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使用花括号将长方程式拆分为两行

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使用花括号将长方程式拆分为两行

下面我粘贴的方程很长。我希望第一个方程（e34a）dw^2/dx^2分成两行，并用一个共同的花括号将方程分成两行。我的代码是

\begin{subequations}\label{e34} 
  The Normal Stress are given by 
  \begin{align}
   q = & \overline{C}_1\overline{C}_3\overline{C}_5\dfrac{\alpha^{*}k_s w}{U\big(1+ \alpha^{*}k_s(\frac{w}{q_u})\big)}- \dfrac{\partial^{2}w}{\partial x^{2}}\Bigg(G_1 H_1+\overline{C}_2(T_p+T_1)\cos\theta+G_2 H_2\overline{C}_1 \label{e34a}\\
&+ G_3 H_3 \overline{C}_1 \overline{C}_3+G_4 H_4 \overline{C}_1 \overline{C}_3\overline{C}_5+ \overline{C}_1\overline{C}_4(T_p+T_2)\cos\theta+
         \overline{C}_1\overline{C}_3\overline{C}_6(T_p+T_3)\cos\theta\Bigg)\nonumber\\
        \intertext{The Mobilised Tension for top reinforcement is}
            \frac{\partial T_1}{\partial x} = & -\Bigg(q+G_1H_1\frac{\partial^{2}w}{\partial x^{2}}\Bigg)\  \overline{D}_1
            - \Bigg(\overline{C}_3\overline{C}_5\dfrac{\alpha^{*}k_s w}{U\big(1+\alpha^{*}k_s(\frac{w}{q_u})\big)} - A_1 \frac{\partial^{2}w}{\partial x^{2}}\Bigg)\  \overline{D}_2\\
        \intertext{The Mobilised Tension for middle reinforcement is}
            \frac{\partial T_2}{\partial x} = & -\Bigg(\frac{1}{\overline{C}_1} \Big(q+A_2\frac{\partial^{2}w}{\partial x^{2}} \Big) +G_2 H_2\frac{\partial^{2}w}{\partial x^{2}} \Bigg)\ \overline{D}_3  \label{e34c} \\
            & - \Bigg( \overline{C}_5 \dfrac{\alpha^{*}k_s w}{U\big(1+\alpha^{*}k_s(\frac{w}{q_u})\big)} - A_3 \frac{\partial^{2}w}{\partial x^{2}} \Bigg)\ \overline{D}_4  \nonumber
            \intertext{The Mobilised Tension for bottom reinforcement is}
            \frac{\partial T_3}{\partial x} = & -\Biggl(\frac{1}{\overline{C}_3}\Bigg(\frac{1}{\overline{C}_1}\Big(q+A_2\frac{\partial^{2}w}{\partial x^{2}} \Big)+A_4\frac{\partial^{2}w}{\partial x^{2}}\Bigg)+G_3 H_3\frac{\partial^{2}w}{\partial x^{2}} \Biggl)\ \overline{D}_5  \label{e34d} \\
            & - \Bigg( \dfrac{\alpha^{*}k_s w}{U\big(1+\alpha^{*}k_s(\frac{w}{q_u})\big)} - G_4H_4 \frac{\partial^{2}w}{\partial x^{2}} \Bigg)\ \overline{D}_6  \nonumber
\end{align}
\end{subequations}

答案1

我发现使用“\big和朋友”是一种快速修复方法，99% 的时间都有效，而那 1% 的时间解决起来相当令人沮丧。我Mico 的回答（感谢他改进了前两个方程的布局）并恢复到旧的值得信赖的\left和\right对，其中 LaTeX 决定如何拉伸/扩展可扩展分隔符()和{}。但是，当拉伸必须在方程中的多行上进行时会出现问题。但为此，\vphantom{...}允许垂直（零宽度）拉伸正确的高度。例如，这是我的代码：

\documentclass{article}
\newcommand{\Cbar}{\overline{C}}
\newcommand{\Dbar}{\overline{D}}
\newcommand{\myfrac}{\dfrac{\partial^{2}w}{\partial x^{2}}}% For height purposes
\usepackage{amsmath}

