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26 KiB
Hence, we have
\begin{array}{l} \sup _ {\mathbf {v} _ {h} \in \mathcal {V} _ {h}} \frac {\int_ {\mathrm{Vol}} \gamma_ {h} \left[ \left(\partial w _ {h} / \partial x\right) - \beta_ {h} \right] d \mathrm{Vol}}{\| \mathbf {v} _ {h} \|} \geq \frac {\int_ {\mathrm{Vol}} \gamma_ {h} \left[ \left(\partial \hat {w} _ {h} / \partial x\right) - \hat {\beta} _ {h} \right] d \mathrm{Vol}}{\| \hat {\mathbf {v}} _ {h} \|} \\ = \sqrt {\int_ {\mathrm{Vol}} \left(\gamma_ {h}\right) ^ {2} d \mathrm{Vol}} \tag {4.222} \\ \end{array}
with \gamma_{h} still a variable. Therefore, for the two-node mixed interpolated beam element we have
\inf _ {\gamma_ {h} \in P _ {h} (D _ {h})} \sup _ {\mathbf {v} _ {h} \in V _ {h}} \frac {\int_ {\mathrm{Vol}} \gamma_ {h} \left[ \left(\partial w _ {h} / \partial x\right) - \beta_ {h} \right] d \mathrm{Vol}}{\| \gamma_ {h} \| \| \mathbf {v} _ {h} \|} \geq 1 \tag {4.223}
and the inf-sup condition is satisfied.
We can also apply the inf-sup eigenvalue test to the two-node beam elements. The equations used are those presented for the elasticity problem, but we use the spaces of the beam elements (see Exercise 4.63). Figure 4.27 shows the results obtained. We note that in (4.207) the smallest nonzero eigenvalue of the pure displacement-based discretization approaches zero as the mesh is refined, whereas the mixed interpolated beam element meshes give an eigenvalue that equals 1.0 for all meshes [which corresponds to the equal sign in (4.223)].
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| log(1/N) | 2-node displacement-based element | 2-node mixed interpolated (linear displacement, linear rotation, constant shear) element | 3-node displacement-based element | | -------- | ---------------------------------- | ---------------------------------------------------------------------------------- | ---------------------------------- | | -1.0 | 0.0 | 0.0 | -1.8 | | -0.8 | 0.0 | 0.0 | -1.6 | | -0.6 | 0.0 | 0.0 | -1.4 | | -0.4 | 0.0 | 0.0 | -1.2 | | -0.2 | 0.0 | 0.0 | -1.0 | | 0.0 | 0.0 | 0.0 | -0.8 |Figure 4.27 Inf-sup test of beam elements (a cantilever beam is considered)
Higher-order mixed interpolated beam elements can be analyzed in the same way as the two-node elements (see Exercise 4.62). Figure 4.27 also shows the results obtained for the three-node pure displacement-based element with the numerical inf-sup test.
4.5.8 Exercises
4.47. Prove that \| \mathbf{u} - \mathbf{u}_h\| \leq \tilde{c} d[\mathbf{u},K_h(0)] is always true, where \mathbf{u}_h is the finite element solution and K_{h}(0) is defined in (4.159). Use that
\exists \alpha > 0 \quad \text { such that } \quad \forall \mathbf {v} _ {h} \in K _ {h} (0), a (\mathbf {v} _ {h}, \mathbf {v} _ {h}) \geq \alpha \| \mathbf {v} _ {h} \| ^ {2}
\exists M > 0 \quad \text { such that } \quad \forall \mathbf {v} _ {h 1}, \mathbf {v} _ {h 2} \in V _ {h}, | a (\mathbf {v} _ {h 2}, \mathbf {v} _ {h 1}) | \leq M \| \mathbf {v} _ {h 1} \| \| \mathbf {v} _ {h 2} \|
and the approach in (4.94). Note that the constant \tilde{c} is independent of the bulk modulus.
