김경종
348 lines
22 KiB
Markdown
348 lines
22 KiB
Markdown
<!-- source-page: 351 -->
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<details>
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<summary>text_image</summary>
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m = 2
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j = 3
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①
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i = 1
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</details>
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Figure 6–18 Element 1 of the discretized plate
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or, in general, A can be obtained equivalently by the nodal coordinate formula of Eq. (6.2.9).
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We will now evaluate ½B , where ½B is given by Eq. (6.2.34), expanded here as
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$$
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[ B ] = \frac {1}{2 A} \left[ \begin{array}{c c c c c c} \beta_ {i} & 0 & \beta_ {j} & 0 & \beta_ {m} & 0 \\ 0 & \gamma_ {i} & 0 & \gamma_ {j} & 0 & \gamma_ {m} \\ \gamma_ {i} & \beta_ {i} & \gamma_ {j} & \beta_ {j} & \gamma_ {m} & \beta_ {m} \end{array} \right] \tag {6.5.4}
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$$
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and, from Eqs. (6.2.10),
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$$
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\beta_ {i} = y _ {j} - y _ {m} = 1 0 - 1 0 = 0
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$$
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$$
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\beta_ {j} = y _ {m} - y _ {i} = 1 0 - 0 = 1 0
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$$
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$$
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\beta_ {m} = y _ {i} - y _ {j} = 0 - 1 0 = - 1 0 \tag {6.5.5}
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$$
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$$
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\gamma_ {i} = x _ {m} - x _ {j} = 0 - 2 0 = - 2 0
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$$
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$$
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\gamma_ {j} = x _ {i} - x _ {m} = 0 - 0 = 0
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$$
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$$
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\gamma_ {m} = x _ {j} - x _ {i} = 2 0 - 0 = 2 0
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$$
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Therefore, substituting Eqs. (6.5.5) into Eq. (6.5.4), we obtain
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$$
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[ B ] = \frac {1}{2 0 0} \left[ \begin{array}{r r r r r r} 0 & 0 & 1 0 & 0 & - 1 0 & 0 \\ 0 & - 2 0 & 0 & 0 & 0 & 2 0 \\ - 2 0 & 0 & 0 & 1 0 & 2 0 & - 1 0 \end{array} \right] \frac {1}{\text { in. }} \tag {6.5.6}
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$$
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For plane stress, the ½D matrix is conveniently expressed here as
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$$
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[ D ] = \frac {E}{(1 - \nu^ {2})} \left[ \begin{array}{c c c} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac {1 - \nu}{2} \end{array} \right] \tag {6.5.7}
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$$
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With n ¼ 0:3 and $E = 3 0 \times 1 0 ^ { 6 } ~ \mathrm { p s i }$ , we obtain
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$$
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[ D ] = \frac {3 0 (1 0 ^ {6})}{0 . 9 1} \left[ \begin{array}{l l l} 1 & 0. 3 & 0 \\ 0. 3 & 1 & 0 \\ 0 & 0 & 0. 3 5 \end{array} \right] \text { psi } \tag {6.5.8}
