김경종
678 lines
30 KiB
Markdown
678 lines
30 KiB
Markdown
<!-- source-page: 441 -->
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and then, using the interpolation functions defined in Fig. 5.4,
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$$
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\left[ \begin{array}{l} \frac {\partial w}{\partial x} \\ \frac {\partial w}{\partial y} \end{array} \right] = \frac {1}{4} \left[ \begin{array}{l l l} \frac {2}{3} & & 0 \\ 0 & & 1 \end{array} \right] \left[ \begin{array}{c c c c} (1 + s) & - (1 + s) & - (1 - s) & (1 - s) \\ (1 + r) & (1 - r) & - (1 - r) & - (1 + r) \end{array} \right] \left[ \begin{array}{l} w _ {1} \\ w _ {2} \\ w _ {3} \\ w _ {4} \end{array} \right]
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$$
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with similar expressions for the derivatives of $\beta_{x}$ and $\beta_{y}$ . Thus, if we use the following notation,
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$$
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\kappa (r, s) = \mathbf {B} _ {\kappa} \hat {\mathbf {u}}
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$$
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$$
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\boldsymbol {\gamma} (r, s) = \mathbf {B} _ {\gamma} \hat {\mathbf {u}}
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$$
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$$
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w (r, s) = \mathbf {H} _ {w} \hat {\mathbf {u}}
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$$
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where $\hat{\mathbf{u}}^T = [w_1\theta_x^1\theta_y^1;w_2\ldots \theta_y^4]$
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we have $\mathbf{B}_{\kappa}=\begin{bmatrix}0&0&-\frac{1}{6}(1+s)&\\0&\frac{1}{4}(1+r)&0&\cdots&0\\0&\frac{1}{6}(1+s)&-\frac{1}{4}(1+r)&\end{bmatrix}$
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$$
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\mathbf {B} _ {\gamma} = \left[ \begin{array}{c c c c c} \frac {1}{6} (1 + s) & 0 & \frac {1}{4} (1 + r) (1 + s) & | & \frac {1}{4} (1 + r) (1 - s) \\ \frac {1}{4} (1 + r) & - \frac {1}{4} (1 + r) (1 + s) & 0 & | & 0 \end{array} \right]
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$$
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$$
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\mathbf {H} _ {w} = \frac {1}{4} [ (1 + r) (1 + s) \quad 0 \quad 0 \quad | \quad \dots \quad 0 ]
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$$
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The element stiffness matrix is then
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$$
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\mathbf {K} = \frac {3}{2} \int_ {- 1} ^ {+ 1} \int_ {- 1} ^ {+ 1} \left(\mathbf {B} _ {\kappa} ^ {T} \mathbf {C} _ {b} \mathbf {B} _ {\kappa} + \mathbf {B} _ {\gamma} ^ {T} \mathbf {C} _ {s} \mathbf {B} _ {\gamma}\right) d r d s \tag {a}
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$$
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and the consistent load vector is
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$$
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\mathbf {R} _ {s} = \frac {3}{2} \int_ {- 1} ^ {+ 1} \int_ {- 1} ^ {+ 1} \mathbf {H} _ {w} ^ {T} p d r d s \tag {b}
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$$
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where the integrals in (a) and (b) could be evaluated in closed form but are usually evaluated using numerical integration (see Section 5.5).
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This pure displacement-based plate element formulation is of value only when higher-order elements are employed. Indeed, the least order of interpolation that should be used is a cubic interpolation, which results in a 16-node quadrilateral element and a 10-node triangular element. However, even these high-order elements still do not display a good predictive capability, particularly when the elements are geometrically distorted and used for stress predictions (see, for example, M. L. Bucalem and K. J. Bathe [A]).
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As in the formulation of isoparametric beam elements, the basic difficulty is that spurious shear stresses are predicted with the displacement-based elements. These spurious shear stresses result in a strong artificial stiffening of the elements as the thickness/length ratio decreases. This effect of shear locking is more pronounced for a low-order element and when the elements are geometrically distorted because, simply, the error in the shear stresses is then larger.
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To arrive at efficient and reliable plate bending elements, the pure displacement-based formulation must be extended, and a successful approach is to use a mixed interpolation of transverse displacement, section rotations, and transverse shear strains.
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<!-- source-page: 442 -->
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We should note here that in the above discussion, we assumed that the integrals for the computation of the element matrices (stiffness and mass matrices and load vectors) are evaluated accurately; hence, throughout our discussion we assumed and shall continue to assume that the error in the numerical integration (that usually is performed in practice; see Section 5.5) is small and certainly does not change the character of the element matrices. A number of authors have advocated the use of simple reduced integration to alleviate the shear locking problem. We discuss such techniques briefly in Section 5.5.6.
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In the following we present a family of plate bending elements that have a good mathematical basis and are reliable and efficient. These elements are referred to as the MITCn elements, where n refers to the number of element nodes and n = 4, 9, 16 for the quadrilateral and n = 7, 12 for the triangular elements (here MITC stands for mixed interpolation of tensorial components), (see K. J. Bathe, M. L. Bucalem, and F. Brezzi [A]). Let us consider the MITC4 element in detail and give the basic interpolations for the other elements in tabular form.
