Total Lagrangian 格式#
在 Total Lagrangian 格式中，以初始构型为参考构型，所有量都在初始构型中度量根据虚功原理 (34)，第 \(t^{n+1}\) 时满足
(35)#\[
\int_{\Omega_{0}} \mathbf{S}^{n+1}:\delta\mathbf{E}^{n+1}\ \mathrm{d}V - \left(\int_{\Omega_{0}}\mathbf{\hat{f}}^{n+1}\cdot\delta\mathbf{u}\ \mathrm{d}V+\oint_{\Gamma_{0}}\mathbf{\hat{t}}^{n+1}\cdot\delta\mathbf{u}\ \mathrm{d}S\right) = 0
\]
增量分解#
对于 Green-Lagrange 应变张量，有
\[\begin{split}
\begin{aligned}
\Delta\mathbf{E}&=\mathbf{E}^{n+1} - \mathbf{E}^{n}\\
&=\frac{1}{2}[\nabla_{0}\mathbf{u}^{n+1}+(\nabla_{0}\mathbf{u}^{n+1})^{T}+(\nabla_{0}\mathbf{u}^{n+1})^{T}\cdot(\nabla_{0}\mathbf{u}^{n+1})]\\
&-\frac{1}{2}[\nabla_{0}\mathbf{u}^{n}+(\nabla_{0}\mathbf{u}^{n})^{T}+(\nabla_{0}\mathbf{u}^{n})^{T}\cdot(\nabla_{0}\mathbf{u}^{n})]
\end{aligned}
\end{split}\]
代入 \(\mathbf{u}^{n+1} = \mathbf{u}^{n} + \Delta\mathbf{u}\)，得到
\[\begin{split}
\begin{aligned}
\Delta\mathbf{E} &= \frac{1}{2}[\nabla_{0}\Delta\mathbf{u}+(\nabla_{0}\Delta\mathbf{u})^{T}+(\nabla_{0}\mathbf{u}^{n})^{T}\cdot(\nabla_{0}\Delta\mathbf{u})\\
&+(\nabla_{0}\Delta\mathbf{u})^{T}\cdot(\nabla_{0}\mathbf{u}^{n})+(\nabla_{0}\Delta\mathbf{u})^{T}\cdot(\nabla_{0}\Delta\mathbf{u})]\\
&:=\Delta\mathbf{E}^{\text{L}}+\Delta\mathbf{E}^{\text{NL}}
\end{aligned}
\end{split}\]
其中
\[\begin{split}
\begin{aligned}
\Delta\mathbf{E}^{\text{L}}&=\frac{1}{2}[\nabla_{0}\Delta\mathbf{u}+(\nabla_{0}\Delta\mathbf{u})^{T}+(\nabla_{0}\mathbf{u}^{n})^{T}\cdot(\nabla_{0}\Delta\mathbf{u})
+(\nabla_{0}\Delta\mathbf{u})^{T}\cdot(\nabla_{0}\mathbf{u}^{n})]\\
\Delta\mathbf{E}^{\text{NL}}&=\frac{1}{2}[(\nabla_{0}\Delta\mathbf{u})^{T}\cdot(\nabla_{0}\Delta\mathbf{u})]
\end{aligned}
\end{split}\]
分别是增量的线性部分和非线性部分
\(\Delta\mathbf{E}^{\text{L}}\) 也可以表示为
\[\begin{split}
\begin{aligned}
\Delta\mathbf{E}^{\text{L}}&=\frac{1}{2}[\nabla_{0}\Delta\mathbf{u}+(\nabla_{0}\Delta\mathbf{u})^{T}+(\nabla_{0}\mathbf{u}^{n})^{T}\cdot(\nabla_{0}\Delta\mathbf{u})
+(\nabla_{0}\Delta\mathbf{u})^{T}\cdot(\nabla_{0}\mathbf{u}^{n})]\\
&=\frac{1}{2}[(\mathbf{I}+(\nabla_{0}\mathbf{u}^{n})^{T})(\nabla_{0}\Delta\mathbf{u})+(\nabla_{0}\Delta\mathbf{u})^{T}(\mathbf{I}+\nabla_{0}\mathbf{u}^{n})]\\
&=\frac{1}{2}[(\mathbf{F}^{n})^{T}(\nabla_{0}\Delta\mathbf{u})+(\nabla_{0}\Delta\mathbf{u})^{T}\mathbf{F}^{n}]
\end{aligned}
\end{split}\]
对于 PK2 应力张量，有
\[
\mathbf{S}^{n+1} = \mathbf{S}^{n} + \Delta\mathbf{S}
\]
由于
\[
\delta\mathbf{E}^{n+1}=\delta(\mathbf{E}^{n}+\Delta\mathbf{E}) = \delta\Delta\mathbf{E}
\]
于是
\[\begin{split}
\begin{aligned}
\int_{\Omega_{0}} \mathbf{S}^{n+1}:\delta\mathbf{E}^{n+1}\ \mathrm{d}V &= \int_{\Omega_{0}}(\mathbf{S}^{n} + \Delta\mathbf{S}):\delta\Delta\mathbf{E}\ \mathrm{d}V\\
