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