김경종
459 lines
21 KiB
Markdown
459 lines
21 KiB
Markdown
<!-- source-page: 511 -->
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where fsg and feg are defined by Eqs. (11.1.3), and the constitutive matrix ½D (see also Appendix C) is now given by
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$$
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[ D ] = \frac {E}{(1 + \nu) (1 - 2 \nu)} \left[ \begin{array}{c c c c c c} 1 - \nu & \nu & \nu & 0 & 0 & 0 \\ & 1 - \nu & \nu & 0 & 0 & 0 \\ & & 1 - \nu & 0 & 0 & 0 \\ & & & \frac {1 - 2 \nu}{2} & 0 & 0 \\ & & & & \frac {1 - 2 \nu}{2} & 0 \\ \text {Symmetry} & & & & & \frac {1 - 2 \nu}{2} \end{array} \right] \tag {11.1.5}
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$$
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# 11.2 Tetrahedral Element
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We now develop the tetrahedral stress element stiffness matrix by again using the steps outlined in Chapter 1. The development is seen to be an extension of the plane element previously described in Chapter 6. This extension was suggested in References [1] and [2].
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# Step 1 Select Element Type
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Consider the tetrahedral element shown in Figure 11–3 with corner nodes 1–4. This element is a four-noded solid. The nodes of the element must be numbered such that when viewed from the last node (say, node 4), the first three nodes are numbered in a counterclockwise manner, such as 1, 2, 3, 4 or 2, 3, 1, 4. This ordering of nodes avoids the calculation of negative volumes and is consistent with the counterclockwise node numbering associated with the CST element in Chapter 6. (Using an isoparametric formulation to evaluate the $\underline { { k } }$ matrix for the tetrahedral element enables us to use the element node numbering in any order. The isoparametric formulation of $\underline { { k } }$ is left
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<details>
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<summary>text_image</summary>
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z, w
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1
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3
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2
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4
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y, v
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x. u
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</details>
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Figure 11–3 Tetrahedral solid element
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<!-- source-page: 512 -->
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to Section 11.3.) The unknown nodal displacements are now given by
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$$
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\{d \} = \left\{ \begin{array}{l} u _ {1} \\ v _ {1} \\ w _ {1} \\ \vdots \\ u _ {4} \\ v _ {4} \\ w _ {4} \end{array} \right\} \tag {11.2.1}
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$$
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Hence, there are 3 degrees of freedom per node, or 12 total degrees of freedom per element.
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# Step 2 Select Displacement Functions
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For a compatible displacement field, the element displacement functions $u , v ,$ and w must be linear along each edge because only two points (the corner nodes) exist along each edge, and the functions must be linear in each plane side of the tetrahedron. We then select the linear displacement functions as
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$$
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u (x, y, z) = a _ {1} + a _ {2} x + a _ {3} y + a _ {4} z
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$$
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$$
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v (x, y, z) = a _ {5} + a _ {6} x + a _ {7} y + a _ {8} z \tag {11.2.2}
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$$
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$$
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w (x, y, z) = a _ {9} + a _ {1 0} x + a _ {1 1} y + a _ {1 2} z
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$$
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In the same manner as in Chapter 6, we can express the $\boldsymbol { a } _ { i } { } ^ { \prime } \boldsymbol { \mathbf { s } }$ in terms of the known nodal coordinates $( x _ { 1 } , y _ { 1 } , z _ { 1 } , \dotsc , z _ { 4 } )$ and the unknown nodal displacements $\left( u _ { 1 } , v _ { 1 } , w _ { 1 } , \ldots , w _ { 4 } \right)$ of the element. Skipping the straightforward but tedious details, we obtain
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$$
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u (x, y, z) = \frac {1}{6 V} \left\{\left(\alpha_ {1} + \beta_ {1} x + \gamma_ {1} y + \delta_ {1} z\right) u _ {1} \right.
