形状関数の勾配

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

$$\frac{ \partial N }{ \partial \{ x \} } _{(0)} = { \left \langle \begin{array}{ccc} 4 b0 L0 - b0 & 4 c0 L0 - c0 & 4 d0 L0 - d0 \end{array} \right \rangle } ^ { T }$$

$$\frac{ \partial N }{ \partial \{ x \} } _{(1)} = { \left \langle \begin{array}{ccc} 4 b1 L1 - b1 & 4 c1 L1 - c1 & 4 d1 L1 - d1 \end{array} \right \rangle } ^ { T }$$

$$\frac{ \partial N }{ \partial \{ x \} } _{(2)} = { \left \langle \begin{array}{ccc} 4 b2 L2 - b2 & 4 c2 L2 - c2 & 4 d2 L2 - d2 \end{array} \right \rangle } ^ { T }$$

$$\frac{ \partial N }{ \partial \{ x \} } _{(3)} = { \left \langle \begin{array}{ccc} 4 b3 L3 - b3 & 4 c3 L3 - c3 & 4 d3 L3 - d3 \end{array} \right \rangle } ^ { T }$$

$$\frac{ \partial N }{ \partial \{ x \} } _{(4)} = { \left \langle \begin{array}{ccc} 4 b1 L0 + 4 b0 L1 & 4 c1 L0 + 4 c0 L1 & 4 d1 L0 + 4 d0 L1 \end{array} \right \rangle } ^ { T }$$

$$\frac{ \partial N }{ \partial \{ x \} } _{(5)} = { \left \langle \begin{array}{ccc} 4 b2 L0 + 4 b0 L2 & 4 c2 L0 + 4 c0 L2 & 4 d2 L0 + 4 d0 L2 \end{array} \right \rangle } ^ { T }$$

$$\frac{ \partial N }{ \partial \{ x \} } _{(6)} = { \left \langle \begin{array}{ccc} 4 b3 L0 + 4 b0 L3 & 4 c3 L0 + 4 c0 L3 & 4 d3 L0 + 4 d0 L3 \end{array} \right \rangle } ^ { T }$$

$$\frac{ \partial N }{ \partial \{ x \} } _{(7)} = { \left \langle \begin{array}{ccc} 4 b2 L1 + 4 b1 L2 & 4 c2 L1 + 4 c1 L2 & 4 d2 L1 + 4 d1 L2 \end{array} \right \rangle } ^ { T }$$

$$\frac{ \partial N }{ \partial \{ x \} } _{(8)} = { \left \langle \begin{array}{ccc} 4 b3 L2 + 4 b2 L3 & 4 c3 L2 + 4 c2 L3 & 4 d3 L2 + 4 d2 L3 \end{array} \right \rangle } ^ { T }$$

$$\frac{ \partial N }{ \partial \{ x \} } _{(9)} = { \left \langle \begin{array}{ccc} 4 b1 L3 + 4 b3 L1 & 4 c1 L3 + 4 c3 L1 & 4 d1 L3 + 4 d3 L1 \end{array} \right \rangle } ^ { T }$$