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$$
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<!-- source-page: 352 -->
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$$
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\text { Then } \quad [ B ] ^ {T} [ D ] = \frac {3 0 (1 0 ^ {6})}{2 0 0 (0 . 9 1)} \left[ \begin{array}{r r r} 0 & 0 & - 2 0 \\ 0 & - 2 0 & 0 \\ 1 0 & 0 & 0 \\ 0 & 0 & 1 0 \\ - 1 0 & 0 & 2 0 \\ 0 & 2 0 & - 1 0 \end{array} \right] \left[ \begin{array}{r r r} 1 & 0. 3 & 0 \\ 0. 3 & 1 & 0 \\ 0 & 0 & 0. 3 5 \end{array} \right] \tag {6.5.9}
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$$
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Simplifying Eq. (6.5.9) yields
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$$
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[ B ] ^ {T} [ D ] = \frac {(0 . 1 5) (1 0 ^ {6})}{0 . 9 1} \left[ \begin{array}{r r r} 0 & 0 & - 7 \\ - 6 & - 2 0 & 0 \\ 1 0 & 3 & 0 \\ 0 & 0 & 3. 5 \\ - 1 0 & - 3 & 7 \\ 6 & 2 0 & - 3. 5 \end{array} \right] \tag {6.5.10}
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$$
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Using Eqs. (6.5.10) and (6.5.6) in Eq. (6.5.3), we have the stiffness matrix for element 1 as
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$$
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\begin{array}{l} [ k ] = (1) (1 0 0) \frac {(0 . 1 5) (1 0 ^ {6})}{0 . 9 1} \left[ \begin{array}{r r r} 0 & 0 & - 7 \\ - 6 & - 2 0 & 0 \\ 1 0 & 3 & 0 \\ 0 & 0 & 3. 5 \\ - 1 0 & - 3 & 7 \\ 6 & 2 0 & - 3. 5 \end{array} \right] \\ \times \frac {1}{2 (1 0 0)} \left[ \begin{array}{r r r r r r} 0 & 0 & 1 0 & 0 & - 1 0 & 0 \\ 0 & - 2 0 & 0 & 0 & 0 & 2 0 \\ - 2 0 & 0 & 0 & 1 0 & 2 0 & - 1 0 \end{array} \right] \tag {6.5.11} \\ \end{array}
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$$
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Finally, simplifying Eq. (6.5.11) yields
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$$
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[ k ] = \frac {7 5 , 0 0 0}{0 . 9 1} \left[ \begin{array}{c c c c c c} u _ {1} & v _ {1} & u _ {3} & v _ {3} & u _ {2} & v _ {2} \\ 1 4 0 & 0 & 0 & - 7 0 & - 1 4 0 & 7 0 \\ 0 & 4 0 0 & - 6 0 & 0 & 6 0 & - 4 0 0 \\ 0 & - 6 0 & 1 0 0 & 0 & - 1 0 0 & 6 0 \\ - 7 0 & 0 & 0 & 3 5 & 7 0 & - 3 5 \\ - 1 4 0 & 6 0 & - 1 0 0 & 7 0 & 2 4 0 & - 1 3 0 \\ 7 0 & - 4 0 0 & 6 0 & - 3 5 & - 1 3 0 & 4 3 5 \end{array} \right] \frac {\mathrm{lb}}{\text {in.}} \tag {6.5.12}
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$$
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where the labels above the columns indicate the nodal order of the degrees of freedom in the element 1 stiffness matrix.
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<!-- source-page: 353 -->
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<details>
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<summary>text_image</summary>
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m = 3
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②
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i = 1
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j = 4
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</details>
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Figure 6–19 Element 2 of the discretized plate
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In Figure 6–19 for element 2, we have $x _ { i } = 0 , y _ { i } = 0 , x _ { j } = 2 0 , y _ { j } = 0 , x _ { m } = 2 0 _ { : }$ and $y _ { m } = 1 0$ . Then, from Eqs. (6.2.10), we have
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$$
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\beta_ {i} = y _ {j} - y _ {m} = 0 - 1 0 = - 1 0
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$$
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$$