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$$
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$$
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+ (\alpha_ {2} + \beta_ {2} x + \gamma_ {2} y + \delta_ {2} z) u _ {2}
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$$
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$$
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+ (\alpha_ {3} + \beta_ {3} x + \gamma_ {3} y + \delta_ {3} z) u _ {3}
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$$
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$$
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\left. + \left(\alpha_ {4} + \beta_ {4} x + \gamma_ {4} y + \delta_ {4} z\right) u _ {4} \right\} \tag {11.2.3}
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$$
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where 6V is obtained by evaluating the determinant
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$$
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6 V = \left| \begin{array}{c c c c} 1 & x _ {1} & y _ {1} & z _ {1} \\ 1 & x _ {2} & y _ {2} & z _ {2} \\ 1 & x _ {3} & y _ {3} & z _ {3} \\ 1 & x _ {4} & y _ {4} & z _ {4} \end{array} \right| \tag {11.2.4}
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$$
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<!-- source-page: 513 -->
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and $V$ represents the volume of the tetrahedron. The coefficients $\alpha_{i},\beta_{i},\gamma_{i}$ , and $\delta_{i}$ ( $i = 1,2,3,4$ ) in Eq. (11.2.3) are given by
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$$
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\alpha_ {1} = \left| \begin{array}{c c c} x _ {2} & y _ {2} & z _ {2} \\ x _ {3} & y _ {3} & z _ {3} \\ x _ {4} & y _ {4} & z _ {4} \end{array} \right| \quad \beta_ {1} = - \left| \begin{array}{c c c} 1 & y _ {2} & z _ {2} \\ 1 & y _ {3} & z _ {3} \\ 1 & y _ {4} & z _ {4} \end{array} \right| \tag {11.2.5}
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$$
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$$
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\gamma_ {1} = \left| \begin{array}{c c c} 1 & x _ {2} & z _ {2} \\ 1 & x _ {3} & z _ {3} \\ 1 & x _ {4} & z _ {4} \end{array} \right| \qquad \delta_ {1} = - \left| \begin{array}{c c c} 1 & x _ {2} & y _ {2} \\ 1 & x _ {3} & y _ {3} \\ 1 & x _ {4} & y _ {4} \end{array} \right|
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$$
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$$
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\alpha_ {2} = - \left| \begin{array}{c c c} x _ {1} & y _ {1} & z _ {1} \\ x _ {3} & y _ {3} & z _ {3} \\ x _ {4} & y _ {4} & z _ {4} \end{array} \right| \quad \beta_ {2} = \left| \begin{array}{c c c} 1 & y _ {1} & z _ {1} \\ 1 & y _ {3} & z _ {3} \\ 1 & y _ {4} & z _ {4} \end{array} \right| \tag {11.2.6}
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$$
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$$
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\gamma_ {2} = - \left| \begin{array}{c c c} 1 & x _ {1} & z _ {1} \\ 1 & x _ {3} & z _ {3} \\ 1 & x _ {4} & z _ {4} \end{array} \right| \qquad \delta_ {2} = \left| \begin{array}{c c c} 1 & x _ {1} & y _ {1} \\ 1 & x _ {3} & y _ {3} \\ 1 & x _ {4} & y _ {4} \end{array} \right|
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$$
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$$
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\alpha_ {3} = \left| \begin{array}{c c c} x _ {1} & y _ {1} & z _ {1} \\ x _ {2} & y _ {2} & z _ {2} \\ x _ {4} & y _ {4} & z _ {4} \end{array} \right| \quad \beta_ {3} = - \left| \begin{array}{c c c} 1 & y _ {1} & z _ {1} \\ 1 & y _ {2} & z _ {2} \\ 1 & y _ {4} & z _ {4} \end{array} \right| \tag {11.2.7}