4.48. Prove that \| \operatorname{div}(\mathbf{v}_1 - \mathbf{v}_2) \|_0 \leq c \| \mathbf{v}_1 - \mathbf{v}_2 \|_V . Here \mathbf{v}_1, \mathbf{v}_2 \in V_h and c is a constant.
4.49. Evaluate P_{h}(\mathrm{div} \mathbf{v}_{h}) for the eight-node element shown assuming a constant pressure field over the element.
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2 5 1 2 6 8 x 3 7 4 2
4.50. Evaluate the stiffness matrix of a general 3/1 triangular u/p element for two-dimensional analysis. Hence, the element has three nodes and a constant discontinuous pressure is assumed. Use the data in Fig. E4.17 and consider plane stress, plane strain, and axisymmetric conditions.
(a) Establish all required matrices using the general procedure for the u / p elements (see Example 4.32) but do not perform any matrix multiplications. Consider the case \kappa finite.
(b) Compare the results obtained in Example 4.17 with the results obtained in part (a).
(c) Give the u / p element matrix when total incompressibility is assumed (hence static condensation on the pressure degree of freedom cannot be performed).
(Note: This element is not a reliable element for practical analysis of (almost) incompressible conditions but is merely used here in an exercise.)
4.51. Consider the 4/1 element in Example 4.32. Show that using the term P_h(\text{div } \mathbf{v}_h) (evaluated in Example 4.34) in (4.179), we obtain the same element stiffness matrix as that found in Example 4.32.
4.52. Consider the 9/3 element in Example 4.36; i.e., assume that Q_{h} = [1, x, y] . Assume that corresponding to \mathbf{v}_h the nodal point displacements are
u _ {1} = 1; \quad u _ {2} = - 1; \quad u _ {3} = - 1; \quad u _ {4} = 1; \quad u _ {6} = - 1; \quad u _ {8} = 1
v _ {1} = 1; \quad v _ {2} = - 1; \quad v _ {3} = - 1; \quad v _ {4} = 1; \quad v _ {6} = - 1; \quad v _ {8} = 1
with all other nodal point displacements zero. Calculate the projection P_{h}(\text{div } \mathbf{v}_{h}) .
4.53. Show that the 8/1 u/p element satisfies the inf-sup condition (and hence discretizations using this element will not display a spurious pressure mode). For the proof refer to Example 4.36.
4.54. Consider the solution of (4.187) and show that the conditions i and ii in (4.188) and (4.189) are necessary and sufficient for a unique solution.
4.55. Consider the ellipticity condition in (4.192). Prove that this condition is satisfied for the 4/1 element in two-dimensional plane stress and plane strain analyses.
4.56. The constant pressure mode, p_0 \in Q_h , in a two-dimensional square plane strain domain of an incompressible material modeled using four 9/3 elements with all boundary displacements set to zero is not a spurious mode (because it physically should exist). Show that this mode is not an element of P_h(D_h) .
4.57. Consider the 4/1 element. Can you construct a two-element model with appropriate boundary conditions that contains a spurious pressure mode? Explain your answer.
4.58. Consider the nine 4/1 elements shown. Assume that all boundary displacements are zero.
(a) Pick a pressure distribution \hat{p}_{h} for which there exists a vector v_{h} such that
\int_ {\mathrm{Vol}} \hat {p} _ {h} \operatorname{div} \mathbf {v} _ {h} d \mathrm{Vol} > 0
(b) Pick a pressure distribution \hat{p}_h for which any displacement distribution \mathbf{v}_h in V_h will give
\int_ {\mathrm{Vol}} \hat {p} _ {h} \operatorname{div} \mathbf {v} _ {h} d \mathrm{Vol} = 0
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Element ① v₁ ④ v₄ ⑦ u₁. u₄ ② v₂ ⑤ v₃ ⑧ u₂ u₃ ③ ⑥ ⑨
4.59. Consider the u/p-c formulation.
(a) Show that the inf-sup condition can be written as in (4.206) but that \mathbf{G}_h = (\mathbf{K}_{up})_h\mathbf{T}_h^{-1}(\mathbf{K}_{pu})_h .