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\beta_ {j} = y _ {m} - y _ {i} = 1 0 - 0 = 1 0
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$$
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$$
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\beta_ {m} = y _ {i} - y _ {j} = 0 - 0 = 0 \tag {6.5.13}
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$$
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$$
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\gamma_ {i} = x _ {m} - x _ {j} = 2 0 - 2 0 = 0
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$$
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$$
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\gamma_ {j} = x _ {i} - x _ {m} = 0 - 2 0 = - 2 0
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$$
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$$
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\gamma_ {m} = x _ {j} - x _ {i} = 2 0 - 0 = 2 0
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$$
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Therefore, using Eqs. (6.5.13) in Eq. (6.5.4) yields
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$$
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[ B ] = \frac {1}{2 0 0} \left[ \begin{array}{c c c c c c} - 1 0 & 0 & 1 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & - 2 0 & 0 & 2 0 \\ 0 & - 1 0 & - 2 0 & 1 0 & 2 0 & 0 \end{array} \right] \frac {1}{\text { in. }} \tag {6.5.14}
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$$
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The ½D matrix is again given by
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$$
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[ D ] = \frac {3 0 (1 0 ^ {6})}{0 . 9 1} \left[ \begin{array}{l l l} 1 & 0. 3 & 0 \\ 0. 3 & 1 & 0 \\ 0 & 0 & 0. 3 5 \end{array} \right] \text { psi } \tag {6.5.15}
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$$
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Then, using Eqs. (6.5.14) and (6.5.15), we obtain
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$$
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[ B ] ^ {T} [ D ] = \frac {3 0 (1 0 ^ {6})}{2 0 0 (0 . 9 1)} \left[ \begin{array}{c c c} - 1 0 & 0 & 0 \\ 0 & 0 & - 1 0 \\ 1 0 & 0 & - 2 0 \\ 0 & - 2 0 & 1 0 \\ 0 & 0 & 2 0 \\ 0 & 2 0 & 0 \end{array} \right] \left[ \begin{array}{c c c} 1 & 0. 3 & 0 \\ 0. 3 & 1 & 0 \\ 0 & 0 & 0. 3 5 \end{array} \right] \tag {6.5.16}
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$$
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<!-- source-page: 354 -->
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Simplifying Eq. (6.5.16) yields
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$$
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[ B ] ^ {T} [ D ] = \frac {(0 . 1 5) (1 0 ^ {6})}{0 . 9 1} \left[ \begin{array}{c c c} - 1 0 & - 3 & 0 \\ 0 & 0 & - 3. 5 \\ 1 0 & 3 & - 7 \\ - 6 & - 2 0 & 3. 5 \\ 0 & 0 & 7 \\ 6 & 2 0 & 0 \end{array} \right] \tag {6.5.17}
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$$
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Finally, substituting Eqs. (6.5.17) and (6.5.14) into Eq. (6.5.3), we obtain the stiffness matrix for element 2 as
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$$
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[ k ] = (1) (1 0 0) \frac {(0 . 1 5) (1 0 ^ {6})}{0 . 9 1} \left[ \begin{array}{r r r} - 1 0 & - 3 & 0 \\ 0 & 0 & - 3. 5 \\ 1 0 & 3 & - 7 \\ - 6 & - 2 0 & 3. 5 \\ 0 & 0 & 7 \\ 6 & 2 0 & 0 \end{array} \right]
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$$
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$$
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\times \frac {1}{2 (1 0 0)} \left[ \begin{array}{c c c c c c} - 1 0 & 0 & 1 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & - 2 0 & 0 & 2 0 \\ 0 & - 1 0 & - 2 0 & 1 0 & 2 0 & 0 \end{array} \right] \tag {6.5.18}
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$$