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$$
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$$
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\gamma_ {3} = \left| \begin{array}{c c c} 1 & x _ {1} & z _ {1} \\ 1 & x _ {2} & z _ {2} \\ 1 & x _ {4} & z _ {4} \end{array} \right| \qquad \delta_ {3} = - \left| \begin{array}{c c c} 1 & x _ {1} & y _ {1} \\ 1 & x _ {2} & y _ {2} \\ 1 & x _ {4} & y _ {4} \end{array} \right|
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$$
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$$
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\alpha_ {4} = - \left| \begin{array}{c c c} x _ {1} & y _ {1} & z _ {1} \\ x _ {2} & y _ {2} & z _ {2} \\ x _ {3} & y _ {3} & z _ {3} \end{array} \right| \quad \beta_ {4} = \left| \begin{array}{c c c} 1 & y _ {1} & z _ {1} \\ 1 & y _ {2} & z _ {2} \\ 1 & y _ {3} & z _ {3} \end{array} \right| \tag {11.2.8}
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$$
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$$
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\gamma_ {4} = - \left| \begin{array}{c c c} 1 & x _ {1} & z _ {1} \\ 1 & x _ {2} & z _ {2} \\ 1 & x _ {3} & z _ {3} \end{array} \right| \qquad \delta_ {4} = \left| \begin{array}{c c c} 1 & x _ {1} & y _ {1} \\ 1 & x _ {2} & y _ {2} \\ 1 & x _ {3} & y _ {3} \end{array} \right|
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$$
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Expressions for $v$ and $w$ are obtained by simply substituting $v_{i}$ 's for all $u_{i}$ 's and then $w_{i}$ 's for all $u_{i}$ 's in Eq. (11.2.3).
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The displacement expression for u given by Eq. (11.2.3), with similar expressions for v and w, can be written equivalently in expanded form in terms of the shape
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<!-- source-page: 514 -->
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functions and unknown nodal displacements as
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$$
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\left\{ \begin{array}{l} u \\ v \\ w \end{array} \right\} = \left[ \begin{array}{c c c c c c c c c c c c c} N _ {1} & 0 & 0 & N _ {2} & 0 & 0 & N _ {3} & 0 & 0 & N _ {4} & 0 & 0 \\ 0 & N _ {1} & 0 & 0 & N _ {2} & 0 & 0 & N _ {3} & 0 & 0 & N _ {4} & 0 \\ 0 & 0 & N _ {1} & 0 & 0 & N _ {2} & 0 & 0 & N _ {3} & 0 & 0 & N _ {4} \end{array} \right] \left\{ \begin{array}{l} u _ {1} \\ v _ {1} \\ w _ {1} \\ \vdots \\ u _ {4} \\ v _ {4} \\ w _ {4} \end{array} \right\} \tag {11.2.9}
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$$
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where the shape functions are given by
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$$
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N _ {1} = \frac {\left(\alpha_ {1} + \beta_ {1} x + \gamma_ {1} y + \delta_ {1} z\right)}{6 V} \quad N _ {2} = \frac {\left(\alpha_ {2} + \beta_ {2} x + \gamma_ {2} y + \delta_ {2} z\right)}{6 V} \tag {11.2.10}
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$$
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$$
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N _ {3} = \frac {\left(\alpha_ {3} + \beta_ {3} x + \gamma_ {3} y + \delta_ {3} z\right)}{6 V} \quad N _ {4} = \frac {\left(\alpha_ {4} + \beta_ {4} x + \gamma_ {4} y + \delta_ {4} z\right)}{6 V}
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$$
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and the rectangular matrix on the right side of Eq. (11.2.9) is the shape function matrix ½N .