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An important feature of the MITC element formulation is the use of tensorial components of shear strains so as to render the resulting element relatively distortion-insensitive. Figure 5.26 shows a generic four-node element with the coordinate systems used.
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<details>
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<summary>text_image</summary>
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z
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y
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2
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t
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s
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r
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θ⁴ᵧ
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w₄
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3
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x
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4
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θ⁴ₓ
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</details>
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View of general element
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<details>
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<summary>text_image</summary>
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y
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2
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s
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1
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A
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B
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D
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r
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3
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C
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4
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x
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2
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</details>
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Special 2 × 2 element in the x, y plane
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<details>
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<summary>text_image</summary>
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y
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2
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s
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1
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A
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B
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D
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r
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3
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C
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4
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x
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y2
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x2
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</details>
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A general element in the $x, y$ plane
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Figure 5.26 Conventions used in formulation of four-node plate bending element
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To circumvent the shear locking problem, we formulate the element stiffness matrix by including the bending and shear effects through different interpolations. For the section curvatures in (5.95) we use the same interpolation as in the displacement-based method, as evaluated using (5.99), but we proceed differently in evaluating the transverse shear strains.
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<!-- source-page: 443 -->
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Consider first the MITC4 element when it is of geometry $2 \times 2$ (for which the x, y coordinates could be taken to be equal to the r, s isoparametric coordinates). For this element we use the interpolation (see K. J. Bathe and E. N. Dvorkin [A])
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$$
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\gamma_ {r z} = \frac {1}{2} (1 + s) \gamma_ {r z} ^ {A} + \frac {1}{2} (1 - s) \gamma_ {r z} ^ {C} \tag {5.100}
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$$
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$$
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\gamma_ {s z} = \frac {1}{2} (1 + r) \gamma_ {s z} ^ {D} + \frac {1}{2} (1 - r) \gamma_ {s z} ^ {B}
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$$
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where $\gamma_{rz}^{A}$ , $\gamma_{rz}^{C}$ , $\gamma_{sz}^{D}$ , and $\gamma_{sz}^{B}$ are the (physical) shear strains at points A, B, C, and D evaluated by the displacement and section rotations in (5.99). Hence,
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$$
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\gamma_ {r z} = \frac {1}{2} (1 + s) \left(\frac {w _ {1} - w _ {2}}{2} + \frac {\theta_ {y} ^ {1} + \theta_ {y} ^ {2}}{2}\right) + \frac {1}{2} (1 - s) \left(\frac {w _ {4} - w _ {3}}{2} + \frac {\theta_ {y} ^ {4} + \theta_ {y} ^ {3}}{2}\right) \tag {5.101}
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$$
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$$
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\gamma_ {s z} = \frac {1}{2} (1 + r) \left(\frac {w _ {1} - w _ {4}}{2} - \frac {\theta_ {x} ^ {1} + \theta_ {x} ^ {4}}{2}\right) + \frac {1}{2} (1 - r) \left(\frac {w _ {2} - w _ {3}}{2} - \frac {\theta_ {x} ^ {2} + \theta_ {x} ^ {3}}{2}\right)
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$$
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With these interpolations given, all strain-displacement interpolation matrices can be directly constructed and the stiffness matrix is formulated in the standard manner. Of course, the same procedure can also be directly employed for any rectangular element.
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Considering next the case of a general quadrilateral four-node element, we use the same basic idea of interpolating the transverse shear strains, but—using the interpolation in (5.100)—we interpolate the covariant tensor components measured in the r, s, z coordinate system. In this way we are directly taking account of the element distortion (from the $2 \times 2$ geometry). Proceeding in this way with the tensor shear strain components, we obtain (see Example 5.30) the following expressions for the $\gamma_{xz}$ and $\gamma_{yz}$ shear strains:
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$$
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\gamma_ {x z} = \gamma_ {r z} \sin \beta - \gamma_ {s z} \sin \alpha \tag {5.102}
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$$
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$$
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\gamma_ {y z} = - \gamma_ {r z} \cos \beta + \gamma_ {s z} \cos \alpha
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$$
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where $\alpha$ and $\beta$ are the angles between the $r$ and $x$ axes and $s$ and $x$ axes, respectively, and
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$$
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\gamma_ {r z} = \frac {\sqrt {\left(C _ {x} + r B _ {x}\right) ^ {2} + \left(C _ {y} + r B _ {y}\right) ^ {2}}}{8 \det \mathbf {J}}
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$$
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$$
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\left\{(1 + s) \left[ \frac {w _ {1} - w _ {2}}{2} + \frac {x _ {1} - x _ {2}}{4} \left(\theta_ {y} ^ {1} + \theta_ {y} ^ {2}\right) - \frac {y _ {1} - y _ {2}}{4} \left(\theta_ {x} ^ {1} + \theta_ {x} ^ {2}\right) \right] \right.
