不同方程组的垂直对齐

Question

我希望以下系统水平对齐（错位是由于某些方程式有分数，而其他方程式没有）。基本上，我希望（例如）所有 $q_1$ 都在同一行上。提前感谢您的帮助！

这是代码

\begin{tcolorbox}[title={2. Calcolo $0+X\cdot a=b$}, colback=white,colframe=black!20,coltitle=black]
\eqal{&\hs{1}0&+X\cdot &\hs{1}a&=&\hs{1}b&\hs{.2}\implies&\hs{1}\delta\mathcal{L}^{ab}_{p/c,i}\\
&\left\{
\begin{aligned}
q_1^0&=0\\
q_2^0&=0\\
q_3^0&=0\\
q_4^0&=-\frac{T}{h}\\
P_1^0&=-\frac{TL}{h}\\
P_2^0&=0\\
P_3^0&=0\\
P_4^0&=\frac{TL}{h}
\end{aligned}
\right.&+X\cdot\hs{.2}
&\left\{
\begin{aligned}
q_1^a&=-\frac{1}{2L}\\
q_2^a&=\frac{1}{2L}\\
q_3^a&=-\frac{1}{2L}\\
q_4^a&=\frac{1}{2L}\\
P_1^a&=1\\
P_2^a&=-1\\    
P_3^a&=1\\
P_4^a&=-1
\end{aligned}
\right.&=\hs{.2}
&\left\{
\begin{aligned}
q_1^b&=-\frac{X}{2L}\\
q_2^b&=\frac{X}{2L}\\
q_3^b&=-\frac{X}{2L}\\
q_4^b&=-\frac{T}{h}+\frac{X}{2L}\\
 P_1^b&=-\frac{TL}{h}+X\\
 P_2^b&=-X\\
 P_3^b&=X\\
 P_4^b&=\frac{TL}{h}-X
 \end{aligned}
 \right.\notag
 &\hs{.2}&\hs{.2}\left\{
 \begin{aligned}
 \delta\mathcal{L}_{c1}^{ab}&=\frac{Lb}{Gs}\left(-\frac{1}{2L}\right)\left(-\frac{X}{2L}\right)\\
 \delta\mathcal{L}_{c2}^{ab}&=\frac{Lh}{Gs}\left(\frac{1}{2L}\frac{X}{2L}\right)\\
\delta\mathcal{L}_{c3}^{ab}&=\frac{Lb}{Gs}\left(-\frac{1}{2L}\right)\left(-\frac{X}{2L}\right)\\
\delta\mathcal{L}_{c4}^{ab}&=\frac{Lh}{Gs}\left(\frac{1}{2L}\right)\left(-\frac{T}{h}+\frac{X}{2L}\right)\\
\delta\mathcal{L}_{p1}^{ab}&=\frac{L}{3EA}\left[1\left(-\frac{TL}{h}+X\right)\right]\\
\delta\mathcal{L}_{p2}^{ab}&=\frac{L}{3EA}\left(-1\right)\left(-X\right)\\
\delta\mathcal{L}_{p3}^{ab}&=\frac{L}{3EA}\left(1\cdot X\right)\\
\delta\mathcal{L}_{p4}^{ab}&=\frac{L}{3EA}\left[-1\left(\frac{TL}{h}-X\right)\right]\\
 \end{aligned}
 \right.
 }
 \end{tcolorbox}

请注意 \hs{X} 是 \hspace{Xcm}（一个新命令*）并且 \eqal{X} 是 \begin{eqnarray} \left{ \begin{aligned} X \end{aligned} \right。 \end{eqnarray}

Answer 1

添加幻影：

\documentclass{article}
\usepackage{amsmath,array}

\newcommand{\phnf}{\vphantom{\left(\frac{A}{B}\right)}}

\begin{document}
\[
\begin{array}{
  @{}c
  @{}>{{}}c<{{}}
  @{}c
  @{}>{{}}c<{{}}
  @{}c
  @{}>{{}}c<{{}}
  @{}c
  @{}>{{}}c<{{}}
  @{}c
}
0 &+& X &\cdot& a&=& b &\implies&\delta\mathcal{L}^{ab}_{p/c,i} \\[4ex]
\begin{cases}
\begin{aligned}
q_1^0&=0 \phnf \\
q_2^0&=0 \phnf \\
q_3^0&=0 \phnf \\
q_4^0&=-\frac{T}{h} \phnf \\
P_1^0&=-\frac{TL}{h} \phnf \\
P_2^0&=0 \phnf \\
P_3^0&=0 \phnf \\
P_4^0&=\frac{TL}{h} \phnf
\end{aligned}
\end{cases}
&+& X &\cdot&
\begin{cases}
\begin{aligned}
q_1^a&=-\frac{1}{2L} \phnf \\
q_2^a&=\frac{1}{2L} \phnf \\
q_3^a&=-\frac{1}{2L} \phnf \\
q_4^a&=\frac{1}{2L} \phnf \\
P_1^a&=1 \phnf \\
P_2^a&=-1 \phnf \\    
P_3^a&=1 \phnf \\
P_4^a&=-1 \phnf
\end{aligned}
\end{cases}
&=&
\begin{cases}
\begin{aligned}
q_1^b&=-\frac{X}{2L} \phnf \\
q_2^b&=\frac{X}{2L} \phnf \\
q_3^b&=-\frac{X}{2L} \phnf \\
q_4^b&=-\frac{T}{h}+\frac{X}{2L} \phnf \\
P_1^b&=-\frac{TL}{h}+X \phnf \\
P_2^b&=-X \phnf \\
P_3^b&=X \phnf \\
P_4^b&=\frac{TL}{h}-X \phnf
\end{aligned}
\end{cases}
 &&
\begin{cases}
\begin{aligned}
\delta\mathcal{L}_{c1}^{ab}&=\frac{Lb}{Gs}\left(-\frac{1}{2L}\right)\left(-\frac{X}{2L}\right) \phnf \\
\delta\mathcal{L}_{c2}^{ab}&=\frac{Lh}{Gs}\left(\frac{1}{2L}\frac{X}{2L}\right) \phnf \\
\delta\mathcal{L}_{c3}^{ab}&=\frac{Lb}{Gs}\left(-\frac{1}{2L}\right)\left(-\frac{X}{2L}\right) \phnf \\
\delta\mathcal{L}_{c4}^{ab}&=\frac{Lh}{Gs}\left(\frac{1}{2L}\right)\left(-\frac{T}{h}+\frac{X}{2L}\right) \phnf \\
\delta\mathcal{L}_{p1}^{ab}&=\frac{L}{3EA}\left[1\left(-\frac{TL}{h}+X\right)\right] \phnf \\
\delta\mathcal{L}_{p2}^{ab}&=\frac{L}{3EA}\left(-1\right)\left(-X\right) \phnf \\
\delta\mathcal{L}_{p3}^{ab}&=\frac{L}{3EA}\left(1\cdot X\right) \phnf \\
\delta\mathcal{L}_{p4}^{ab}&=\frac{L}{3EA}\left[-1\left(\frac{TL}{h}-X\right)\right] \phnf
\end{aligned}
\end{cases}
\end{array}
\]

\end{document}

永远不要使用eqnarray。

您需要具有合适的线宽来包含该内容。