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# Step 3 Define the Strain= Displacement and Stress= Strain Relationships
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The element strains for the three-dimensional stress state are given by
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$$
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\{\varepsilon \} = \left\{ \begin{array}{l} \varepsilon_ {x} \\ \varepsilon_ {y} \\ \varepsilon_ {z} \\ \gamma_ {x y} \\ \gamma_ {y z} \\ \gamma_ {z x} \end{array} \right\} = \left\{ \begin{array}{c} \frac {\partial u}{\partial x} \\ \frac {\partial v}{\partial y} \\ \frac {\partial w}{\partial z} \\ \frac {\partial u}{\partial y} + \frac {\partial v}{\partial x} \\ \frac {\partial v}{\partial z} + \frac {\partial w}{\partial y} \\ \frac {\partial w}{\partial x} + \frac {\partial u}{\partial z} \end{array} \right\} \tag {11.2.11}
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$$
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Using Eq. (11.2.9) in Eq. (11.2.11), we obtain
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$$
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\{\varepsilon \} = [ B ] \{d \} \tag {11.2.12}
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$$
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where $\left[ B \right] = \left[ { \underline { { B } } } _ { 1 } \quad { \underline { { B } } } _ { 2 } \quad { \underline { { B } } } _ { 3 } \quad { \underline { { B } } } _ { 4 } \right]$ ð11:2:13Þ
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<!-- source-page: 515 -->
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The submatrix $\underline { { B } } _ { 1 }$ in Eq. (11.2.13) is defined by
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$$
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\underline {{{B}}} _ {1} = \left[ \begin{array}{c c c} N _ {1, x} & 0 & 0 \\ 0 & N _ {1, y} & 0 \\ 0 & 0 & N _ {1, z} \\ N _ {1, y} & N _ {1, x} & 0 \\ 0 & N _ {1, z} & N _ {1, y} \\ N _ {1, z} & 0 & N _ {1, x} \end{array} \right] \tag {11.2.14}
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$$
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where, again, the comma after the subscript indicates differentation with respect to the variable that follows. Submatrices $\underline { { B } } _ { 2 } , \underline { { B } } _ { 3 }$ , and ${ \underline { { B _ { 4 } } } }$ are defined by simply indexing the subscript in Eq. (11.2.14) from 1 to 2, 3, and then 4, respectively. Substituting the shape functions from Eqs. (11.2.10) into Eq. (11.2.14), $\underline { { B } } _ { 1 }$ is expressed as
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$$
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\underline {{{B}}} _ {1} = \frac {1}{6 V} \left[ \begin{array}{c c c} \beta_ {1} & 0 & 0 \\ 0 & \gamma_ {1} & 0 \\ 0 & 0 & \delta_ {1} \\ \gamma_ {1} & \beta_ {1} & 0 \\ 0 & \delta_ {1} & \gamma_ {1} \\ \delta_ {1} & 0 & \beta_ {1} \end{array} \right] \tag {11.2.15}
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$$
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with similar expressions for $\underline { { B } } _ { 2 } , \underline { { B } } _ { 3 }$ , and ${ \underline { { B _ { 4 } } } }$ .
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The element stresses are related to the element strains by
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$$
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\{\sigma \} = [ D ] \{\varepsilon \} \tag {11.2.16}
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$$
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where the constitutive matrix for an elastic material is now given by Eq. (11.1.5).
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# Step 4 Derive the Element Stiffness Matrix and Equations
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The element stiffness matrix is given by
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$$
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[ k ] = \iiint_ {V} [ B ] ^ {T} [ D ] [ B ] d V \tag {11.2.17}
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$$
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Because both matrices $[ B ]$ and $[ D ]$ are constant for the simple tetrahedral element, Eq. (11.2.17) can be simplified to
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$$
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[ k ] = [ B ] ^ {T} [ D ] [ B ] V \tag {11.2.18}
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$$
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where, again, V is the volume of the element. The element stiffness matrix is now of order $1 2 \times 1 2$ .
