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形状関数の勾配

形状関数の勾配 $ \frac{ \partial N }{ \partial \{ x \} } _{(In)}$ は、


$\displaystyle \frac{ \partial N }{ \partial \{ x \} } _{(0)}$ $\textstyle =$ $\displaystyle { \left \langle \begin{array}{cc} 4 b0 L0 - b0 & 4 c0 L0 - c0 \end{array} \right \rangle } ^ { T }$  
$\displaystyle \frac{ \partial N }{ \partial \{ x \} } _{(1)}$ $\textstyle =$ $\displaystyle { \left \langle \begin{array}{cc} 4 b1 L1 - b1 & 4 c1 L1 - c1 \end{array} \right \rangle } ^ { T }$  
$\displaystyle \frac{ \partial N }{ \partial \{ x \} } _{(2)}$ $\textstyle =$ $\displaystyle { \left \langle \begin{array}{cc} 4 b2 L2 - b2 & 4 c2 L2 - c2 \end{array} \right \rangle } ^ { T }$  
$\displaystyle \frac{ \partial N }{ \partial \{ x \} } _{(3)}$ $\textstyle =$ $\displaystyle { \left \langle \begin{array}{cc} 4 b1 L0 + 4 b0 L1 & 4 c1 L0 + 4 c0 L1 \end{array} \right \rangle } ^ { T }$  
$\displaystyle \frac{ \partial N }{ \partial \{ x \} } _{(4)}$ $\textstyle =$ $\displaystyle { \left \langle \begin{array}{cc} 4 b2 L1 + 4 b1 L2 & 4 c2 L1 + 4 c1 L2 \end{array} \right \rangle } ^ { T }$  
$\displaystyle \frac{ \partial N }{ \partial \{ x \} } _{(5)}$ $\textstyle =$ $\displaystyle { \left \langle \begin{array}{cc} 4 b0 L2 + 4 b2 L0 & 4 c0 L2 + 4 c2 L0 \end{array} \right \rangle } ^ { T }$ (1.41)



Hiroshi KAWAI 平成15年8月11日