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$$
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$$
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+ (1 - s) \left[ \frac {w _ {4} - w _ {3}}{2} + \frac {x _ {4} - x _ {3}}{4} \left(\theta_ {y} ^ {4} + \theta_ {y} ^ {3}\right) - \frac {y _ {4} - y _ {3}}{4} \left(\theta_ {x} ^ {4} + \theta_ {x} ^ {3}\right) \right] \Bigg \} \tag {5.103}
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$$
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$$
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\gamma_ {s z} = \frac {\sqrt {\left(A _ {x} + s B _ {x}\right) ^ {2} + \left(A _ {y} + s B _ {y}\right) ^ {2}}}{8 \det \mathbf {J}}
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$$
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$$
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\left\{(1 + r) \left[ \frac {w _ {1} - w _ {4}}{2} + \frac {x _ {1} - x _ {4}}{4} \left(\theta_ {y} ^ {1} + \theta_ {y} ^ {4}\right) - \frac {y _ {1} - y _ {4}}{4} \left(\theta_ {x} ^ {1} + \theta_ {x} ^ {4}\right) \right] \right.
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$$
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$$
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\left. + (1 - r) \left[ \frac {w _ {2} - w _ {3}}{2} + \frac {x _ {2} - x _ {3}}{4} \left(\theta_ {y} ^ {2} + \theta_ {y} ^ {3}\right) - \frac {y _ {2} - y _ {3}}{4} \left(\theta_ {x} ^ {2} + \theta_ {x} ^ {3}\right) \right] \right\}
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$$
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In equations (5.103) we have
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$$
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\det \mathbf {J} = \det \left[ \begin{array}{l l} \frac {\partial x}{\partial r} & \frac {\partial y}{\partial r} \\ \frac {\partial x}{\partial s} & \frac {\partial y}{\partial s} \end{array} \right] \tag {5.104}
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$$
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<!-- source-page: 444 -->
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and
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$$
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A _ {x} = x _ {1} - x _ {2} - x _ {3} + x _ {4}
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$$
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$$
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B _ {x} = x _ {1} - x _ {2} + x _ {3} - x _ {4}
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$$
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$$
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C _ {x} = x _ {1} + x _ {2} - x _ {3} - x _ {4} \tag {5.105}
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$$
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$$
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A _ {y} = y _ {1} - y _ {2} - y _ {3} + y _ {4}
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$$
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$$
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B _ {y} = y _ {1} - y _ {2} + y _ {3} - y _ {4}
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$$
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$$
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C _ {y} = y _ {1} + y _ {2} - y _ {3} - y _ {4}
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$$
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We further consider the above relationships in the following example.
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EXAMPLE 5.30: Derive the transverse shear strain interpolations of the general MITC4 plate bending element.
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In the natural coordinate system of the plate bending element, the covariant base vectors are defined as
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$$
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\mathbf {g} _ {r} = \frac {\partial \mathbf {x}}{\partial r}; \quad \mathbf {g} _ {s} = \frac {\partial \mathbf {x}}{\partial s} \tag {a}
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$$
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$$
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\mathbf {g} _ {z} = \frac {h}{2} \mathbf {e} _ {z}
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$$
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where x is the vector of coordinates, and $e_{x}$ , $e_{y}$ , $e_{z}$ are the base vectors of the Cartesian system.
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Let us recall that in the natural coordinate system, the strain tensor can be expressed using covariant tensor components and contravariant base vectors (see Section 2.4)
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$$
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\epsilon = \tilde {\epsilon} _ {i j} \mathbf {g} ^ {i} \mathbf {g} ^ {j}; \quad i, j \equiv r, s, z
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$$
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where the tilde ( $\sim$ ) indicates that the tensor components are measured in the natural coordinate system.
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To obtain the shear tensor components we now use the equivalent of (5.100),
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$$
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\tilde {\epsilon} _ {r z} = \frac {1}{2} (1 + s) \tilde {\epsilon} _ {r z} ^ {A} + \frac {1}{2} (1 - s) \tilde {\epsilon} _ {r z} ^ {C} \tag {b}
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$$
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$$
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\tilde {\epsilon} _ {s z} = \frac {1}{2} (1 + r) \tilde {\epsilon} _ {s z} ^ {D} + \frac {1}{2} (1 - r) \tilde {\epsilon} _ {s z} ^ {B} \tag {c}
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$$
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where $\tilde{\epsilon}_{rz}^{A}$ , $\tilde{\epsilon}_{rz}^{C}$ , $\tilde{\epsilon}_{sz}^{D}$ , and $\tilde{\epsilon}_{sz}^{B}$ are the shear tensor components at points A, B, C, and D evaluated from the displacement interpolations. To obtain these components we use the linear terms of the general relation for the strain components in terms of the base vectors (see Example 2.28),
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$$
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{ } _ { 0 } ^ { 1 } \tilde { \boldsymbol { \epsilon } } _ { i j } = \frac { 1 } { 2 } \left[ { } ^ { 1 } \mathbf { g } _ { i } \cdot { } ^ { 1 } \mathbf { g } _ { j } - { } ^ { 0 } \mathbf { g } _ { i } \cdot { } ^ { 0 } \mathbf { g } _ { j } \right]