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# Body Forces
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The element body force matrix is given by
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$$
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\{f _ {b} \} = \iiint_ {V} [ N ] ^ {T} \{X \} d V \tag {11.2.19}
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$$
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<!-- source-page: 516 -->
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where ½N is given by the $3 \times 1 2$ matrix in Eq. (11.2.9), and
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$$
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\{X \} = \left\{ \begin{array}{l} X _ {b} \\ Y _ {b} \\ Z _ {b} \end{array} \right\} \tag {11.2.20}
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$$
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For constant body forces, the nodal components of the total resultant body forces can be shown to be distributed to the nodes in four equal parts. That is,
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$$
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\left\{f _ {b} \right\} = \frac {1}{4} \left[ X _ {b} Y _ {b} Z _ {b} X _ {b} Y _ {b} Z _ {b} X _ {b} Y _ {b} Z _ {b} X _ {b} Y _ {b} Z _ {b} \right. \left. X _ {b} Y _ {b} Z _ {b} X _ {b} Y _ {b} Z _ {b} X _ {b} Y _ {b} Z _ {b} \right] ^ {T}
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$$
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The element body force is then a $1 2 \times 1$ matrix.
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# Surface Forces
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Again, the surface forces are given by
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$$
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\{f _ {s} \} = \iint_ {S} \left[ N _ {s} \right] ^ {T} \{T \} d S \tag {11.2.21}
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$$
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where $[ N _ { s } ]$ is the shape function matrix evaluated on the surface where the surface traction occurs.
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For example, consider the case of uniform pressure $p$ acting on the face with nodes 1–3 of the element shown in Figure 11–3 or 11–4. The resulting nodal forces become
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$$
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\left\{f _ {s} \right\} = \iint_ {S} [ N ] ^ {T} \left| _ {\text { evaluated on surface 1,2,3 }} \left\{ \begin{array}{l} p _ {x} \\ p _ {y} \\ p _ {z} \end{array} \right. \right\} d S \tag {11.2.22}
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$$
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where $p _ { x } , p _ { y } ,$ and $p _ { z }$ are the $x , y ,$ and z components, respectively, of $p .$ Simplifying and integrating Eq. (11.2.22), we can show that
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$$
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\left\{f _ {s} \right\} = \frac {S _ {1 2 3}}{3} \left\{ \begin{array}{l} p _ {x} \\ p _ {y} \\ p _ {z} \\ p _ {x} \\ p _ {y} \\ p _ {z} \\ p _ {x} \\ p _ {y} \\ p _ {z} \\ 0 \\ 0 \\ 0 \end{array} \right\} \tag {11.2.23}
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$$
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where $S _ { 1 2 3 }$ is the area of the surface associated with nodes 1–3. The use of volume coordinates, as explained in Reference [8], facilitates the integration of Eq. (11.2.22).
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<!-- source-page: 517 -->
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Evaluate the matrices necessary to determine the stiffness matrix for the tetrahedral element shown in Figure 11–4. Let $E = 30 \times 10^{6}$ psi and v = 0.30. The coordinates are shown in the figure in units of inches.
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<details>
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<summary>text_image</summary>
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z
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1 (1, 1, 2)
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(0, 0, 0)
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2
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3
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(0, 2, 0)
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y
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x
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4 (2, 1, 0)
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</details>
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Figure 11-4 Tetrahedral element
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To evaluate the element stiffness matrix, we first determine the element volume $V$ and all $\alpha$ 's, $\beta$ 's, $\gamma$ 's, and $\delta$ 's from Eqs. (11.2.4)-(11.2.8). From Eq. (11.2.4), we have
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$$
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6 V = \left| \begin{array}{c c c c} 1 & 1 & 1 & 2 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 2 & 0 \\ 1 & 2 & 1 & 0 \end{array} \right| = 8 \text { in } ^ {3} \tag {11.2.24}
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$$
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From Eqs. (11.2.5), we obtain
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$$
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\alpha_ {1} = \left| \begin{array}{l l l} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 1 & 0 \end{array} \right| = 0 \quad \beta_ {1} = - \left| \begin{array}{l l l} 1 & 0 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & 0 \end{array} \right| = 0 \tag {11.2.25}
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$$
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and similarly,
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$$
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\gamma_ {1} = 0 \quad \delta_ {1} = 4
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$$
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From Eqs. (11.2.6)-(11.2.8), we obtain
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$$
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\begin{array}{l} \alpha_ {2} = 8 \quad \beta_ {2} = - 2 \quad \gamma_ {2} = - 4 \quad \delta_ {2} = - 1 \\ \alpha_ {3} = 0 \quad \beta_ {3} = - 2 \quad \gamma_ {3} = 4 \quad \delta_ {3} = - 1 \tag {11.2.26} \\ \alpha_ {4} = 0 \quad \beta_ {4} = 4 \quad \gamma_ {4} = 0 \quad \delta_ {4} = - 2 \\ \end{array}
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$$
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Note that $\alpha$ 's typically have units of cubic inches or cubic meters, where $\beta$ 's, $\gamma$ 's, and $\delta$ 's have units of square inches or square meters.