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Equation (6.5.18) simplifies to
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$$
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[ k ] = \frac {7 5 , 0 0 0}{0 . 9 1} \left[ \begin{array}{c c c c c c} u _ {1} & v _ {1} & u _ {4} & v _ {4} & u _ {3} & v _ {3} \\ 1 0 0 & 0 & - 1 0 0 & 6 0 & 0 & - 6 0 \\ 0 & 3 5 & 7 0 & - 3 5 & - 7 0 & 0 \\ - 1 0 0 & 7 0 & 2 4 0 & - 1 3 0 & - 1 4 0 & 6 0 \\ 6 0 & - 3 5 & - 1 3 0 & 4 3 5 & 7 0 & - 4 0 0 \\ 0 & - 7 0 & - 1 4 0 & 7 0 & 1 4 0 & 0 \\ - 6 0 & 0 & 6 0 & - 4 0 0 & 0 & 4 0 0 \end{array} \right] \frac {\mathrm{lb}}{\mathrm{in.}} \tag {6.5.19}
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$$
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where the degrees of freedom in the element 2 stiffness matrix are shown above the columns in Eq. (6.5.19). Rewriting the element stiffness matrices, Eqs. (6.5.12) and (6.5.19), expanded to the order of, and rearranged according to, increasing nodal degrees of freedom of the total K matrix (where we have factored out a constant 5), we obtain
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<!-- source-page: 355 -->
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# Element 1
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$$
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[ k ] = \frac {3 7 5 , 0 0 0}{0 . 9 1} \left[ \begin{array}{c c c c c c c c} u _ {1} & v _ {1} & u _ {2} & v _ {2} & u _ {3} & v _ {3} & u _ {4} & v _ {4} \\ 2 8 & 0 & - 2 8 & 1 4 & 0 & - 1 4 & 0 & 0 \\ 0 & 8 0 & 1 2 & - 8 0 & - 1 2 & 0 & 0 & 0 \\ - 2 8 & 1 2 & 4 8 & - 2 6 & - 2 0 & 1 4 & 0 & 0 \\ 1 4 & - 8 0 & - 2 6 & 8 7 & 1 2 & - 7 & 0 & 0 \\ 0 & - 1 2 & - 2 0 & 1 2 & 2 0 & 0 & 0 & 0 \\ - 1 4 & 0 & 1 4 & - 7 & 0 & 7 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right] \frac {\mathrm{lb}}{\mathrm{in.}} \tag {6.5.20}
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$$
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# Element 2
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$$
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[ k ] = \frac {3 7 5 , 0 0 0}{0 . 9 1} \left[ \begin{array}{c c c c c c c c} u _ {1} & v _ {1} & u _ {2} & v _ {2} & u _ {3} & v _ {3} & u _ {4} & v _ {4} \\ 2 0 & 0 & 0 & 0 & 0 & - 1 2 & - 2 0 & 1 2 \\ 0 & 7 & 0 & 0 & - 1 4 & 0 & 1 4 & - 7 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & - 1 4 & 0 & 0 & 2 8 & 0 & - 2 8 & 1 4 \\ - 1 2 & 0 & 0 & 0 & 0 & 8 0 & 1 2 & - 8 0 \\ - 2 0 & 1 4 & 0 & 0 & - 2 8 & 1 2 & 4 8 & - 2 6 \\ 1 2 & - 7 & 0 & 0 & 1 4 & - 8 0 & - 2 6 & 8 7 \end{array} \right] \frac {\mathrm{lb}}{\text {in.}} \tag {6.5.21}
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$$
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Using superposition of the element stiffness matrices, Eqs. (6.5.20) and (6.5.21), now that the orders of the degrees of freedom are the same, we obtain the total global stiffness matrix as
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$$
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[ K ] = \frac {3 7 5 , 0 0 0}{0 . 9 1} \left[ \begin{array}{c c c c c c c c} u _ {1} & v _ {1} & u _ {2} & v _ {2} & u _ {3} & v _ {3} & u _ {4} & v _ {4} \\ 4 8 & 0 & - 2 8 & 1 4 & 0 & - 2 6 & - 2 0 & 1 2 \\ 0 & 8 7 & 1 2 & - 8 0 & - 2 6 & 0 & 1 4 & - 7 \\ - 2 8 & 1 2 & 4 8 & - 2 6 & - 2 0 & 1 4 & 0 & 0 \\ 1 4 & - 8 0 & - 2 6 & 8 7 & 1 2 & - 7 & 0 & 0 \\ 0 & - 2 6 & - 2 0 & 1 2 & 4 8 & 0 & - 2 8 & 1 4 \\ - 2 6 & 0 & 1 4 & - 7 & 0 & 8 7 & 1 2 & - 8 0 \\ - 2 0 & 1 4 & 0 & 0 & - 2 8 & 1 2 & 4 8 & - 2 6 \\ 1 2 & - 7 & 0 & 0 & 1 4 & - 8 0 & - 2 6 & 8 7 \end{array} \right] \frac {\mathrm{lb}}{\text {in.}} \tag {6.5.22}
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$$
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[Alternatively, we could have applied the direct stiffness method to Eqs. (6.5.12) and (6.5.19) to obtain Eq. (6.5.22).] Substituting [K] into $\{F\} = [K]\{d\}$ of Eq. (6.5.2), we