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$$
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where the left superscript of the base vectors is equal to 1 for the deformed configuration and equal to 0 for the initial configuration. Substituting from (5.99) and (a), we obtain
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$$
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\tilde {\epsilon} _ {r z} ^ {A} = \frac {1}{4} \left[ \frac {h}{2} \left(w _ {1} - w _ {2}\right) + \frac {h}{4} \left(x _ {1} - x _ {2}\right) \left(\theta_ {y} ^ {1} + \theta_ {y} ^ {2}\right) - \frac {h}{4} \left(y _ {1} - y _ {2}\right) \left(\theta_ {x} ^ {1} + \theta_ {x} ^ {2}\right) \right]
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$$
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and $\tilde{\epsilon}_{rz}^{C} = \frac{1}{4}\left[\frac{h}{2} (w_{4} - w_{3}) + \frac{h}{4} (x_{4} - x_{3})(\theta_{y}^{4} + \theta_{y}^{3}) - \frac{h}{4} (y_{4} - y_{3})(\theta_{x}^{4} + \theta_{x}^{3})\right]$
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<!-- source-page: 445 -->
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Therefore, using (b), we obtain
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$$
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\begin{array}{l} \tilde {\epsilon} _ {r z} = \frac {1}{8} (1 + s) \left[ \frac {h}{2} \left(w _ {1} - w _ {2}\right) + \frac {h}{4} \left(x _ {1} - x _ {2}\right) \left(\theta_ {y} ^ {1} + \theta_ {y} ^ {2}\right) - \frac {h}{4} \left(y _ {1} - y _ {2}\right) \left(\theta_ {x} ^ {1} + \theta_ {x} ^ {2}\right) \right] \\ + \frac {1}{8} (1 - s) \left[ \frac {h}{2} \left(w _ {4} - w _ {3}\right) + \frac {h}{4} \left(x _ {4} - x _ {3}\right) \left(\theta_ {y} ^ {4} + \theta_ {y} ^ {3}\right) - \frac {h}{4} \left(y _ {4} - y _ {3}\right) \left(\theta_ {x} ^ {4} + \theta_ {x} ^ {3}\right) \right] \\ \end{array}
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$$
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and in the same way, using (c),
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$$
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\tilde {\epsilon} _ {s z} = \frac {1}{8} (1 + r) \left[ \frac {h}{2} \left(w _ {1} - w _ {4}\right) + \frac {h}{4} \left(x _ {1} - x _ {4}\right) \left(\theta_ {y} ^ {1} + \theta_ {y} ^ {4}\right) - \frac {h}{4} \left(y _ {1} - y _ {4}\right) \left(\theta_ {x} ^ {1} + \theta_ {x} ^ {4}\right) \right]
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$$
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$$
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+ \frac {1}{8} (1 - r) \left[ \frac {h}{2} \left(w _ {2} - w _ {3}\right) + \frac {h}{4} \left(x _ {2} - x _ {3}\right) \left(\theta_ {y} ^ {2} + \theta_ {y} ^ {3}\right) - \frac {h}{4} \left(y _ {2} - y _ {3}\right) \left(\theta_ {x} ^ {2} + \theta_ {x} ^ {3}\right) \right]
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$$
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Next we use
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$$
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\tilde {\epsilon} _ {i j} \mathbf {g} ^ {i} \mathbf {g} ^ {j} = \epsilon_ {k l} \mathbf {e} _ {k} \mathbf {e} _ {l} \tag {d}
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$$
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where the $\epsilon_{kl}$ are the components of the strain tensor measured in the Cartesian coordinate system. From (d) we obtain
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$$
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\gamma_ {x z} = 2 \tilde {\epsilon} _ {r z} \left(\mathbf {g} ^ {r} \cdot \mathbf {e} _ {x}\right) \left(\mathbf {g} ^ {z} \cdot \mathbf {e} _ {z}\right) + 2 \tilde {\epsilon} _ {s z} \left(\mathbf {g} ^ {s} \cdot \mathbf {e} _ {x}\right) \left(\mathbf {g} ^ {z} \cdot \mathbf {e} _ {z}\right)
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$$
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$$
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\gamma_ {y z} = 2 \tilde {\epsilon} _ {r z} \left(\mathbf {g} ^ {r} \cdot \mathbf {e} _ {y}\right) \left(\mathbf {g} ^ {z} \cdot \mathbf {e} _ {z}\right) + 2 \tilde {\epsilon} _ {s z} \left(\mathbf {g} ^ {s} \cdot \mathbf {e} _ {y}\right) \left(\mathbf {g} ^ {z} \cdot \mathbf {e} _ {z}\right) \tag {e}
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$$
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However (using the standard procedure described in Section 2.4),
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$$
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\mathbf {g} ^ {r} = \sqrt {g ^ {r r}} (\sin \beta \mathbf {e} _ {x} - \cos \beta \mathbf {e} _ {y})
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$$
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$$
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\mathbf {g} ^ {s} = \sqrt {g ^ {s s}} (- \sin \alpha \mathbf {e} _ {x} + \cos \alpha \mathbf {e} _ {y})
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$$
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$$
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\mathbf {g} ^ {z} = \sqrt {g ^ {z z}} \mathbf {e} _ {z}
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$$
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where $\alpha$ and $\beta$ are the angles between the r and x axes and s and x axes, respectively, and
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$$
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g ^ {r r} = \frac {\left(C _ {x} + r B _ {x}\right) ^ {2} + \left(C _ {y} + r B _ {y}\right) ^ {2}}{1 6 (\det J) ^ {2}}
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$$
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$$
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g ^ {s s} = \frac {\left(A _ {x} + s B _ {x}\right) ^ {2} + \left(A _ {y} + s B _ {y}\right) ^ {2}}{1 6 (\det \mathbf {J}) ^ {2}}
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$$
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where $A_{x}, B_{x}, C_{x}, A_{y}, B_{y}$ , and $C_{y}$ are defined in (5.105) and
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$$
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g ^ {z z} = \frac {4}{h ^ {2}}
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$$
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Substituting into (e), the relations in (5.102) are obtained.
