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Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.4 Isoparametric Formulation of the Plane Quadrilateral Element3(2.25, 1.5)4(1.25, 1)y1(1, 0)2(2, 0)xFigure 9.7 Dimensions are in inches.Axes are shown for orientation only.■ SolutionThe mapping functions arex (r, s) =1[(1 − r )(1 − s)(1) + (1 + r )(1 − s)(2) + (1 + r )(1 + s)(2.25)4+ (1 − r )(1 + s)(1.25)]y(r, s) =1[(1 − r )(1 − s)(0) + (1 + r )(1 − s)(0) + (1 + r )(1 + s)(1.5)4+ (1 − r )(1 + s)(1)]and the terms of the Jacobian matrix are∂x∂r∂y=∂r∂x=∂s∂y=∂sJ11 =J12J21J22121= (0.5 − 0.5s)41=21= (2.5 − 0.5r )4=and the determinant is| J | = J11 J22 − J12 J21 =1(4 − r + s)16Therefore, the geometric matrix [G ] of Equation 9.79 is known in terms of ratios ofmonomials in r and s as2.5 − 0.5r40[G] =4−r +s−2−(0.5 − 0.5s)020−22.5 − 0.5r02−(0.5 − 0.5s)For plane stress with the values given, the material property matrix is1 0.300  psi[D] = 32.97(10) 6  0.3 100 0.35353Hutton: Fundamentals ofFinite Element Analysis3549.

Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid MechanicsNext, we note that, since the matrix of partial derivatives [P] as defined in Equation 9.81is also composed of monomials in r and s,s−11−s1r−1−(1+ r)[P] = 04 0001+s1+r00−(1 + s)1−r000000s−11−sr − 1 −(1 + r)001+s1+r00−(1 + s) 1−rthe stiffness matrix of Equation 9.86 is no more than quadratic in the natural coordinates.Hence, we select four integration points given by√ri = s j = ±33and weighting factorsW i = W j = 1.0per Table 6.1.The element stiffness matrix is then given byk (e) = t22W i W j [B(r i , s j )] T [D][ B(r i , s j )]| J (r i , s j )|i =1 j =1The numerical results for this example are obtained via a computer program written inMATLAB using the built-in matrix functions of that software package. The stiffnessmatrix is calculated to be2305 −1759 −61772 −1759 1957471−669 −617471166−19 (e)  −669 −19616 72k= 798−52 −214 −533 −152 −522 −41633 −2141457143−432560116 −244798−52−214−5331453−169−389−895−152 −214 −432−52214560 4157116 633143 −244  310 lb/in.−169 −389 −895 99345−869 45104240 −869 2401524EXAMPLE 9.4A classic example of plane stress analysis is shown in Figure 9.8a.

A uniform thin platewith a central hole of radius a is subjected to uniaxial stress ␴0 . Use the finite elementmethod to determine the stress concentration factor given the physical data ␴0 = 1000 psi ,a = 0.5 in., h = 3 in., w = 6 in., E = 10(10) 6 psi, and Poisson’s ratio = 0.3.■ SolutionThe solution for this example is obtained using commercial finite element software withplane quadrilateral elements. The initial (coarse) element mesh, shown in Figure 9.8b, iscomposed of 33 elements.

Note that the symmetry conditions have been used to reducethe model to quarter-size and the corresponding boundary conditions are as shown on thefigure. For this model, the maximum stress (as expected) is calculated to occur at node 1(at the top of the hole) and has a magnitude of 3101 psi.Hutton: Fundamentals ofFinite Element Analysis9.

Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.4 Isoparametric Formulation of the Plane Quadrilateral Element␴02h␴0a2w(a)41920212223391853743244440381725332629273241 28344645 47 306 131 781312 11 10 92354834236161514(b)Figure 9.8(a) A uniformly loaded plate in plane stress with a central hole ofradius a. (b) A coarse finite element mesh using quadrilateralelements.

