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子方程对齐

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子方程对齐

有人能帮我解决这个问题吗？在此处输入图片描述 实际上，没有必要有 1(b)、1(d)、1(f) 和 1(h)，因为它们是从最后一行派生出来的。我可以只用四个子方程吗？

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{subequations}
\begin{alignat}{4}
E_{11}&=Q_{11}^2\ln\sqrt{B'_{11}} +Q_{12}^2\ln\sqrt{B'_{22}}+Q_{13}^2\ln\sqrt{B'_{33}}\\
&=Q_{11}^2\ln\sqrt{B'_{11}}+Q_{21}^2\ln\sqrt{B'_{33}}\\
E_{12}&=Q_{11}Q_{21}\ln\sqrt{B'_{11}}     +Q_{12}Q_{22}\ln\sqrt{B'_{22}}+Q_{13}Q_{23}\ln\sqrt{B'_{33}} \\
&=Q_{11}Q_{21}\ln\sqrt{B'_{11}}-Q_{11}Q_{21}\ln\sqrt{B'_{33}}\\
E_{22}&=Q_{21}^2\ln\sqrt{B'_{11}} +Q_{22}^2\ln\sqrt{B'_{22}}+Q_{23}^2\ln\sqrt{B'_{33}}\\
&=Q_{21}^2\ln\sqrt{B'_{11}} +Q_{11}^2\ln\sqrt{B'_{33}}\\
E_{33}&=Q_{31}^2\ln\sqrt{B'_{11}} +Q_{32}^2\ln\sqrt{B'_{22}}+Q_{33}^2\ln\sqrt{B'_{33}}\\
&=Q_{32}^2\ln\sqrt{B'_{22}}
\end{alignat}
\label{straincomponent}
\end{subequations}

\end{document}

答案1

另一种变化是，每对方程的编号垂直放置在两条线之间：

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{subequations}
\begin{align}
\begin{split}
E_{11}&=Q_{11}^2\ln\sqrt{B'_{11}} +Q_{12}^2\ln\sqrt{B'_{22}}+Q_{13}^2\ln\sqrt{B'_{33}}\\
&=Q_{11}^2\ln\sqrt{B'_{11}}+Q_{21}^2\ln\sqrt{B'_{33}}
\end{split}\\
\begin{split}
E_{12}&=Q_{11}Q_{21}\ln\sqrt{B'_{11}}     +Q_{12}Q_{22}\ln\sqrt{B'_{22}}+Q_{13}Q_{23}\ln\sqrt{B'_{33}} \\
&=Q_{11}Q_{21}\ln\sqrt{B'_{11}}-Q_{11}Q_{21}\ln\sqrt{B'_{33}}
\end{split}\\
\begin{split}
E_{22}&=Q_{21}^2\ln\sqrt{B'_{11}} +Q_{22}^2\ln\sqrt{B'_{22}}+Q_{23}^2\ln\sqrt{B'_{33}}\\
&=Q_{21}^2\ln\sqrt{B'_{11}} +Q_{11}^2\ln\sqrt{B'_{33}}
\end{split}\\
\begin{split}
E_{33}&=Q_{31}^2\ln\sqrt{B'_{11}} +Q_{32}^2\ln\sqrt{B'_{22}}+Q_{33}^2\ln\sqrt{B'_{33}}\\
&=Q_{32}^2\ln\sqrt{B'_{22}}
\end{split}
\end{align}
\label{straincomponent}
\end{subequations}
\end{document}

示例代码的输出

答案2

\nonumber在您不想使用标签的行中使用。

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{subequations}
\begin{alignat}{4}
E_{11}&=Q_{11}^2\ln\sqrt{B'_{11}} +Q_{12}^2\ln\sqrt{B'_{22}}+Q_{13}^2\ln\sqrt{B'_{33}}\\
      &=Q_{11}^2\ln\sqrt{B'_{11}}+Q_{21}^2\ln\sqrt{B'_{33}}\nonumber\\            
E_{12}&=Q_{11}Q_{21}\ln\sqrt{B'_{11}}     +Q_{12}Q_{22}\ln\sqrt{B'_{22}}+Q_{13}Q_{23}\ln\sqrt{B'_{33}} \\ 
      &=Q_{11}Q_{21}\ln\sqrt{B'_{11}}-Q_{11}Q_{21}\ln\sqrt{B'_{33}}\nonumber\\ 
E_{22}&=Q_{21}^2\ln\sqrt{B'_{11}} +Q_{22}^2\ln\sqrt{B'_{22}}+Q_{23}^2\ln\sqrt{B'_{33}}\\ 
      &=Q_{21}^2\ln\sqrt{B'_{11}} +Q_{11}^2\ln\sqrt{B'_{33}}\nonumber\\
E_{33}&=Q_{31}^2\ln\sqrt{B'_{11}} +Q_{32}^2\ln\sqrt{B'_{22}}+Q_{33}^2\ln\sqrt{B'_{33}}\\ 
      &=Q_{32}^2\ln\sqrt{B'_{22}}\nonumber    
\end{alignat}
\label{straincomponent}
\end{subequations}

\end{document}

在此处输入图片描述

答案3

一个变体，具有单个对齐和 4 个对齐环境。我在方程组之间添加了一些垂直间距：

\documentclass{article}
\usepackage{amsmath}

\begin{document}
\begin{subequations}
  \begin{align}%{4}
    E_{11} & \!\begin{aligned}[t] & = Q_{11}^2\ln\sqrt{B'_{11}} +Q_{12}^2\ln\sqrt{B'_{22}}+Q_{13}^2\ln\sqrt{B'_{33}} \\
    &=Q_{11}^2\ln\sqrt{B'_{11}}+Q_{21}^2\ln\sqrt{B'_{33}}
    \end{aligned}\\[0.75ex]
    E_{12} & \!\begin{aligned}[t] & =Q_{11}Q_{21}\ln\sqrt{B'_{11}} +Q_{12}Q_{22}\ln\sqrt{B'_{22}}+Q_{13}Q_{23}\ln\sqrt{B'_{33}} \\
    & =Q_{11}Q_{21}\ln\sqrt{B'_{11}}-Q_{11}Q_{21}\ln\sqrt{B'_{33}}
    \end{aligned}\\[0.75ex]
    E_{22} & \!\begin{aligned}[t] & =Q_{21}^2\ln\sqrt{B'_{11}} +Q_{22}^2\ln\sqrt{B'_{22}}+Q_{23}^2\ln\sqrt{B'_{33}} \\
    &=Q_{21}^2\ln\sqrt{B'_{11}} +Q_{11}^2\ln\sqrt{B'_{33}}
    \end{aligned}\\[0.85ex]
    E_{33} & \!\begin{aligned}[t] & =Q_{31}^2\ln\sqrt{B'_{11}} +Q_{32}^2\ln\sqrt{B'_{22}}+Q_{33}^2\ln\sqrt{B'_{33}} \\
    &=Q_{32}^2\ln\sqrt{B'_{22}}
    \end{aligned}
  \end{align}
  \label{straincomponent}
\end{subequations}

\end{document}

在此处输入图片描述

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