(b) Also, show that, alternatively, the eigenproblem
\mathbf {G} _ {h} ^ {\prime} \mathbf {Q} _ {h} = \lambda^ {\prime} \mathbf {T} _ {h} \mathbf {Q} _ {h} \tag {a}
can be considered, where \mathbf{G}_h' = (\mathbf{K}_{pu})_h\mathbf{S}_h^{-1}(\mathbf{K}_{up})_h , and that the smallest nonzero eigenvalues of (a) and (4.207) are the same.
Here T_{h} is the matrix of the L^{2} -norm of p_{h} ; that is, for any vector of nodal pressures P_{h} , we have \|p_{h}\| = P_{h}^{T}T_{h}P_{h} ; hence T_{h} = -\kappa(K_{pp})_{h} .
4.60. Consider the analysis of the cantilever plate in plane strain conditions shown. Assume that the 3/1 u/p element is to be used in a sequence of uniform mesh refinements. Let n_{u} be the number
of nodal point displacements and n_{p} the number of pressure variables. Show that as the mesh is refined, the ratio n_{u}/n_{p} approaches 1. (This clearly indicates solution difficulties.)
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L L
Young's modulus E Poisson's ratio v = 0.499 Plane strain conditions
Calculate the same ratio when the 9/3 and 9/8-c elements are used (the 9/8-c element is defined in Table 4.8) and discuss your result.
4.61. Show that the mixed interpolated two-, three-, and four-node beam elements satisfy the ellipticity condition. The two-node element was considered in Section 4.5.7, and the three- and four-node elements are discussed in Section 5.4.1 (see also Exercise 4.62).
4.62. Show analytically that the inf-sup condition is not satisfied for the three- and four-node displacement-based beam elements and that the condition is satisfied for the mixed interpolated beam elements with \gamma_{h} varying, respectively, linearly and parabolically (see Section 5.4.1).
4.63. Establish the eigenvalue problem of the numerical inf-sup test for the beam elements considered in Section 4.5.7. Use the form (4.207) and define all matrices in detail.
4.64. Consider the problem in Fig. 4.24 and the elements mentioned in Table 4.8. Calculate, for each of these elements, the constraint ratio defined as the number of displacement degrees of freedom divided by the number of pressure degrees of freedom as the mesh is refined, that is, as h \rightarrow 0 . Hence note that this constraint ratio alone does not show whether or not the inf-sup condition is satisfied.
Formulation and Calculation of Isoparametric Finite Element Matrices
5.1 INTRODUCTION
A very important phase of a finite element solution is the calculation of the finite element matrices. In Chapter 4 we discussed the formulation and calculation of generalized coordinate finite element models. The aim in the presentation of the generalized coordinate finite elements was primarily to enhance our understanding of the finite element method. We have already pointed out that in most practical analyses the use of isoparametric finite elements is more effective. For the original developments of these elements, see I. C. Taig [A] and B. M. Irons [A, B].
Our objective in this chapter is to present the formulation of isoparametric finite elements and describe effective implementations. In the derivation of generalized coordinate finite element models, we used local element coordinate systems x, y, z and assumed the element displacements u(x, y, z) , v(x, y, z) , and w(x, y, z) (and in the case of mixed methods also the element stress and strain variables) in the form of polynomials in x, y, and z with undetermined constant coefficients \alpha_{i} , \beta_{i} , and \gamma_{i} , i = 1, 2, \ldots , identified as generalized coordinates. It was not possible to associate a priori a physical meaning with the generalized coordinates; however, on evaluation we found that the generalized coordinates determining the displacements are linear combinations of the element nodal point displacements. The principal idea of the isoparametric finite element formulation is to achieve the relationship between the element displacements at any point and the element nodal point displacements directly through the use of interpolation functions (also called shape functions). This means that the transformation matrix A^{-1} [see (4.57)] is not evaluated; instead, the element matrices corresponding to the required degrees of freedom are obtained directly.