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<!-- source-page: 356 -->
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have
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$$
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\left\{ \begin{array}{c} R _ {1 x} \\ R _ {1 y} \\ R _ {2 x} \\ R _ {2 y} \\ 5 0 0 0 \\ 0 \\ 5 0 0 0 \\ 0 \end{array} \right\} = \frac {3 7 5 , 0 0 0}{0 . 9 1} \left[ \begin{array}{c c c c c c c c} 4 8 & 0 & - 2 8 & 1 4 & 0 & - 2 6 & - 2 0 & 1 2 \\ 0 & 8 7 & 1 2 & - 8 0 & - 2 6 & 0 & 1 4 & - 7 \\ - 2 8 & 1 2 & 4 8 & - 2 6 & - 2 0 & 1 4 & 0 & 0 \\ 1 4 & - 8 0 & - 2 6 & 8 7 & 1 2 & - 7 & 0 & 0 \\ 0 & - 2 6 & - 2 0 & 1 2 & 4 8 & 0 & - 2 8 & 1 4 \\ - 2 6 & 0 & 1 4 & - 7 & 0 & 8 7 & 1 2 & - 8 0 \\ - 2 0 & 1 4 & 0 & 0 & - 2 8 & 1 2 & 4 8 & - 2 6 \\ 1 2 & - 7 & 0 & 0 & 1 4 & - 8 0 & - 2 6 & 8 7 \end{array} \right] \left\{ \begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ d _ {3 x} \\ d _ {3 y} \\ d _ {4 x} \\ d _ {4 y} \end{array} \right\} \tag {6.5.23}
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$$
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Applying the support or boundary conditions by eliminating rows and columns corresponding to displacement matrix rows and columns equal to zero [namely, rows and columns 1–4 in Eq. (6.5.23)], we obtain
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$$
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\left\{ \begin{array}{c} 5 0 0 0 \\ 0 \\ 5 0 0 0 \\ 0 \end{array} \right\} = \frac {3 7 5 , 0 0 0}{0 . 9 1} \left[ \begin{array}{c c c c} 4 8 & 0 & - 2 8 & 1 4 \\ 0 & 8 7 & 1 2 & - 8 0 \\ - 2 8 & 1 2 & 4 8 & - 2 6 \\ 1 4 & - 8 0 & - 2 6 & 8 7 \end{array} \right] \left\{ \begin{array}{c} d _ {3 x} \\ d _ {3 y} \\ d _ {4 x} \\ d _ {4 y} \end{array} \right\} \tag {6.5.24}
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$$
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Premultiplying both sides of Eq. (6.5.24) by $\underline { { K } } ^ { - 1 }$ , we have
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$$
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\left\{ \begin{array}{l} d _ {3 x} \\ d _ {3 y} \\ d _ {4 x} \\ d _ {4 y} \end{array} \right\} = \frac {0 . 9 1}{3 7 5 , 0 0 0} \left[ \begin{array}{c c c c} 4 8 & 0 & - 2 8 & 1 4 \\ 0 & 8 7 & 1 2 & - 8 0 \\ - 2 8 & 1 2 & 4 8 & - 2 6 \\ 1 4 & - 8 0 & - 2 6 & 8 7 \end{array} \right] ^ {- 1} \left\{ \begin{array}{c} 5 0 0 0 \\ 0 \\ 5 0 0 0 \\ 0 \end{array} \right\} \tag {6.5.25}
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$$
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Solving for the displacements in Eq. (6.5.25), we obtain
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$$
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\left\{ \begin{array}{l} d _ {3 x} \\ d _ {3 y} \\ d _ {4 x} \\ d _ {4 y} \end{array} \right\} = \frac {0 . 9 1}{7 5} \left\{ \begin{array}{l} 0. 0 5 0 2 4 \\ 0. 0 0 0 3 4 \\ 0. 0 5 4 7 0 \\ 0. 0 0 8 7 8 \end{array} \right\} \tag {6.5.26}
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$$
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Simplifying Eq. (6.5.26), the final displacements are given by
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$$
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\left\{ \begin{array}{l} d _ {3 x} \\ d _ {3 y} \\ d _ {4 x} \\ d _ {4 y} \end{array} \right\} = \left\{ \begin{array}{c} 6 0 9. 6 \\ 4. 2 \\ 6 6 3. 7 \\ 1 0 4. 1 \end{array} \right\} \times 1 0 ^ {- 6} \text { in. } \tag {6.5.27}
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$$
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Comparing the finite element solution to an analytical solution, as a first approximation, we have the axial displacement given by
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$$
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\delta = \frac {P L}{A E} = \frac {(1 0 , 0 0 0) 2 0}{1 0 (3 0 \times 1 0 ^ {6})} = 6 7 0 \times 1 0 ^ {- 6} \text { in }.