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<!-- source-page: 518 -->
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Next, the shape functions are determined using Eqs. (11.2.10) and the results from Eqs. (11.2.25) and (11.2.26) as
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$$
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N _ {1} = \frac {4 z}{8} \quad N _ {2} = \frac {8 - 2 x - 4 y - z}{8} \tag {11.2.27}
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$$
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$$
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N _ {3} = \frac {- 2 x + 4 y - z}{8} \quad N _ {4} = \frac {4 x - 2 z}{8}
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$$
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Note that $N _ { 1 } + N _ { 2 } + N _ { 3 } + N _ { 4 } = 1$ is again satisfied.
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The $6 \times 3$ submatrices of the matrix $\underline { { B } } ,$ Eq. (11.2.13), are now evaluated using Eqs. (11.2.14) and (11.2.27) as
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$$
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\underline {{{B}}} _ {1} = \left[ \begin{array}{l l l} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \frac {1}{2} \\ 0 & 0 & 0 \\ 0 & \frac {1}{2} & 0 \\ \frac {1}{2} & 0 & 0 \end{array} \right] \quad \underline {{{B}}} _ {2} = \left[ \begin{array}{c c c} - \frac {1}{4} & 0 & 0 \\ 0 & - \frac {1}{2} & 0 \\ 0 & 0 & - \frac {1}{8} \\ - \frac {1}{2} & - \frac {1}{4} & 0 \\ 0 & - \frac {1}{8} & - \frac {1}{2} \\ - \frac {1}{8} & 0 & - \frac {1}{4} \end{array} \right] \tag {11.2.28}
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$$
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$$
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\underline {{B}} _ {3} = \left[ \begin{array}{c c c} - \frac {1}{4} & 0 & 0 \\ 0 & \frac {1}{2} & 0 \\ 0 & 0 & - \frac {1}{8} \\ \frac {1}{2} & - \frac {1}{4} & 0 \\ 0 & - \frac {1}{8} & \frac {1}{2} \\ - \frac {1}{8} & 0 & - \frac {1}{4} \end{array} \right] \quad \underline {{B}} _ {4} = \left[ \begin{array}{c c c} \frac {1}{2} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & - \frac {1}{4} \\ 0 & \frac {1}{2} & 0 \\ 0 & - \frac {1}{4} & 0 \\ - \frac {1}{4} & 0 & \frac {1}{2} \end{array} \right]
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$$
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Next, the matrix $\underline { { \boldsymbol { D } } }$ is evaluated using Eq. (11.1.5) as
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$$
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[ D ] = \frac {3 0 \times 1 0 ^ {6}}{(1 + 0 . 3) (1 - 0 . 6)} \left[ \begin{array}{c c c c c c} 0. 7 & 0. 3 & 0. 3 & 0 & 0 & 0 \\ & 0. 7 & 0. 3 & 0 & 0 & 0 \\ & & 0. 7 & 0 & 0 & 0 \\ & & & 0. 2 & 0 & 0 \\ & & & & 0. 2 & 0 \\ \text { Symmetry } & & & & & 0. 2 \end{array} \right] \tag {11.2.29}
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$$
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Finally, substituting the results from Eqs. (11.2.24), (11.2.28), and (11.2.29) into Eq. (11.2.18), we obtain the element stiffness matrix. The resulting $1 2 \times 1 2$ matrix, being cumbersome to obtain by longhand calculations, is best left for the computer to evaluate.