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The MITC4 plate bending element is in rectangular or parallelogram geometric configurations identical or closely related to other four-node plate bending elements (see T. J. R. Hughes and T. E. Tezduyar [A] and R. H. MacNeal [A]). However, an important attribute of the MITC plate element is that it is a special case of a general shell element for linear and nonlinear analysis, see E. N. Dvorkin and K. J. Bathe [A] and K. J. Bathe and P. S. Lee [A]; and see for further developments and applications D.N. Kim and K. J. Bathe [A], T. Sussman and K. J. Bathe [D] and Z. Kazanci and K. J. Bathe [A].
|
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|
||
<!-- source-page: 446 -->
|
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|
||
Some observations pertaining to the MITC4 element are the following.
|
||
|
||
The element behaves like the two-node mixed interpolated isoparametric beam element (discussed in the previous section) when used in the analysis of two-dimensional beam action.
|
||
|
||
The element can be derived from the Hu-Washizu variational principle (see Example 4.30).
|
||
|
||
The element passes the patch test (for an analytical proof see K. J. Bathe and E. N. Dvorkin [B]).
|
||
|
||
A mathematical convergence analysis for the transverse displacement and section rotations has been provided by K. J. Bathe and F. Brezzi [A] (assuming uniform meshes, i.e., that the element assemblage consists of square elements of sides h).
|
||
|
||
This analysis gives the results
|
||
|
||
$$
|
||
\| \boldsymbol {\beta} - \boldsymbol {\beta} _ {h} \| _ {1} \leq c _ {1} h; \| \nabla w - \nabla w _ {h} \| _ {0} \leq c _ {2} h \tag {5.106}
|
||
$$
|
||
|
||
where $\beta$ and w are the exact solutions, $\beta_{h}$ and $w_{h}$ are the finite element solutions corresponding to a mesh of elements with sides h, and $c_{1}$ , $c_{2}$ are constants independent of h. A convergence analysis of the transverse shear strains gave the result that the $L^{2}$ -norm of the error is not bounded independent of the plate thickness (see F. Brezzi, M. Fortin, and R. Stenberg [A]).
|
||
|
||
The essence of the results of these analytical convergence studies is also seen in practice for uniform and distorted meshes. The element predicts the transverse displacements and bending strains quite well, but the transverse shear strain predictions may not be satisfactory, particularly when very thin plates are analyzed.
|
||
|
||
A most important observation in the mathematical analysis of the MITC4 element was that this element, in its mathematical basis, is an analog of the 4/1 element of the u/p element family presented in Section 4.4.3: in the u/p formulation the displacements and pressure are interpolated to satisfy the constraint of (almost) incompressibility, $e_{v} \simeq 0$ , whereas in the MITC4 element formulation the transverse displacement, section rotations and transverse shear strains are interpolated to satisfy the thin plate condition, $\gamma \simeq 0$ . This analogy between the incompressibility constraint in solid mechanics and the zero transverse shear strain constraint in the Reissner-Mindlin plate theory resulted in the development of a mathematical basis aimed at the construction of new plate bending elements (see K. J. Bathe and F. Brezzi [B]). Since these elements are all based on the mixed interpolation of the transverse displacement, section rotations and transverse shear strains and for general geometries use the tensorial components (as for the MITC4 element), we refer to these elements as MITC elements with n nodes (i.e., MITCn elements).
|
||
|
||
The basic difficulty is choosing the orders of interpolations of transverse displacement, section rotations, and transverse shear strains which together result in nonlocking behavior and optimal convergence of the element. The mathematical considerations for choosing the appropriate interpolations were summarized by K. J. Bathe and F. Brezzi [B], K. J. Bathe, M. L. Bucalem, and F. Brezzi [A], and F. Brezzi, K. J. Bathe, and M. Fortin [A], who presented the elements in Fig. 5.27 as well as additional ones, and also gave numerical results.