&=\int_{\Omega_{0}}(S^{n}_{ij} + \Delta S_{ij})(\delta\Delta E_{ij})\ \mathrm{d}V\\
&=\int_{\Omega_{0}}(S^{n}_{ij}(\delta\Delta E^{\text{L}}_{ij}+\delta\Delta E^{\text{NL}}_{ij}) + \Delta S_{ij}(\delta\Delta E_{ij}))\ \mathrm{d}V
\end{aligned}
\end{split}\]
记
\[
\delta R^{n+1}=\int_{\Omega_{0}}\mathbf{\hat{f}}^{n+1}\cdot\delta\mathbf{u}\ \mathrm{d}V+\oint_{\Gamma_{0}}\mathbf{\hat{t}}^{n+1}\cdot\delta\mathbf{u}\ \mathrm{d}S
\]
于是 (35) 可以写为
(36)#\[
\int_{\Omega_{0}} \Delta S_{ij}(\delta\Delta E_{ij})+S^{n}_{ij}(\delta\Delta E^{\text{L}}_{ij})+S^{n}_{ij}(\delta\Delta E^{\text{NL}}_{ij})\ \mathrm{d}V - \delta R^{n+1} = 0
\]
这是关于 \(\Delta \mathbf{u}\) 的非线性方程
线性化#
增量形式的本构方程满足
\[
\Delta\mathbf{S}\approx\mathbb{C}:\Delta\mathbf{E}\quad \text{or}\quad \Delta S_{ij} \approx C_{ijkl}\Delta E_{kl}
\]
其中，\(\mathbb{C}\) 是材料模型的瞬时刚度模量，对于线弹性模型，退化为 Hooke 定律
代入本构关系得到
(37)#\[
\int_{\Omega_{0}} C_{ijkl}\Delta E_{kl}(\delta\Delta E_{ij})+S^{n}_{ij}(\delta\Delta E^{\text{L}}_{ij})+S^{n}_{ij}(\delta\Delta E^{\text{NL}}_{ij})\ \mathrm{d}V - \delta R^{n+1} = 0
\]
由于 \(\Delta\mathbf{E}^{\text{NL}}\) 是 \(\Delta\mathbf{u}\) 的高阶无穷小量，因此
\[
\Delta E_{kl} \rightarrow \Delta E^{\text{L}}_{kl}
\]
于是方程 (37) 写为
(38)#\[
\int_{\Omega_{0}} C_{ijkl}\Delta E_{kl}^{\text{L}}(\delta\Delta E_{ij}^{\text{L}})+S^{n}_{ij}(\delta\Delta E^{\text{L}}_{ij})+S^{n}_{ij}(\delta\Delta E^{\text{NL}}_{ij})\ \mathrm{d}V - \delta R^{n+1} = 0
\]
变分计算#
\[\begin{split}
\begin{aligned}
\delta\Delta\mathbf{E}^{\text{L}}&=\frac{1}{2}\delta[\nabla_{0}\Delta\mathbf{u}+(\nabla_{0}\Delta\mathbf{u})^{T}+(\nabla_{0}\mathbf{u}^{n})^{T}\cdot(\nabla_{0}\Delta\mathbf{u})
+(\nabla_{0}\Delta\mathbf{u})^{T}\cdot(\nabla_{0}\mathbf{u}^{n})]\\
&=\frac{1}{2}[\nabla_{0}\delta\Delta\mathbf{u}+(\nabla_{0}\delta\Delta\mathbf{u})^{T}+(\nabla_{0}\mathbf{u}^{n})^{T}\cdot(\nabla_{0}\delta\Delta\mathbf{u})
+(\nabla_{0}\delta\Delta\mathbf{u})^{T}\cdot(\nabla_{0}\mathbf{u}^{n})]\\
\delta\Delta\mathbf{E}^{\text{NL}}&=\frac{1}{2}\delta[(\nabla_{0}\Delta\mathbf{u})^{T}\cdot(\nabla_{0}\Delta\mathbf{u})]\\
&=\frac{1}{2}[(\nabla_{0}\delta\Delta\mathbf{u})^{T}\cdot(\nabla_{0}\Delta\mathbf{u}) + (\nabla_{0}\Delta\mathbf{u})^{T}\cdot(\nabla_{0}\delta\Delta\mathbf{u})]
\end{aligned}
\end{split}\]
由于 \(\Delta \mathbf{E}^{\text{L}}\) 是关于 \(\mathbf{u}\) 的线性函数，因此 \(\delta\Delta \mathbf{E}^{\text{L}}\) 中仅包含 \(\delta\Delta\mathbf{u}\) 而不包含 \(\Delta\mathbf{u}\)，将自由度项置于左端，最终得到
(39)#\[
\int_{\Omega_{0}} C_{ijkl}\Delta E_{kl}^{\text{L}}(\delta\Delta E_{ij}^{\text{L}})+S^{n}_{ij}(\delta\Delta E^{\text{NL}}_{ij})\ \mathrm{d}V = \delta R^{n+1} - \int_{\Omega_{0}}S^{n}_{ij}(\delta\Delta E^{\text{L}}_{ij})\ \mathrm{d}V
\]