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<!-- source-page: 519 -->
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# 11.3 Isoparametric Formulation
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We now describe the isoparametric formulation of the stiffness matrix for some threedimensional hexahedral elements.
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# Linear Hexahedral Element
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The basic (linear) hexahedral element [Figure 11–5(a)] now has eight corner nodes with isoparametric natural coordinates given by $s , t ,$ and $z ^ { \prime }$ as shown in Figure 11–5(b). The element faces are now defined by $s , t , z ^ { \prime } = \pm 1$ . (We use s; t, and $z ^ { \prime }$ for the coordinate axes because they are probably simpler to use than Greek letters $\xi , \eta _ { ; }$ , and z).
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The formulation of the stiffness matrix follows steps analogous to the isoparametric formulation of the stiffness matrix for the plane element in Chapter 10.
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The function use to describe the element geometry for x in terms of the generalized degrees of freedom $\boldsymbol { a } _ { i } { } ^ { \prime } \boldsymbol { \mathbf { s } }$ is
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$$
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x = a _ {1} + a _ {2} s + a _ {3} t + a _ {4} z ^ {\prime} + a _ {5} s t + a _ {6} t z ^ {\prime} + a _ {7} z ^ {\prime} s + a _ {8} s t z ^ {\prime} \tag {11.3.1}
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$$
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The same form as Eq. (11.3.1) is used for y and z as well. Just start with a9 through $a _ { 1 6 }$ for y and $a _ { 1 7 }$ through $a _ { 2 4 }$ for z.
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First, we expand Eq. (10.3.4) to include the z coordinate as follows:
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$$
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\left\{ \begin{array}{l} x \\ y \\ z \end{array} \right\} = \sum_ {i = 1} ^ {8} \left(\left[ \begin{array}{c c c} N _ {i} & 0 & 0 \\ 0 & N _ {i} & 0 \\ 0 & 0 & N _ {i} \end{array} \right] \left\{ \begin{array}{l} x _ {i} \\ y _ {i} \\ z _ {i} \end{array} \right\}\right) \tag {11.3.2}
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$$
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where the shape functions are now given by
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$$
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N _ {i} = \frac {(1 + s s _ {i}) (1 + t t _ {i}) (1 + z ^ {\prime} z _ {i} ^ {\prime})}{8} \tag {11.3.3}
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$$
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with $s _ { i } , t _ { i } , z _ { i } ^ { \prime } = \pm 1$ and $i = 1 , 2 , \dots , 8$ . For instance,
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$$
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N _ {1} = \frac {(1 + s s _ {1}) (1 + t t _ {1}) (1 + z ^ {\prime} z _ {1} ^ {\prime})}{8} \tag {11.3.4}
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$$
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<details>
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<summary>text_image</summary>
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|
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(x₃, y₃, z₃)
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3
|
||
(x₇, y₇, z₇)
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7
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||
(x₄, y₄, z₄)
|