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$$
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<!-- source-page: 357 -->
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for a one-dimensional bar subjected to tensile force. Hence, the nodal x displacement components of Eq. (6.5.27) for the two-dimensional plate appear to be reasonably correct, considering the coarseness of the mesh and the directional stiffness bias of the model. (For more on this subject see Section 7.5.) The y displacement would be expected to be downward at the top (node 3) and upward at the bottom (node 4) as a result of the Poisson effect. However, the directional stiffness bias due to the coarse mesh accounts for this unexpected poor result.
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We now determine the stresses in each element by using Eq. (6.2.36):
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$$
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\{\sigma \} = [ D ] [ B ] \{d \} \tag {6.5.28}
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$$
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In general, for element 1, we then have
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$$
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\{\sigma \} = \frac {E}{(1 - v ^ {2})} \left[ \begin{array}{c c c} 1 & v & 0 \\ v & 1 & 0 \\ 0 & 0 & \frac {1 - v}{2} \end{array} \right] \times \left(\frac {1}{2 A}\right) \left[ \begin{array}{c c c c c c} \beta_ {1} & 0 & \beta_ {3} & 0 & \beta_ {2} & 0 \\ 0 & \gamma_ {1} & 0 & \gamma_ {3} & 0 & \gamma_ {2} \\ \gamma_ {1} & \beta_ {1} & \gamma_ {3} & \beta_ {3} & \gamma_ {2} & \beta_ {2} \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ d _ {3 x} \\ d _ {3 y} \\ d _ {2 x} \\ d _ {2 y} \end{array} \right\} \tag {6.5.29}
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$$
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Substituting numerical values for $[ B ] _ { : }$ , given by Eq. (6.5.6); for $[ D ] ,$ , given by $\operatorname { E q }$ . (6.5.8); and the appropriate part of $\{ d \}$ , given by Eq. (6.5.27), we obtain
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$$
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\begin{array}{l} \{\sigma \} = \frac {3 0 (1 0 ^ {6}) (1 0 ^ {- 6})}{0 . 9 1 (2 0 0)} \left[ \begin{array}{c c c} 1 & 0. 3 & 0 \\ 0. 3 & 1 & 0 \\ 0 & 0 & 0. 3 5 \end{array} \right] \\ \times \left[ \begin{array}{r r r r r r} 0 & 0 & 1 0 & 0 & - 1 0 & 0 \\ 0 & - 2 0 & 0 & 0 & 0 & 2 0 \\ - 2 0 & 0 & 0 & 1 0 & 2 0 & - 1 0 \end{array} \right] \left\{ \begin{array}{c} 0 \\ 0 \\ 6 0 9. 6 \\ 4. 2 \\ 0 \\ 0 \end{array} \right\} \tag {6.5.30} \\ \end{array}
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$$
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Simplifying Eq. (6.5.30), we obtain
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$$
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\left\{ \begin{array}{l} \sigma_ {x} \\ \sigma_ {y} \\ \tau_ {x y} \end{array} \right\} = \left\{ \begin{array}{c} 1 0 0 5 \\ 3 0 1 \\ 2. 4 \end{array} \right\} \text { psi } \tag {6.5.31}
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$$
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<!-- source-page: 358 -->
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In general, for element 2, we have
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$$
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\{\sigma \} = \frac {E}{(1 - v ^ {2})} \left(\frac {1}{2 A}\right) \left[ \begin{array}{c c c} 1 & v & 0 \\ v & 1 & 0 \\ 0 & 0 & \frac {1 - v}{2} \end{array} \right] \times \left[ \begin{array}{c c c c c c} \beta_ {1} & 0 & \beta_ {4} & 0 & \beta_ {3} & 0 \\ 0 & \gamma_ {1} & 0 & \gamma_ {4} & 0 & \gamma_ {3} \\ \gamma_ {1} & \beta_ {1} & \gamma_ {4} & \beta_ {4} & \gamma_ {3} & \beta_ {3} \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ d _ {4 x} \\ d _ {4 y} \\ d _ {3 x} \\ d _ {3 y} \end{array} \right\} \tag {6.5.32}
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$$
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Substituting numerical values into Eq. (6.5.32), we obtain
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$$
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\begin{array}{l} \{\sigma \} = \frac {3 0 (1 0 ^ {6}) (1 0 ^ {- 6})}{0 . 9 1 (2 0 0)} \left[ \begin{array}{c c c} 1 & 0. 3 & 0 \\ 0. 3 & 1 & 0 \\ 0 & 0 & 0. 3 5 \end{array} \right] \\ \times \left[ \begin{array}{c c c c c c} - 1 0 & 0 & 1 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & - 2 0 & 0 & 2 0 \\ 0 & - 1 0 & - 2 0 & 1 0 & 2 0 & 0 \end{array} \right] \left\{ \begin{array}{c} 0 \\ 0 \\ 6 6 3. 7 \\ 1 0 4. 1 \\ 6 0 9. 6 \\ 4. 2 \end{array} \right\} \tag {6.5.33} \\ \end{array}