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|
||
<!-- source-page: 447 -->
|
||
|
||
Figure 5.27 and Table 5.3 summarize the interpolations of 9- and 16-node quadrilateral elements and 7- and 12-node triangular elements and give the rates of convergence. In Fig. 5.27 the interpolations are given for the elements in geometrically nondistorted form, and we use tensorial components, as for the MITC4 element, to generalize the interpolations to geometrically distorted elements. Let us illustrate the use of the interpolations given in Fig. 5.27 in an example.
|
||
|
||
MITC4 element: 4 nodes for interpolation of section rotations and transverse displacement (2 × 2 Gauss integration)
|
||
|
||
|
||
|
||
<details>
|
||
<summary>natural_image</summary>
|
||
|
||
Simple geometric diagram showing a square with vertices marked by dots, and a 3D Cartesian coordinate system (x, y, z) with origin labeled at origin.
|
||
</details>
|
||
|
||
- Nodes for $\beta_{x}, \beta_{y}$ , and $w$ interpolation
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
A
|
||
B
|
||
C
|
||
D
|
||
</details>
|
||
|
||
$\gamma_{xz} = a_1 + b_1y$ ; tying at points $A$ and $C$ $\gamma_{yz} = a_2 + b_2x$ ; tying at points $B$ and $D$
|
||
|
||
MITC9 element: 9 nodes for interpolation of section rotations and 8 nodes for interpolation of transverse displacement (3 × 3 Gauss integration)
|
||
|
||
|
||
|
||
<details>
|
||
<summary>natural_image</summary>
|
||
|
||
Simple geometric diagram showing a square with vertices marked by dots, and a 3D Cartesian coordinate system with x, y, z axes (no text or labels)
|
||
</details>
|
||
|
||
\- Nodes for $\beta_{x}, \beta_{y}$ , and $w$ interpolation
|
||
|
||
$⑤$ Node for $\beta_{x},\beta_{y}$ interpolation
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
B
|
||
A
|
||
C
|
||
H
|
||
a
|
||
a = 1/√3
|
||
D
|
||
G
|
||
E
|
||
F
|
||
</details>
|
||
|
||
$\gamma_{xz} = a_1 + b_1x + c_1y + d_1xy + e_1y^2$ ; tying at points $A, B, E, F$ $\gamma_{yz} = a_2 + b_2x + c_2y + d_2xy + e_2x^2$ ; tying at points $C, D, G, H$ plus integral tying $\int_A (\nabla w - \beta - \gamma) dA = 0$
|
||
Figure 5.27 Plate bending elements; square and equilateral triangles of side lengths 2 units are considered.
|
||
|
||
<!-- source-page: 448 -->
|
||
|
||
MITC16 element:
|
||
16 nodes for interpolation of section rotations and 13 nodes for interpolation of transverse displacement (4 × 4 Gauss integration)
|
||
|
||
|
||
<details>
|
||
<summary>natural_image</summary>
|
||
|
||
Simple 3D coordinate system with x, y, z axes and a square grid containing five circles (no text or labels)
|
||
</details>
|
||
|
||
- Nodes for $\beta_{x}, \beta_{y}$ , and $w$ interpolation
|
||
© Nodes for $\beta_{x}, \beta_{y}$ interpolation
|
||
○ Node for w interpolation
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
C B A
|
||
D L
|
||
b
|
||
E K
|
||
F J
|
||
G H I
|
||
b = \u221a 3/5
|
||
</details>
|
||
|
||
$$
|
||
\gamma_ {x z} = a _ {1} + b _ {1} x + c _ {1} y + d _ {1} x ^ {2} + \theta_ {1} x y +
|
||
$$
|
||
|
||
$$
|
||
f _ {1} y ^ {2} + g _ {1} x ^ {2} y + h _ {1} x y ^ {2} + i _ {1} y ^ {3};
|
||
$$
|
||
|
||
tying at points A, B, C, G, H, I;
|
||
|
||
$$
|
||
\gamma_ {y z} = a _ {2} + b _ {2} x + c _ {2} y + d _ {2} x ^ {2} + e _ {2} x y +
|
||
$$
|
||
|
||
$$
|
||
f _ {2} y ^ {2} + g _ {2} x ^ {2} y + h _ {2} x y ^ {2} + i _ {2} x ^ {3};
|
||
$$
|
||
|
||
tying at points D, E, F, J, K, L;
|
||
|
||
plus integral tying $\int_{A}(\nabla w - \beta -\gamma)dA =$
|
||
|
||
$$
|
||
\int_ {A} (\nabla w - \beta - \gamma) x d A = \int_ {A} (\nabla w - \beta - \gamma) y d A = 0
|
||
$$
|
||
|
||
MITC7 element:
|
||
7 nodes for interpolation of section rotations and 6 nodes for interpolation of transverse displacement (7-point Gauss integration)