\begin{document}
\begin{subequations}\label{e34} 
\noindent
The Normal Stress $q$ is given by 
\begin{align}
\begin{split}
q = &\phantom{=} \Cbar_1\Cbar_3\Cbar_5\dfrac{\alpha^{*}k_s w}{U\left(1+ \alpha^{*}k_s 
(w/q_u)\right)}
- \myfrac\left\{\vphantom{\myfrac}G_1 H_1+\Cbar_2(T_p+T_1)\cos\theta \right.
\label{e34a}\\
&+G_2 H_2\Cbar_1+ G_3 H_3 \Cbar_1 \Cbar_3+G_4 H_4 \Cbar_1 \Cbar_3\Cbar_5 \\
&+\>\Cbar_1\Cbar_4(T_p+T_2)\cos\theta+\Cbar_1\Cbar_3\Cbar_6(T_p+T_3)\cos\theta
\left.\vphantom{\myfrac}\!\right\}\,.\\
\end{split}
\intertext{The Mobilised Tension for top reinforcement is}
\begin{split}
\frac{\partial T_1}{\partial x} 
= & -\bigg(q+G_1H_1\frac{\partial^{2}w}{\partial x^{2}}\bigg)\  \Dbar_1 \\
&- \bigg(\Cbar_3\Cbar_5\dfrac{\alpha^{*}k_s w}{U\big(1+\alpha^{*}k_s(w/q_u)\big)} 
- A_1 \frac{\partial^{2}w}{\partial x^{2}}\bigg)\  \Dbar_2\\
\end{split}
\intertext{The Mobilised Tension for middle reinforcement is}
\begin{split}
\frac{\partial T_2}{\partial x} 
= & -\left(\frac{1}{\Cbar_1} 
\left(q+A_2\frac{\partial^{2}w}{\partial x^{2}} \right) 
+G_2 H_2\frac{\partial^{2}w}{\partial x^{2}} \right)\ \Dbar_3  \label{e34c} \\
&- \left( \Cbar_5 \dfrac{\alpha^{*}k_s w}{U\left(1+\alpha^{*}k_s(w/q_u)\right)} 
- A_3 \dfrac{\partial^{2}w}{\partial x^{2}} \right)\ \Dbar_4\\
\end{split}
\intertext{The Mobilised Tension for bottom reinforcement is}
\begin{split}
\frac{\partial T_3}{\partial x} 
= & -\left(\frac{1}{\Cbar_3}\left(\frac{1}{\Cbar_1}\left(q+A_2\frac{\partial^{2}w}
{\partial x^{2}} \right)
+A_4\frac{\partial^{2}w}{\partial x^{2}}\right)+G_3 H_3\frac{\partial^{2}w}{\partial
x^{2}} \right)\ 
\Dbar_5  \label{e34d} \\
&-\left( \dfrac{\alpha^{*}k_s w}{U\left(1+\alpha^{*}k_s(w/q_u)\right)} - G_4H_4 
\frac{\partial^{2}w}{\partial x^{2}} \right)\ \Dbar_6 \\
\end{split}
\end{align}
\end{subequations}

\end{document}

使用可扩展分隔符分隔的多个方程式

介绍宏\newcommand{\myfrac}{\dfrac{\partial^{2}w}{\partial x^{2}}}只是为了简洁起见。

有关如何在 LaTeX 中使用数学运算的更多想法，赫伯特·沃斯的mathmode文件是一个无价之宝。

答案2

我建议使用以下代码，它使用包split定义的环境amsmath来对齐每个子方程。请注意，我还尝试通过定义命令\Cbar和\Dbar，并用替换各种\Bigg指令来使代码更具可读性\bigg。（Bigg 似乎太大了……）

\documentclass{article}
\newcommand{\Cbar}{\overline{C}}
\newcommand{\Dbar}{\overline{D}}
\usepackage{amsmath}

\begin{document}
\begin{subequations}\label{e34} 
\noindent
The Normal Stress $q$ is given by 
\begin{align}
\begin{split}
q = &\phantom{=} \Cbar_1\Cbar_3\Cbar_5\dfrac{\alpha^{*}k_s w}{U\big(1+ \alpha^{*}k_s 
(w/q_u)\big)}
- \dfrac{\partial^{2}w}{\partial x^{2}}\bigg\{G_1 H_1+\Cbar_2(T_p+T_1)\cos\theta 
\label{e34a}\\
&+G_2 H_2\Cbar_1+ G_3 H_3 \Cbar_1 \Cbar_3+G_4 H_4 \Cbar_1 \Cbar_3\Cbar_5\\
&+\Cbar_1\Cbar_4(T_p+T_2)\cos\theta+\Cbar_1\Cbar_3\Cbar_6(T_p+T_3)\cos\theta
\bigg\}\,.\\
\end{split}
\intertext{The Mobilised Tension for top reinforcement is}
\begin{split}
\frac{\partial T_1}{\partial x} 
= & -\bigg(q+G_1H_1\frac{\partial^{2}w}{\partial x^{2}}\bigg)\  \Dbar_1 \\
&- \bigg(\Cbar_3\Cbar_5\dfrac{\alpha^{*}k_s w}{U\big(1+\alpha^{*}k_s(w/q_u)\big)} 
- A_1 \frac{\partial^{2}w}{\partial x^{2}}\bigg)\  \Dbar_2\\
\end{split}
\intertext{The Mobilised Tension for middle reinforcement is}
\begin{split}
\frac{\partial T_2}{\partial x} 
= & -\bigg(\frac{1}{\Cbar_1} 
\Big(q+A_2\frac{\partial^{2}w}{\partial x^{2}} \Big) 
+G_2 H_2\frac{\partial^{2}w}{\partial x^{2}} \bigg)\ \Dbar_3  \label{e34c} \\
&- \bigg( \Cbar_5 \dfrac{\alpha^{*}k_s w}{U\big(1+\alpha^{*}k_s(w/q_u)\big)} 
- A_3 \dfrac{\partial^{2}w}{\partial x^{2}} \bigg)\ \Dbar_4\\
\end{split}
\intertext{The Mobilised Tension for bottom reinforcement is}
\begin{split}
\frac{\partial T_3}{\partial x} 
= & -\biggl(\frac{1}{\Cbar_3}\bigg(\frac{1}{\Cbar_1}\Big(q+A_2\frac{\partial^{2}w}
{\partial x^{2}} \Big)
+A_4\frac{\partial^{2}w}{\partial x^{2}}\bigg)+G_3 H_3\frac{\partial^{2}w}{\partial
x^{2}} \biggr)\ 
\Dbar_5  \label{e34d} \\
&-\bigg( \dfrac{\alpha^{*}k_s w}{U\big(1+\alpha^{*}k_s(w/q_u)\big)} - G_4H_4 
\frac{\partial^{2}w}{\partial x^{2}} \bigg)\ \Dbar_6 \\
\end{split}
\end{align}
\end{subequations}

\end{document}

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