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$$
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Simplifying Eq. (6.5.33), we obtain
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$$
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\left\{ \begin{array}{l} \sigma_ {x} \\ \sigma_ {y} \\ \tau_ {x y} \end{array} \right\} = \left\{ \begin{array}{l} 9 9 5 \\ - 1. 2 \\ - 2. 4 \end{array} \right\} \text { psi } \tag {6.5.34}
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$$
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The principal stresses can now be determined from Eq. (6.1.2), and the principal angle made by one of the principal stresses can be determined from Eq. (6.1.3). (The other principal stress will be directed 90
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from the first.) We determine these principal stresses for element 2 (those for element 1 will be similar) as
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$$
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\begin{array}{l} \sigma_ {1} = \frac {\sigma_ {x} + \sigma_ {y}}{2} + \left[ \left(\frac {\sigma_ {x} - \sigma_ {y}}{2}\right) ^ {2} + \tau_ {x y} ^ {2} \right] ^ {1 / 2} \\ \sigma_ {1} = \frac {9 9 5 + (- 1 . 2)}{2} + \left[ \left(\frac {9 9 5 - (- 1 . 2)}{2}\right) ^ {2} + (- 2. 4) ^ {2} \right] ^ {1 / 2} \tag {6.5.35} \\ \sigma_ {1} = 4 9 7 + 4 9 8 = 9 9 5 \mathrm{psi} \\ \sigma_ {2} = \frac {9 9 5 + (- 1 . 2)}{2} - 4 9 8 = - 1. 1 \mathrm{psi} \\ \end{array}
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$$
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The principal angle is then
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$$
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\theta_ {p} = \frac {1}{2} \tan^ {- 1} \left[ \frac {2 \tau_ {x y}}{\sigma_ {x} - \sigma_ {y}} \right]
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$$
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ð6:5:36Þ or $\theta _ { p } = \frac { 1 } { 2 } \tan ^ { - 1 } \left[ \frac { 2 ( - 2 . 4 ) } { 9 9 5 - ( - 1 . 2 ) } \right] = 0 ^ { \circ }$ tan1
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Owing to the uniform stress of 1000 psi acting only in the x direction on the edge of the plate, we would expect the stress $\sigma _ { x } ( = \sigma _ { 1 } )$ to be near 1000 psi in each element. Thus, the results from Eqs. (6.5.31) and (6.5.34) for $\sigma _ { x }$ are quite good. We would expect the stress $\sigma _ { y }$ to be very small (at least near the free edge). The restraint of element 1 at nodes 1 and 2 causes a relatively large element stress $\sigma _ { y { \mathrm { : } } }$ , whereas the restraint of element 2 at only one node causes a very small stress $\sigma _ { y }$ . The shear stresses $\tau _ { x y }$ remain close to zero, as expected. Had the number of elements been increased, with smaller ones used near the support edge, even more realistic results would have been obtained. However, a finer discretization would result in a cumbersome longhand solution and thus was not used here. Use of a computer program is recommended for a detailed solution to this plate problem and certainly for solving more complex stress/strain problems. 9
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The maximum distortion energy theory [4] (also called the von Mises or von Mises-Hencky theory) for ductile materials subjected to static loading predicts that a material will fail if the von Mises stress (also called equivalent or effective stress) reaches the yield strength, $S _ { y } { _ { ; } }$ , of the material. The von Mises stress as derived in [4], for instance, is given in terms of the three principal stresses by
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$$
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\sigma_ {v m} = \frac {1}{\sqrt {2}} \left[ \left(\sigma_ {1} - \sigma_ {2}\right) ^ {2} + \left(\sigma_ {2} - \sigma_ {3}\right) ^ {2} + \left(\sigma_ {3} - \sigma_ {1}\right) ^ {2} \right] ^ {1 / 2} \tag {6.5.37a}
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$$
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or equivalently in terms of the x-y-z components as
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$$
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\sigma_ {v m} = \frac {1}{\sqrt {2}} \left[ \left(\sigma_ {x} - \sigma_ {y}\right) ^ {2} + \left(\sigma_ {y} - \sigma_ {z}\right) ^ {2} + \left(\sigma_ {z} - \sigma_ {x}\right) ^ {2} + 6 \left(\tau_ {x y} ^ {2} + \tau_ {y z} ^ {2} + \tau_ {z x} ^ {2}\right) \right] \tag {6.5.37b}