5.2 ISOPARAMETRIC DERIVATION OF BAR ELEMENT STIFFNESS MATRIX
Consider the example of a bar element to illustrate the procedure of an isoparametric stiffness formulation. In order to simplify the explanation, assume that the bar lies in the global X-coordinate axis, as shown in Fig. 5.1. The first step is to relate the actual global coordinates X to a natural coordinate system with variable r, -1 \leq r \leq 1 (Fig. 5.1). This transformation is given by
X = \frac {1}{2} (1 - r) X _ {1} + \frac {1}{2} (1 + r) X _ {2} \tag {5.1}
or
X = \sum_ {i = 1} ^ {2} h _ {i} X _ {i} \tag {5.2}
where h_{1} = \frac{1}{2}(1 - r) and h_{2} = \frac{1}{2}(1 + r) are the interpolation or shape functions. Note that (5.2) establishes a unique relationship between the coordinates X and r on the bar.
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Y X2 X1 U1 U2 Z r = -1 r r = +1 X, U
Figure 5.1 Element in global and natural coordinate system
The bar global displacements are expressed in the same way as the global coordinates:
U = \sum_ {i = 1} ^ {2} h _ {i} U _ {i} \tag {5.3}
where in this case a linear displacement variation is specified. The interpolation of the element coordinates and element displacements using the same interpolation functions, which are defined in a natural coordinate system, is the basis of the isoparametric finite element formulation.
For the calculation of the element stiffness matrix we need to find the element strains \epsilon = dU / dX . Here we use
\epsilon = \frac {d U}{d r} \frac {d r}{d X} \tag {5.4}
where, from (5.3), \frac{dU}{dr} = \frac{U_2 - U_1}{2} (5.5)
and using (5.2), we obtain
\frac {d X}{d r} = \frac {X _ {2} - X _ {1}}{2} = \frac {L}{2} \tag {5.6}
where L is the length of the bar. Hence, as expected, we have
\epsilon = \frac {U _ {2} - U _ {1}}{L} \tag {5.7}
The strain-displacement transformation matrix corresponding to (4.32) is therefore
\mathbf {B} = \frac {1}{L} [ - 1 \quad 1 ] \tag {5.8}
In general, the strain-displacement transformation matrix is a function of the natural coordinates, and we therefore evaluate the stiffness matrix volume integral in (4.33) by integrating over the natural coordinates. Following this general procedure, although in this example it is not necessary, we have
\mathbf {K} = \frac {A E}{L ^ {2}} \int_ {- 1} ^ {1} \left[ \begin{array}{c} - 1 \\ 1 \end{array} \right] \left[ \begin{array}{l l} - 1 & 1 \end{array} \right] J d r \tag {5.9}
where the bar area A and modulus of elasticity E have been assumed constant and J is the Jacobian relating an element length in the global coordinate system to an element length in the natural coordinate system; i.e.,
d X = J d r \tag {5.10}
From (5.6) we have
J = \frac {L}{2} \tag {5.11}
Then, evaluating (5.9), we obtain the well-known matrix
\mathbf {K} = \frac {A E}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \tag {5.12}
As stated in the introduction, the isoparametric formulation avoids the construction of the transformation matrix A^{-1} . In order to compare this formulation with the generalized coordinate formulation, we need to solve from (5.1) for r and then substitute for r into (5.3). We obtain
r = \frac {X - [ (X _ {1} + X _ {2}) / 2 ]}{L / 2} \tag {5.13}
and then
U = \alpha_ {0} + \alpha_ {1} X \tag {5.14}
where
\left. \begin{array}{l} \alpha_ {0} = \frac {1}{2} (U _ {1} + U _ {2}) - \frac {X _ {1} + X _ {2}}{2 L} (U _ {2} - U _ {1}) \\ \alpha_ {1} = \frac {1}{L} (U _ {2} - U _ {1}) \end{array} \right\} \tag {5.15}
or
\boldsymbol {\alpha} = \left[ \begin{array}{c c} \frac {1}{2} + \frac {X _ {1} + X _ {2}}{2 L} & \frac {1}{2} - \frac {X _ {1} + X _ {2}}{2 L} \\ - \frac {1}{L} & \frac {1}{L} \end{array} \right] \mathbf {U} \tag {5.16}
where
\boldsymbol {\alpha} ^ {T} = [ \alpha_ {0} \quad \alpha_ {1} ]; \quad \mathbf {U} ^ {T} = [ U _ {1} \quad U _ {2} ] \tag {5.17}
and the matrix relating \alpha to U in (5.16) is A^{-1} . It should be noted that in this example the generalized coordinates \alpha_{0} and \alpha_{1} relate the global element displacement to the global element coordinate [see (5.14)].