Node numbers are as shown (31 elements).To examine the solution convergence, a refined model is shown in Figure 9.8c, using101 elements. For this model, the maximum stress also occurs at node 1 and has a calculated magnitude of 3032 psi. Hence, between the two models, the maximum stress valueschanged on the order of 2.3 percent. It is interesting to note that the maximum displacementgiven by the two models is essentially the same.

This observation reinforces the need toexamine the derived variables for convergence, not simply the directly computed variables.As a final step in examining the convergence, the model shown in Figure 9.8d containing 192 elements is also solved. (The node numbers are eliminated for clarity.) Themaximum computed stress, again at node 1, is 3024 psi, a miniscule change relative tothe previous model, so we conclude that convergence has been attained. (The change inmaximum displacement is essentially nil.) Hence, we conclude that the stress concentration factor K t = ␴max /␴0 = 3024 /1000 = 3.024 is applicable to the geometry and loading of this example.

It is interesting to note that the theoretical (hence, the subscript t)355Hutton: Fundamentals ofFinite Element Analysis3569. Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid Mechanics(c)(d)Figure 9.8 (Continued)(c) Refined mesh of 101 elements. Node numbers are removed forclarity. (d) An additional refined mesh with 192 elements.stress concentration factor for this problem as computed by the mathematical theory ofelasticity is exactly 3.

The same result is shown in many texts on machine design andstress analysis [2].9.5 AXISYMMETRIC STRESS ANALYSISThe concept of axisymmetry is discussed in Chapter 6 in terms of general interpolation functions. Here, we specialize the axisymmetric concept to problemsof elastic stress analysis. To satisfy the conditions for axisymmetric stress, theproblem must be such that1.

The solid body under stress must be a solid of revolution; by convention, theaxis of revolution is the z axis in a cylindrical coordinate system (r, ␪, z ).2. The loading of the body is symmetric about the z axis.Hutton: Fundamentals ofFinite Element Analysis9. Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.5Axisymmetric Stress Analysisz(r ⫹u) d␪r d␪u⫹dzrud␪⭸udr⭸rrdr(a)(b)(c)Figure 9.9(a) Cross section of an axisymmetric body. (b) Differential element inan rz plane.

(c) Differential element in an r-␪ plane illustrating tangentialdeformation. Dashed lines represent deformed positions.3. All boundary (constraint) conditions are symmetric about the z axis.4. Materials properties are also symmetric (automatically satisfied by a linearlyelastic, homogeneous, isotropic material).If these conditions are satisfied, the displacement field is independent of thetangential coordinate ␪, and hence the stress analysis is mathematically twodimensional, even though the physical problem is three-dimensional.

To developthe axisymmetric equations, we examine Figure 9.9a, representing a solid of revolution that satisfies the preceding requirements. Figure 9.9b is a differentialelement of the body in the rz plane; that is, any section through the body forwhich ␪ is constant. We cannot ignore the tangential coordinate completely,however, since as depicted in Figure 9.9c, there is strain in the tangential direction (recall the basic definition of hoop stress in thin-walled pressure vesselsfrom mechanics of materials).

Note that, in the radial direction, the elementundergoes displacement, which introduces increase in circumference and associated circumferential strain.We denote the radial displacement as u, the tangential (circumferential) displacement as v, and the axial displacement as w. From Figure 9.9c, the radialstrain is1∂u∂uεr =u+dr − u =(9.88)dr∂r∂rThe axial strain isεz =1∂w∂ww+dz − w =dz∂z∂z(9.89)and these relations are as expected, since the rz plane is effectively the same as arectangular coordinate system.

In the circumferential direction, the differentialelement undergoes an expansion defined by considering the original arc length357Hutton: Fundamentals ofFinite Element Analysis3589. Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid Mechanicsversus the deformed arc length. Prior to deformation, the arc length is ds = r d␪,while after deformation, arc length is ds = (r + u) d␪.

