有限元基本公式6

redufa發表於2024-10-07

1、伽遼金法

沒有分佈載荷

\[\frac{d}{d\hat{x}}(AE\frac{d\hat{u}}{d\hat{x}}) \]

伽遼金法:

\[\int_{0}^{L}\frac{d}{d\hat{x}}(AE\frac{d\hat{u}}{d\hat{x}}){N}_{i}d\hat{x} \]

分佈積分

\[({N}_{i}AE\frac{d\hat{u}}{d\hat{x}})\bigg|_{0}^{L}-\int_{0}^{L}AE\frac{d\hat{u}}{d\hat(x)} \frac{d{N}_{i}}{d\hat{x}}d\hat{x} \]

分佈積分中引入了邊界條件。

由於 \(\hat{u}=[N]{\hat{d}}\) , 因此

\[\frac{d\hat{u}}{d\hat{x}} = [ -\frac{1}{L} \quad \frac{1}{L}] \left\{ \begin{array} \hat{d}_{1x} \\ \hat{d}_{2x} \end{array} \right\} \]

將上述公式代入分佈積分後的公式

\[AE\int_{0}^{L}\frac{d{N}_{i}}{d\hat(x)}[-\frac{1}{L} \quad \frac{1}{L}]d\hat{x} \left\{ \begin{array} \hat{d}_{1x} \\ \hat{d}_{2x} \end{array} \right\} =\bigg({N}_{i}AE\frac{d\hat{u}}{d\hat{x}} \bigg)\bigg|_{0}^{L} \]

上述方程是兩個方程,一個\({N}_{i}={N}_{1}\) ,一個\({N}_{i}={N}_{2}\)

利用\({N}_{i}={N}_{1}\) 得出

\[AE\int_{0}^{L}\frac{d{N}_{1}}{d\hat(x)}[-\frac{1}{L} \quad \frac{1}{L}]d\hat{x} \left\{ \begin{array} \hat{d}_{1x} \\ \hat{d}_{2x} \end{array} \right\} =\bigg({N}_{1}AE\frac{d\hat{u}}{d\hat{x}} \bigg)\bigg|_{0}^{L} \]

替換\(d{N}_{i}/d\hat(x)\), 得出

\[AE\int_{0}^{L} \bigg[-\frac{1}{L}\bigg] \bigg[-\frac{1}{L} \quad \frac{1}{L} \bigg] d\hat{x} \left\{ \begin{array} \hat{d}_{1x} \\ \hat{d}_{2x} \end{array} \right\} =\hat{f}_{1x} \]

式中$ \hat{f}_{1x}=AE(d\hat{u}/d\hat{x})$, 因為 \(x=0\) 時,\({N}_{1}=1\), \(x=L\) 時,\({N}_{1}=0\)

\[\frac{AE}{L}\int_{0}^{L} \big[\frac{1}{L} \big] \big[ -\frac{1}{L} \quad \frac{1}{L} \big] \left\{ \begin{array}{c} \hat{d}_{1x} \\ \hat{d}_{2x} \end{array} \right\} =\bigg({N}_{2} AE \frac{d\hat{u}}{d\hat{x}} \bigg)\bigg|_{0}^{L} \]

簡化方程

\[\frac{AE}{L}(\hat{d}_{2x}-\hat{d}_{1x})=\hat{f}_{2x} \]

參考文獻 :

1、有限元法基礎教程 p102

2、直接剛度法

節點力與節點位移的關係,此關係是剛度矩陣

\[\left\{ \begin{array} \hat{f}_{1x} \\\hat{f}_{2x} \end{array} \right\} \bigg[ \begin{array} {c} {k}_{21} \quad {k}_{22} \\ {k}_{21} \quad {k}_{22} \end{array} \bigg] \]

位移函式

\[\hat{u}={a}_{1}+{a}_{2}\hat{x} \]

矩陣形式

\[\hat{u}=[1 \quad \hat{x}] \left\{ \begin{array}{c} {a}_{1} \\ {a}_{2} \end{array} \right\} \]

位移用節點位移表示

\[\hat{u}(0)=\hat{d}_{1x}={a}_{1} \\ \hat{u}(L)=\hat{d}_{2x}={a}_{2}L+\hat{d}_{1x} \]