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
y
|
||
z
|
||
x
|
||
</details>
|
||
|
||
- Nodes for $\beta_{x}, \beta_{y}$ , and $w$ interpolation
|
||
© Node for $\beta_{x}, \beta_{y}$ interpolation
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
a
|
||
A
|
||
F
|
||
τ
|
||
a=1/√3
|
||
B
|
||
E
|
||
C
|
||
D
|
||
</details>
|
||
|
||
$$
|
||
\gamma_ {x z} = a _ {1} + b _ {1} x + c _ {1} y + y (d x + e y);
|
||
$$
|
||
|
||
$$
|
||
\gamma_ {y z} = a _ {2} + b _ {2} x + c _ {2} y - x (d x + e y);
|
||
$$
|
||
|
||
tying of $\gamma \cdot \tau$ at $A, B, C, D, E, F$
|
||
|
||
plus integral tying $\int_{A}(\nabla w - \beta -\gamma)dA = 0$
|
||
|
||
Figure 5.27 (continued)
|
||
|
||
<!-- source-page: 449 -->
|
||
|
||
MITC12 element:
|
||
|
||
12 nodes for interpolation of section rotations and 10 nodes for interpolation of transverse displacement (13-point Gauss integration)
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
y
|
||
z
|
||
x
|
||
</details>
|
||
|
||
- Nodes for $\beta_{x}, \beta_{y}$ , and $w$ interpolation
|
||
© Nodes for $\beta_{x}, \beta_{y}$ interpolation
|
||
○ Node for w interpolation
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
b
|
||
A
|
||
τ
|
||
B
|
||
H
|
||
b = \u221a3/5
|
||
C
|
||
D
|
||
E
|
||
F
|
||
G
|
||
</details>
|
||
|
||
$$
|
||
\gamma_ {x z} = a _ {1} + b _ {1} x + c _ {1} y + d _ {1} x ^ {2} + e _ {1} x y +
|
||
$$
|
||
|
||
$$
|
||
f _ {1} y ^ {2} + y (g x ^ {2} + h x y + i y ^ {2});
|
||
$$
|
||
|
||
$$
|
||
\gamma_ {y z} = a _ {2} + b _ {2} x + c _ {2} y + d _ {2} x ^ {2} + e _ {2} x y +
|
||
$$
|
||
|
||
$$
|
||
f _ {2} y ^ {2} - x (g x ^ {2} + h x y + i y ^ {2});
|
||
$$
|
||
|
||
$$
|
||
\text { tying of } \gamma \cdot \tau \text { at points } A, B, C, D, E, F, G, H, I
|
||
$$
|
||
|
||
$$
|
||
\text { plus integral tying } \int_ {A} (\nabla w - \beta - \gamma) d A = \int_ {A} (\nabla w - \beta - \gamma) x d A =
|
||
$$
|
||
|
||
$$
|
||
\int_ {A} (\nabla w - \beta - \gamma) y d A = 0
|
||
$$
|
||
|
||
Figure 5.27 (continued)
|
||
TABLE 5.3 Interpolation spaces and theoretically predicted error estimates for plate bending elements
|
||
<table><tr><td>Element</td><td>Spaces used for section rotations and transverse displacement $^{\dagger}$ </td><td>Error estimates</td></tr><tr><td>MITC4</td><td> $\beta_{h} \in Q_{1} \times Q_{1}$ $w_{h} \in Q_{1}$ </td><td> $\|\beta - \beta_{h}\|_{1} \leq ch$ $\|\nabla_{w} - \nabla w_{h}\|_{0} \leq ch$ </td></tr><tr><td>MITC9</td><td> $\beta_{h} \in Q_{2} \times Q_{2}$ $w_{h} \in Q_{2} \cap P_{3}$ </td><td> $\|\beta - \beta_{h}\|_{1} \leq ch^{2}$ $\|\nabla_{w} - \nabla w_{h}\|_{0} \leq ch^{2}$ </td></tr><tr><td>MITC16</td><td> $\beta_{h} \in Q_{3} \times Q_{3}$ $w_{h} \in Q_{3} \cap P_{4}$ </td><td> $\|\beta - \beta_{h}\|_{1} \leq ch^{3}$ $\|\nabla_{w} - \nabla w_{h}\|_{0} \leq ch^{3}$ </td></tr><tr><td>MITC7</td><td> $\beta_{h} \in (P_{2} \oplus \{L_{1}L_{2}L_{3}\}) \times (P_{2} \oplus \{L_{1}L_{2}L_{3}\})$ $w_{h} \in P_{2}$ </td><td> $\|\beta - \beta_{h}\|_{1} \leq ch^{2}$ $\|\nabla_{w} - \nabla w_{h}\|_{0} \leq ch^{2}$ </td></tr><tr><td>MITC12</td><td> $\beta_{h} \in (P_{3} \oplus \{L_{1}L_{2}L_{3}\} P_{1}) \times (P_{3} \oplus \{L_{1}L_{2}L_{3}\} P_{1})$ $w_{h} \in P_{3}$ </td><td> $\|\beta - \beta_{h}\|_{1} \leq ch^{3}$ $\|\nabla_{w} - \nabla w_{h}\|_{0} \leq ch^{3}$ </td></tr></table>
|
||
|
||
$^{\dagger}$ For notation used, see Section 4.3.
|
||
|
||
EXAMPLE 5.31: Show how to establish the strain interpolation matrices for the stiffness matrix of the MITC9 element shown in Fig. E5.31.
|
||
|
||
The geometry of the element is the same as for the four-node element considered in Fig. E5.29; hence the Jacobian matrix is the same.
