̄ A[ B̄] T [D][ B̄](9.100)Of course, the accuracy of the approximation improves as element size isdecreased.Referring to a previous observation, formulation of the [B] matrix is troublesome if r = 0 is included in the domain. In this occurrence, three terms of Equation 9.97 “blow up,” owing to division by zero.
If the stiffness matrix is evaluatedusing the centroidal approximation of Equation 9.100, the problem is avoided,since the radial coordinate of the centroid of any element cannot be zero in anaxisymmetric finite element model. Nevertheless, radial and tangential strain andstress components cannot be evaluated at nodes for which r = 0.
Physically, weknow that the radial and tangential displacements at r = 0 in an axisymmetricproblem must be zero. Mathematically, the observation is not accounted for inthe general finite element formulation, which is for an arbitrary domain. Onetechnique for avoiding the problem is to include a hole, coinciding with the z axisand having a small, but finite radius [4].9.5.2 Element LoadsAxisymmetric problems often involve surface forces in the form of internal orexternal pressure and body forces arising from rotation of the body (centrifugalHutton: Fundamentals ofFinite Element Analysis9.
Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.5Axisymmetric Stress Analysis3613pzprpzzr21(a)dSpr(b)Figure 9.10(a) Axisymmetric element. (b) Differential lengthof the element edge.force) and gravity. In each case, the external influences are reduced to nodalforces using the work equivalence concept previously introduced.The triangular axisymmetric element shown in Figure 9.10a is subjected topressures pr and pz in the radial and axial directions, respectively. The equivalentnodal forces are determined by analogy with Equation 9.39, with the notableexception depicted in Figure 9.10b, showing a differential length dS of the element edge in question.
As dS is located a radial distance r from the axis of symmetry, the area on which the pressure components act is 2r dS. The nodalforces are given by ( p) ( p) f rprT=2r dSf=[N](9.101)( p)pzfzSand the path of integration S is the element edge. In this expression, [N ]T is asdefined by Equation 9.40.EXAMPLE 9.5Calculate the nodal forces corresponding to a uniform radial pressure pr = 10 psi actingas shown on the axisymmetric element in Figure 9.11.■ SolutionAs we have pressure on one face only and no axial pressure, we immediately observe thatf r 2 = f z1 = f z2 = f z3 = 0The nonzero terms aref r 1 = 2N 1 pr r dSSf r 3 = 2N 3 pr r dSSHutton: Fundamentals ofFinite Element Analysis3629.
Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid Mechanics3 (3, 1)10 psizr1 (3, 0)2 (4, 0)Figure 9.11 Uniform radial pressure.Dimensions are in inches.Using Equation 9.28 with r, z in place of x, y, the interpolation functions areN1 = 4 − r − zN2 = r − 3N3 = zand along the integration path (r = 3) , we haveN1 = 1 − zN2 = 0N3 = zIf the integration path is from node 1 to node 3, then dS = dz and1f r 1 = 2(10)(3)z dz = 30 lb01f r 3 = 2(10)(3)(1 − z) dz = 30 lb0Note that, if the integration path is taken in the opposite sense (i.e., from node 3 tonode 2), then dS = −dz and the same results are obtained.Body forces acting on axisymmetric elements are accounted for in a mannersimilar to that discussed for the plane stress element, while taking into consideration the geometric differences.
If body forces (force per unit mass) R B and Z Bact in the radial and axial directions, respectively, the equivalent nodal forces arecalculated as ( B) RBf= 2[N ] Tr dr dz(9.102)ZBA(e)For the three-node triangular element, [N ]T would again be as given in Equation 9.40. Extension to other element types is similar.Hutton: Fundamentals ofFinite Element Analysis9.
Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.5Axisymmetric Stress Analysis363Generally, radial body force arises from rotation of an axisymmetric bodyabout the z axis. For constant angular velocity , the radial body force component RB is equal to the magnitude of the normal acceleration component r 2 anddirected in the positive radial direction.EXAMPLE 9.6The axisymmetric element of Figure 9.11 is part of a body rotating with angular velocity10 rad/s about the z axis and subjected to gravity in the negative z direction. Compute theequivalent nodal forces.
Density is 7.3(10)−4 lb-s2/in.4■ SolutionFor the stated conditions, we haveR B = r 2 = 100r in./s 2Z B = −g = −386.4 in./s 2Using the interpolation functions as given in Example 9.5,f r 1 = 2 4 4−rN 1 R B r dr dz = 2 (100)A3f r 2 = 2(4 − r − z)r 2 dz dr = 0.84 lb0 4 4−rN 2 R B r dr dz = 2 (100)(r − 3)r 2 dz dr = 0.98 lbA3 4 4−rf r 3 = 20N 3 R B r dr dz = 2 (100)Azr 2 dz dr = 0.84 lb30f z1 = 2N 1 Z B r dr dz = −2 (386.4) 4 4−r(4 − r − z)r dz dr = −1.00 lbA3 4 4−rf z2 = 20N 2 Z B r dr dz = −2 (386.4)Af z3 = 2(r − 3)r dz dr = −1.08 lb30 4 4−rN 3 Z B r dr dz = −2 (386.4)Azr dz dr = −1.00 lb30The integrations required to obtain the given results are straightforward but algebraicallytedious.