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$$
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Thus for yielding to occur, the von Mises stress must become equal to or greater than the yield strength of the material as given by
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$$
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\sigma_ {v m} \geq S _ {y} \tag {6.5.38}
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$$
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We can see from Eqs. (6.5.37a or 6.5.37b) that the von Mises stress is a scalar that measures the intensity of the entire stress state as it includes the three principal stresses or the three normal stresses in the x, y, and z directions, along with the shear stresses on the $x , y ,$ and z planes. Other stresses, such as the maximum principal one, do not provide the most accurate way of predicting failure.
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Most computer programs incorporate this failure theory and, as an optional result, the user can request a plot of the von Mises stress throughout the material model being analyzed. If the von Mises stress value is equal to or greater than the yield strength of the material being considered, then another material with greater yield strength can be selected or other design changes can be made.
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For brittle materials, such as glass and cast iron, with different tension and compression properties, it is recommended to use the Coulomb-Mohr theory to predict failure. For more on this theory consult [4].
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# CST Element Defects
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The CST element has its limitations. In bending problems, the mesh of CST elements will produce a model that is stiffer than the actual problem. As we will observe from the results shown for a beam-bending problem modeled by CST and LST (to be described in Chapter 8) elements, the CST model converges very slowly to the exact solution. This is partly due to the element predicting only constant stress within each element, when for a bending problem, the stress actually varies linearly through the depth of the beam. This problem is rectified by using the LST element as described in Chapter 8.
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As shown in [3] for a beam subjected to pure bending, the CST has a spurious or false shear stress and hence a spurious shear strain in parts of the model that should not have any shear stress or shear strain. This spurious shear strain absorbs energy; therefore, some of the energy that should go into bending is lost. The CST is then too stiff in bending, and the resulting deformation is smaller than actually should be. This phenomenon of excessive stiffness developing in one or more modes of deformation is sometimes described as shear locking or parasitic shear.
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Furthermore, in problems where plane strain conditions exist (recall this means when $\varepsilon _ { z } = 0 )$ and the Poisson’s ratio approaches 0.5, a mesh can actually lock, which means the mesh then cannot deform at all.
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This brief description of some limitations in using the CST element does not stop us from using it to model plane stress and plane strain problems. It just requires us to use a fine mesh as opposed to a coarse one, particularly where bending occurs and where in general large stress gradients will result. Also, we must make sure our program can handle Poisson’s ratios that approach 0.5 if that is desired, such as in rubber-like materials. For common materials, such as metals, Poisson’s ratio is around 0.3 and so locking should not be of concern.
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# References
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[1] Timoshenko, S., and Goodier, J., Theory of Elasticity, 3rd ed., McGraw-Hill, New York 1970.
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[2] Gere, J. M., Mechanics of Materials, 5th ed., Brooks/Cole Publishers, Pacific Grove, CA, 2001.
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[3] Cook, R. D., Malkus, D. S., Plesha, M. E., and Witt, R. J., Concepts and Applications of Finite Element Analysis, 4th ed., Wiley, New York, 2002.
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[4] Shigley, J. E., Mischke, C. R., and Budynas, R. G., Mechanical Engineering Design, 7th ed., McGraw-Hill, New York, 2004.
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