|
||
|
||
Since the transverse displacements are determined by the eight-node interpolations, which are given in Fig. 5.4, we have
|
||
|
||
<!-- source-page: 450 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
z
|
||
2 cm
|
||
3
|
||
6
|
||
2
|
||
y
|
||
7
|
||
9
|
||
s
|
||
r
|
||
5
|
||
3 cm
|
||
w1
|
||
4
|
||
8
|
||
1
|
||
x
|
||
h
|
||
θx1
|
||
θy1
|
||
</details>
|
||
|
||
Figure E5.31 A nine-node plate bending element
|
||
|
||
$$
|
||
\left[ \begin{array}{l} \frac {\partial w}{\partial x} \\ \frac {\partial w}{\partial y} \end{array} \right] = \frac {1}{4} \left[ \begin{array}{l l} \frac {2}{3} & 0 \\ 0 & 1 \end{array} \right] \left[ \begin{array}{l l} (1 + 2 r) (1 + s) - (1 - s ^ {2}) & \\ (1 + 2 s) (1 + r) - (1 - r ^ {2}) & \end{array} \right] \dots \left[ \begin{array}{l} w _ {1} \\ w _ {2} \\ \vdots \\ w _ {8} \end{array} \right] \tag {a}
|
||
$$
|
||
|
||
The section rotations are determined by the nine-node interpolation functions, which are also given in Fig. 5.4, and we have
|
||
|
||
$$
|
||
\left[ \begin{array}{l} \frac {\partial \beta_ {x}}{\partial x} \\ \frac {\partial \beta_ {x}}{\partial y} \end{array} \right] = - \frac {1}{4} \left[ \begin{array}{l l} \frac {2}{3} & 0 \\ 0 & 1 \end{array} \right] \left[ \begin{array}{l l} (1 + 2 r) (1 + s) - (1 + 2 r) (1 - s ^ {2}) & \\ (1 + 2 s) (1 + r) - (1 + 2 s) (1 - r ^ {2}) & \end{array} \right] \dots \left[ \begin{array}{l} \theta_ {y} ^ {1} \\ \theta_ {y} ^ {2} \\ \vdots \\ \dot {\theta} _ {y} ^ {9} \end{array} \right] \tag {b}
|
||
$$
|
||
|
||
$$
|
||
\left[ \begin{array}{l} \frac {\partial \beta_ {y}}{\partial x} \\ \frac {\partial \beta_ {y}}{\partial y} \end{array} \right] = \frac {1}{4} \left[ \begin{array}{l l} \frac {2}{3} & 0 \\ 0 & 1 \end{array} \right] \left[ \begin{array}{l l} (1 + 2 r) (1 + s) - (1 + 2 r) (1 - s ^ {2}) & | \\ (1 + 2 s) (1 + r) - (1 + 2 s) (1 - r ^ {2}) & | \end{array} \dots \right] \left[ \begin{array}{l} \theta_ {x} ^ {1} \\ \theta_ {x} ^ {2} \\ \cdot \\ \cdot \\ \dot {\theta} _ {x} ^ {9} \end{array} \right] \tag {c}
|
||
$$
|
||
|
||
Let us use the following ordering of nodal point displacements and rotations
|
||
|
||
$$
|
||
\hat {\mathbf {u}} ^ {T} = \left[ w _ {1} \quad \theta_ {x} ^ {1} \quad \theta_ {y} ^ {1} \quad \mid \quad \dots \quad \mid \quad w _ {8} \quad \theta_ {x} ^ {8} \quad \theta_ {y} ^ {8} \quad \mid \quad \theta_ {x} ^ {9} \quad \theta_ {y} ^ {9} \right]
|
||
$$
|
||
|
||
The transverse displacement interpolation matrix $H_{w}$ is then given by
|
||
|
||
$$
|
||
\mathbf {H} _ {w} = \left[ \begin{array}{c c c c c c c c c c c c c c c c} h _ {1} & 0 & 0 & | & h _ {2} & 0 & 0 & | & \dots & | & h _ {8} & 0 & 0 & | & 0 & 0 \end{array} \right]
|
||
$$
|
||
|
||
where the $h_1$ to $h_8$ are given in Fig. 5.4 and correspond to an eight-node element.
|
||
|
||
The curvature interpolation matrix $B_{\kappa}$ is obtained directly from the relations (b) and (c),
|
||
|
||
$$
|
||
\mathbf {B} _ {\kappa} = \left[ \begin{array}{c c} 0 & 0 \\ 0 & \frac {1}{4} [ (1 + 2 s) (1 + r) - (1 + 2 s) (1 - r ^ {2}) ] \\ 0 & \frac {1}{6} [ (1 + 2 r) (1 + s) - (1 + 2 r) (1 - s ^ {2}) ] \end{array} \right.
|
||
$$
|
||
|
||
$$
|
||
\left. \begin{array}{c c} - \frac {1}{6} [ (1 + 2 r) (1 + s) - (1 + 2 r) (1 - s ^ {2}) ] & \\ 0 & \dots \\ - \frac {1}{4} [ (1 + 2 s) (1 + r) - (1 + 2 s) (1 - r ^ {2}) ] & \end{array} \right]
|
||
$$
|