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4
|
||
(x₈, y₈, z₈)
|
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8
|
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2
|
||
6
|
||
(x₂, y₂, z₂)
|
||
1
|
||
(x₁, y₁, z₁)
|
||
5
|
||
(x₅, y₅, z₅)
|
||
6
|
||
(x₆, y₆, z₆)
|
||
z
|
||
y
|
||
x
|
||
</details>
|
||
|
||
(a)
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
(-1, 1, -1)
|
||
(-1, 1, 1)
|
||
(-1, 1, 1)
|
||
(-1, -1, -1)
|
||
(-1, -1, 1)
|
||
1
|
||
3
|
||
4
|
||
(1, 1, 1)
|
||
8
|
||
2
|
||
5
|
||
(1, -1, 1)
|
||
7
|
||
6
|
||
s
|
||
t
|
||
z'
|
||
</details>
|
||
|
||
(b)
|
||
Figure 11–5 Linear hexahedral element (a) in a global-coordinate system and (b) element mapped into a cube of two unit sides placed symmetrically with natural or intrinsic coordinates s, t, and $z ^ { \prime }$
|
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|
||
<!-- source-page: 520 -->
|
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|
||
and when, from Figure $1 1 - 5 , s _ { 1 } = - 1 , t _ { 1 } = - 1$ , and $z _ { 1 } ^ { \prime } = + 1$ are used in Eq. (11.3.4), we obtain
|
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|
||
$$
|
||
N _ {1} = \frac {(1 - s) (1 - t) (1 + z ^ {\prime})}{8} \tag {11.3.5}
|
||
$$
|
||
|
||
Explicit forms of the other shape functions follow similarly. The shape functions in Eq. (11.3.3) map the natural coordinates $( s , t , z ^ { \prime } )$ of any point in the element to any point in the global coordinates $( x , y , z )$ when used in Eq. (11.3.2). For instance, when we let $i = 8$ and substitute $s _ { 8 } = 1 , t _ { 8 } = 1 , z _ { 8 } ^ { \prime } = 1$ into Eq. (11.3.3) for $N _ { 8 } .$ , we obtain
|
||
|
||
$$
|
||
N _ {8} = \frac {(1 + s) (1 + t) (1 + z ^ {\prime})}{8}
|
||
$$
|
||
|
||
Similar expressions are obtained for the other shape functions. Then evaluating all shape functions at node 8, we obtain $N _ { 8 } = 1$ , and all other shape functions equal zero at node 8. [From Eq. (11.3.5), we see that $N _ { 1 } = 0$ when $s = 1$ or when $t = 1 .$ Therefore, using Eq. (11.3.2), we obtain
|
||
|
||
$$
|
||
x = x _ {8} \quad y = y _ {8} \quad z = z _ {8}
|
||
$$
|
||
|
||
We see that indeed Eq. (11.3.2) maps any point in the natural-coordinate system to one in the global-coordinate system.
|
||
|
||
The displacement functions in terms of the generalized degrees of freedom are of the same form as used to describe the element geometry given by Eq. (11.3.1). Therefore we use the same shape functions as used to describe the geometry (Eq. (11.3.3)). The displacement functions now include w such that
|
||
|
||
$$
|
||
\left\{ \begin{array}{l} u \\ v \\ w \end{array} \right\} = \sum_ {i = 1} ^ {8} \left(\left[ \begin{array}{c c c} N _ {i} & 0 & 0 \\ 0 & N _ {i} & 0 \\ 0 & 0 & N _ {i} \end{array} \right] \left\{ \begin{array}{l} u _ {i} \\ v _ {i} \\ w _ {i} \end{array} \right\}\right) \tag {11.3.6}
|
||
$$
|
||
|
||
with the same shape functions as defined by Eq. (11.3.3) and the size of the shape function matrix now $3 \times 2 4$ .
|
||
|
||
The Jacobian matrix [Eq. (10.3.10)] is now expanded to
|
||
|
||
$$
|
||
[ J ] = \left[ \begin{array}{c c c} \frac {\partial x}{\partial s} & \frac {\partial y}{\partial s} & \frac {\partial z}{\partial s} \\ \frac {\partial x}{\partial t} & \frac {\partial y}{\partial t} & \frac {\partial z}{\partial t} \\ \frac {\partial x}{\partial z ^ {\prime}} & \frac {\partial y}{\partial z ^ {\prime}} & \frac {\partial z}{\partial z ^ {\prime}} \end{array} \right] \tag {11.3.7}
|
||
$$
|
||
|
||
Because the strain/displacement relationships, given by Eq. (11.2.11) in terms of global coordinates, include differentiation with respect to z, we expand Eq. (10.3.9)
|