The tangential strain isε␪ =(r + u)(d␪) − r d␪u=r d␪r(9.90)and we observe that, even though the problem is independent of the tangentialcoordinate, the tangential strain must be considered in the problem formulation.Note that, if r = 0, the preceding expression for the tangential strain is troublesome mathematically, since division by zero is indicated. The situation occurs,for example, if we examine stresses in a rotating solid body, in which case thestresses are induced by centrifugal force (normal acceleration).

Additional discussion of this problem is included later when we discuss element formulation.Additionally, the shear strain components are∂u∂w␥r z =+∂z∂r(9.91)␥r␪ = 0␥␪z = 0If we substitute the strain components into the generalized stress-strain relationsof Appendix B (and, in this case, we utilize ␪ = y ), we obtain␴r =E[(1 − ␯)εr + ␯(ε␪ + ε z )](1 + ␯)(1 − 2␯)␴␪ =E[(1 − ␯)ε␪ + ␯(εr + ε z )](1 + ␯)(1 − 2␯)E␴z =[(1 − ␯)ε z + ␯(εr + ε␪ )](1 + ␯)(1 − 2␯)␶r z =(9.92)E␥r z = G ␥r z2(1 + ␯)For convenience in finite element development, Equation 9.92 is expressed inmatrix form as1−␯␯␯0 εr ␴r1−␯␯0   ␯Eε␪␴␪=(9.93)␯␯1−␯0  εz  ␴z  (1 + ␯)(1 − 2␯) 1 − 2␯ ␥ez␶r z0002in which we identify the material property matrix for axisymmetric elasticity as1−␯␯␯0 ␯1−␯␯0 E␯␯1−␯0 [D] =(9.94)(1 + ␯)(1 − 2␯) 1 − 2␯ 0002Hutton: Fundamentals ofFinite Element Analysis9. Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.5Axisymmetric Stress Analysis9.5.1 Finite Element FormulationRecall from the general discussion of interpolation functions in Chapter 6 that essentially any two-dimensional element can be used to generate an axisymmetricelement.

As there is, by definition, no dependence on the ␪ coordinate and no circumferential displacement, the displacement field for the axisymmetric stressproblem can be expressed asMu(r, z) =N i (r, z)u ii=1(9.95)Mw(r, z) =N i (r, z)wii=1with ui and wi representing the nodal radial and axial displacements, respectively.For illustrative purposes, we now assume the case of a three-node triangularelement.The strain components becomeεr =ε␪ =∂u=∂ru=r3i=13i=1∂ Niui∂rNiuir(9.96)∂wεz ==∂z␥r z =3i=1∂ Niwi∂z∂u∂w+=∂z∂r3i=1∂ Niui +∂z3i=1∂ Niwi∂rand these are conveniently expressed in the matrix form∂ N1 ∂ N2 ∂ N3000∂r∂r∂r  NN2N3ε   1000 r  ε␪ rrr=ε  ∂ N1 ∂ N2 ∂ N3 z   000␥r z∂z∂z∂z ∂ N1 ∂ N2 ∂ N3 ∂ N1 ∂ N2 ∂ N3∂z∂z∂z∂r∂r∂r u1  u2   u3w1  w2  w3(9.97)In keeping with previous developments, Equation 9.97 is denoted {ε} = [B]{␦}with [B] representing the 4 × 6 matrix involving the interpolation functions.Thus total strain energy of the elements, as described by Equation 9.15 or 9.58359Hutton: Fundamentals ofFinite Element Analysis3609.

Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid Mechanicsand the stiffness matrix, isk(e)=[B] T [D][ B] dV (e)(9.98)V (e)While Equation 9.98 is becoming rather familiar, a word or two of caution isappropriate. Recall in particular that, although the interpolation functions usedhere are two dimensional, the axisymmetric element is truly three dimensional(toroidal). Second, the element is not a constant strain element, owing to theinverse variation of ε␪ with radial position, so the integrand in Equation 9.98 isnot constant.

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