彈簧示意圖

螢幕截圖 2024-06-15 184018

\[{a}_{1}=\frac{ \hat{d}_{2x}- \hat{d}_{1x} }{L} \]

\[\hat{u}=\big(\frac{ \hat{d}_{2x}- \hat{d}_{1x}} {L} \big)+ \hat{d}_{1x} \]

寫成矩陣形式

\[\hat{u}= \big[1- \frac{\hat{x}}{L} \quad \frac{\hat{x}}{L} \big] \left\{ \begin{array}{c} \hat{d}_{1x}\\ \hat{d}_{2x} \end{array} \right\} \]

或者

\[\hat{u}=[{N}_{1} \quad {N}_{2}] \left\{ \begin{array}{c} \hat{d}_{1x}\\ \hat{d}_{2x} \end{array} \right\} \]

其中

\[{N}_{1}=1- \frac{\hat(x)}{L} \\ {N}_{2}= \frac{\hat(x)}{L} \]

其中$ {N}_{1}$ 叫形函式

彈簧變形

\[\delta=\hat{d}_{2x}-{d}_{1x} \]

應變應力關係用力位移關係代替

\[T=k\delta \]

代入

\[T=k(\hat{d}_{2x}-{d}_{1x}) \]

力的平衡

\[\hat{f}_{1x}=-T \quad\quad \hat{f}_{2x}=T \]

聯合得出

\[T=-\hat{f}_{1x}=k(\hat{d}_{2x}-\hat{d}_{1x}) \\ T=\hat{f}_{2x}=k(\hat{d}_{2x}-\hat{d}_{1x}) {}{} \]

重寫方程

\[\hat{f}_{1x}=k(\hat{d}_{1x}-\hat{d}_{2x}) \\ \hat{f}_{2x}=k(\hat{d}_{2x}-\hat{d}_{1x}) \]

矩陣形式

\[\underline{\hat{k}}= \bigg[ \begin{array}{c} \quad k \quad -k \\ -k \quad\quad k \end{array} \bigg] \]

線性彈簧的剛度矩陣

\[\underline{K}=[K]=\sum_{e=1}^{N}\underline{k}^{e} \\ \underline{F}=[F]=\sum_{e=1}^{N}\underline{f}^{e} \]

參考文獻:

1、有限元方法基礎教程 p20

3、勢能法

總勢能

\[\pi_{p}=\frac{1}{2}k(\hat{d}_{2x}- \hat{d}_{1x}) ^2 -\hat{f}_{1x}{}\hat{d}_{1x} -\hat{f}_{2x}{}\hat{d}_{2x} \]

式中$ \hat{d}{2x}-\hat{d} $ 是彈簧的變形

方程的第一項是彈簧的應變能,簡化方程

\[\pi_{p}=\frac{1}{2}k(\hat{d}_{2x}^{2}- 2\hat{d}_{2x}\hat{d}_{1x}+ \hat{d}_{1x}^{2}) -\hat{f}_{1x}{}\hat{d}_{1x} -\hat{f}_{2x}{}\hat{d}_{2x} \]

\(\pi_{p}\)對每個節點位移取最小值,

\[\frac{\partial \pi_{p}}{\partial \hat{d}_{1x}}= \frac{1}{2}k(-2\hat{d}_{2x}+2\hat{d}_{1x})-\hat{f}_{1x} \\ \frac{\partial \pi_{p}}{\partial \hat{d}_{2x}}= \frac{1}{2}k(2\hat{d}_{2x}-2\hat{d}_{1x})-\hat{f}_{2x} \]

有限元基本公式6

簡化方程

\[k(-\hat{d}_{2x}+\hat{d}_{1x})=\hat{f}_{1x} \\ k(\hat{d}_{2x}-\hat{d}_{1x})=\hat{f}_{2x} \]

矩陣形式

\[\underline{\hat{k}}= \bigg[ \begin{array}{c} \quad k \quad -k \\ -k \quad\quad k \end{array} \bigg] \]

參考文獻:

1、有限元方法基礎教程 p56

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