5.3 FORMULATION OF CONTINUUM ELEMENTS
For a continuum finite element, it is in most cases effective to calculate directly the element matrices corresponding to the global degrees of freedom. However, we shall first present the formulation of the matrices that correspond to the element local degrees of freedom because additional considerations may be necessary when the element matrices that correspond to the global degrees of freedom are calculated directly (see Section 5.3.4). In the following we consider the derivation of the element matrices of straight truss elements; two-dimensional plane stress, plane strain, and axisymmetric elements; and three-dimensional elements that all have a variable number of nodes. Typical elements are shown in Fig. 5.2.
We direct our discussion to the calculation of displacement-based finite element matrices. However, the same procedures are also used in the calculation of the element matrices of mixed formulations, and in particular of the displacement/pressure-based formulations for incompressible analysis, as briefly discussed in Section 5.3.5.
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Two identical diagonal lines with circular endpoints, no text or symbols present
(a) Truss and cable elements
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Simple geometric diagram of a rectangle with vertices marked by dots (no text or symbols)
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Simple curved line diagram with marked points (no text or symbols)
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Simple geometric diagram of a triangle with vertices marked by dots (no text or labels)
(b) Two-dimensional elements
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Simple line drawing of a 3D rectangular prism with solid and dashed lines indicating visible and hidden edges (no text or symbols)
(c) Three-dimensional elements
Figure 5.2 Some typical continuum elements
5.3.1 Quadrilateral Elements
The basic procedure in the isoparametric finite element formulation is to express the element coordinates and element displacements in the form of interpolations using the natural coordinate system of the element. This coordinate system is one-, two-, or three-dimensional, depending on the dimensionality of the element. The formulation of the element matrices is the same whether we deal with a one, two-, or three-dimensional element. For this reason we use in the following general presentation the equations of a three-dimensional element. However, the one- and two-dimensional elements are included by simply using only the relevant coordinate axes and the appropriate interpolation functions.
Considering a general three-dimensional element, the coordinate interpolations are
\boxed {x = \sum_ {i = 1} ^ {q} h _ {i} x _ {i}; \quad y = \sum_ {i = 1} ^ {q} h _ {i} y _ {i}; \quad z = \sum_ {i = 1} ^ {q} h _ {i} z _ {i}} \tag {5.18}
where x, y, and z are the coordinates at any point of the element (here local coordinates) and x_{i} , y_{i} , z_{i} , i = 1, \ldots, q , are the coordinates of the q element nodes. The interpolation functions h_{i} are defined in the natural coordinate system of the element, which has variables r, s, and t that each vary from -1 to +1. For one- or two-dimensional elements, only the relevant equations in (5.18) would be employed, and the interpolation functions would depend only on the natural coordinate variables r and r, s, respectively.