Another approach that can be used and is increasingly accurate for decreasingelement size is to evaluate the body forces and the integrand at the centroid of the crosssection of the element area as an approximation. Using this approximation, it can beshown thatN i ( r̄ , z̄) r̄ dz dr =Ar̄ A3i = 1, 3Hutton: Fundamentals ofFinite Element Analysis3649. Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid Mechanicsso the body forces are allocated equally to each node.
For the present example, the result isf r 1 = f r 2 = f r 3 = 0.88 lbf z1 = f z2 = f z3 = −1.03 lbNote that, within the numerical accuracy used here, the total radial force and the totalaxial force are the same for the two methods.9.6 GENERAL THREE-DIMENSIONALSTRESS ELEMENTSWhile the conditions of plane stress, plane strain, and axisymmetry are frequentlyencountered, more often than not the geometry of a structure and the applied loadsare such that a general three-dimensional state of stress exists.
In the general case,there are three displacement components u, v, and w in the directions of the x, y,and z axes, respectively, and six strain components given by (Appendix B)∂u∂x∂v ∂yεx∂wεy ∂zεz{ε} == ∂u(9.103)∂v ␥x y + ∂y ␥ ∂x xz ␥yz∂u∂w+ ∂z∂x∂w∂v+∂z∂yFor convenience of presentation, the strain-displacement relations of Equation 9.103 can be expressed as∂00 ∂x∂ 00 ∂y∂ 0 0u∂z u=[L]{ε} = ∂vv(9.104)∂ w0w ∂y ∂x ∂∂ 0 ∂z∂x ∂∂ 0∂z ∂ yand matrix [L ] is the 6 × 3 matrix of derivative operators.Hutton: Fundamentals ofFinite Element Analysis9.
Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.6 General Three-Dimensional Stress ElementsThe stress-strain relations, Equation B.12, are expressed in matrix form as{} =x yzx y xzy z1− E 0=(1 + )(1 − 2) 001−1−000001 − 2200000001 − 2200000000 {ε}= [D]{ε}0 1 − 2 2(9.105)Note that, for the general case, the material property matrix [D] is a 6 × 6 matrixinvolving only the elastic modulus and Poisson’s ratio (we continue to restrictthe presentation to linear elasticity). Also note that the displacement componentsare continuous functions of the Cartesian coordinates.9.6.1 Finite Element FormulationFollowing the general procedure established in the context of two-dimensionalelements, a three-dimensional elastic stress element having M nodes is formulated by first discretizing the displacement components asMu(x , y, z) =N i (x , y, z)u ii=1Mv(x , y, z) =N i (x , y, z)vi(9.106)i=1Mw(x , y, z) =N i (x , y, z)wii=1As usual, the Cartesian nodal displacements are u i , vi , and wi and N i (x , y, z) isthe interpolation function associated with node i.
At this point, we make no assumption regarding the element shape or number of nodes. Instead, we simplynote that the interpolation functions may be any of those discussed in Chapter 6for three-dimensional elements.Introducing the vector (column matrix) of nodal displacements,{␦} = [ u 1u2···uMv1v2···vMw1w2···w M ]T(9.107)365Hutton: Fundamentals ofFinite Element Analysis3669. Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid Mechanicsthe discretized representation of the displacement field can be written in matrixform as [N ] [0] [0]uv = [0] [N ] [0] {␦} = [N3 ]{␦}(9.108)w[0] [0] [N ]In the last equation, each submatrix [N ] is the 1 × M row matrix of interpolationfunctions[N ] = [N 1N2···(9.109)NM]so the matrix we have chosen to denote as [N 3 ] is a 3 × 3M matrix composed ofthe interpolation functions and many zero values. (Before proceeding, weemphasize that the order of nodal displacements in Equation 9.107 is convenientfor purposes of development but not efficient for computational purposes.
Muchhigher computational efficiency is obtained in the model solution phase if thedisplacement vector is defined as {␦} = [u 1 v1 w1 u 2 v2 w2 · · · u M v M w M ]T .)Recalling Equations 9.10 and 9.19, total potential energy of an element canbe expressed as1{ε} T [D]{ε} dV − {␦} T { f } = Ue − W =(9.110)2VThe element nodal force vector is defined in the column matrix{ f } = [ f 1xf 2x···fMxf 1yf 2y···fMyf 1zf 2z···f M z ]T(9.111)and may include the effects of concentrated forces applied at the nodes, nodalequivalents to body forces, and nodal equivalents to applied pressure loadings.Considering the foregoing developments, Equation 9.110 can be expressed(using Equations 9.104, 9.105, and 9.108), as1 = Ue − W =␦T [L ] T [N 3 ] T [D][L ][ N 3 ]{␦} dV − {␦} T { f }(9.112)2VAs the nodal displacement components are independent of the integration overthe volume, Equation 9.112 can be written as1 = U e − W = {␦} T[L ] T [N 3 ] T [D][L ][ N 3 ] dV {␦} − {␦} T { f }(9.113)2Vwhich is in the form1 = U e − W = {␦} T2[B] T [D][ B] dV {␦} − {␦} T { f }V(9.114)Hutton: Fundamentals ofFinite Element Analysis9.
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