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The equation that I have pasted below is very long. I want the first equation (e34a) after dw^2/dx^2 to be split in two lines with a common curly brace for the equation in two lines. My code is

\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}
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Isn't this a continuation of this question: stackoverflow.com/questions/6963414/… ? –  Karuna-bdc Aug 7 '11 at 10:24
    
@Reshma if you want to thank someone for having answered your question, upvote his/her answer (if you can) and accept it (if it did answer your question). Please see How does accepting an answer work?. (This site is not a forum.) –  Mat Aug 7 '11 at 10:27
    
No its not the same question. But it is a continuation. The equation is longer than that in the previous question. And I want curly braces for the two split equations –  Reshma B Aug 7 '11 at 10:41
    
The answer seems the same. –  Svante Aug 7 '11 at 11:45
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migrated from stackoverflow.com Aug 7 '11 at 12:13

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2 Answers

I suggest the following code, which uses the split environment, defined by the amsmath package, to align each of the sub-equatons. Note that I've also attempted to make the code more readable by defining the commands \Cbar and \Dbar, and by replacing the various \Bigg instructions with \bigg. (Bigg seems just too big...)

\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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I find the use of "\big and friends" to be a quick fix that works 99% of the time, with that 1% being quite frustrating to solve. I took Mico's answer (credit to him for the improved layout of the first two equations) and reverted back to the old trustworthy \left and \right pairs, where LaTeX decides how to stretch/extend the extensible delimiters () and {}. There is problem when stretching has to occur across more than one line in an equation though. But for that, \vphantom{...} allows a vertical (zero-width) stretch of the correct height. By example, here's my code:

\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}

Multiple equations that are split with extensible delimiters

The macro \newcommand{\myfrac}{\dfrac{\partial^{2}w}{\partial x^{2}}} is introduced just for the sake of brevity.

For more ideas on how to work with math in LaTeX, Herbert Voss' mathmode document is an invaluable source.

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1  
I like the additional fixes -- automating the sizing of the parentheses, and using a brief macro for the recurring second partial derivative -- you've applied to Reshma's original code. It remains to be seen whether Reshma has anything more to say (including accepting one of the proposed answers)... –  Mico Aug 8 '11 at 10:25
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