The unknown quantities in (5.18) are so far the interpolation functions h_{i} . The fundamental property of the interpolation function h_{i} is that its value in the natural coordinate system is unity at node i and zero at all other nodes. Using these conditions, the functions h_{i} corresponding to a specific nodal point layout could be solved for in a systematic manner. However, it is convenient to construct them by inspection, which is demonstrated in the following simple example.
EXAMPLE 5.1: Construct the interpolation functions corresponding to the three-node truss element in Fig. E5.1.
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Node 1 0.3L 0.7L Node 2 r x r Node 3 r = -1 r = 0 r = +1 x = 0 x = 0.3L x = L
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| r | Value | |-------|-------| | -1 | +1 | | 0 | 0 | | +1 | 0 |
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r = -1 r = 0 +1 r = +1 h₂ = 1 - r²
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r = -1 r = 0 r = +1 h₃ = ½(1 + r) - ½(1 - r²) +1
Figure E5.1 One-dimensional interpolation functions of a truss element
A first observation is that for the three-node truss element we want interpolation polynomials that involve r^{2} as the highest power of r; in other words, the interpolation functions shall be parabolas. The function h_{2} can thus be constructed without much effort. Namely, the parabola that satisfies the conditions to be equal to zero at r = \pm 1 and equal to 1 at r = 0 is given by (1 - r^{2}) . The other two interpolation functions h_{1} and h_{3} are constructed by superimposing a linear function and a parabola. Consider the interpolation function h_{3} . Using \frac{1}{2}(1 + r) , the conditions that the function shall be zero at r = -1 and 1 at r = +1 are satisfied. To ensure that h_{3} is also zero at r = 0, we need to use h_{3} = \frac{1}{2}(1 + r) - \frac{1}{2}(1 - r^{2}) . The interpolation function h_{1} is obtained in a similar manner.
The procedure used in Example 5.1 of constructing the final required interpolation functions suggests an attractive formulation of an element with a variable number of nodes. This formulation is achieved by constructing first the interpolations corresponding to a basic two-node element. The addition of another node then results in an additional interpolation function and a correction to be applied to the already existing interpolation functions. Figure 5.3 gives the interpolation functions of the one-dimensional element considered in Example 5.1, with an additional fourth node possible. As shown, the element can have from two to four nodes. We should note that nodes 3 and 4 are now interior nodes because nodes 1 and 2 are used to define the two-node element.
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Node 1 0.3L 0.5L 0.2L Node 3 Node 4 Node 2 r = -1 r = 0 r = -1 r = -1/3 r = +1/3 r = +1 ... 3-node case r = +1 ... 4-node case
(a) 2 to 4 variable-number-nodes truss element
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Include only if node 3 is present Include only if nodes 3 and 4 are present h₁ = ½(1 - r)..... -½(1 - r²)..... +½(-9r³ + r² + 9r - 1) h₂ = ½(1 + r)..... -½(1 - r²)..... +½(9r³ + r² - 9r - 1) h₃ = (1 - r²)..... +½(27r³ + 7r² - 27r - 7) h₄ = ½(-27r³ - 9r² + 27r + 9)
(b) Interpolation functions
Figure 5.3 Interpolation functions of two to four variable-number-nodes one-dimensional element
This procedure of constructing the element interpolation functions for one-dimensional analysis can be directly generalized for use in two and three dimensions. Figure 5.4 shows the interpolation functions of a four to nine variable-number-nodes two-dimensional element, and Fig. 5.5 gives the interpolation functions for three-dimensional 8- to 20-node elements. The two- and three-dimensional interpolations have been established in a manner analogous to the one-dimensional interpolations, where the basic functions used are, in fact, those already employed in Fig. 5.3. We consider in Figs. 5.4 and 5.5 at most parabolic interpolation, but variable-number-nodes elements with interpolations of higher order could be derived in an analogous way.