Modifica la “baseline” dei disegni:
`\begin{tikzpicture}[>=latex,baseline=(current bounding box.center)]`
e scrivi le equazioni come
`&
$\displaystyle
a = \frac{Fh^3}{3EI} \frac{3}{\alpha^3 h^3}(\tan \alpha h-\alpha h)
$
\\`
Ovviamente, poi, hai qualcosa di troppo. Esempio completo, ho aggiunto un’impostazione locale di [tt]\arraystretch[/tt] per distanziare un po’ le due righe.
`\documentclass[a4paper,10pt]{article}
\usepackage[big]{layaureo}
\usepackage{multicol}
\usepackage[latin1]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[italian]{babel}
\usepackage{tikz}
\usetikzlibrary{calc,intersections,patterns}
\usepackage{pgfplots}
\usepackage{siunitx} % Per le tabelle numeriche
\sisetup{output-decimal-marker={,}}
\usepackage{amsmath,mathcomp}
\colorlet{verde}{green!80!black}
\newcommand{\abs}[1]{\left|#1\right|}
%opening
\title{Formulario C.A.\&C.A.P.}
\begin{document}
\begin{center}
\renewcommand{\arraystretch}{15}
\begin{tabular}{ll}
\begin{tikzpicture}[>=latex,baseline=(current bounding box.center)]
%%% struttura
%campiture:
\tikzset{
ground/.style={fill,pattern=north east lines,draw=none,
minimum width=1.2cm,minimum height=0.2cm}
}
% disegno della struttura e del telaio
\draw[thick] (0,0) — (0,5) node[pos=0.5,above right]{$l$};
\draw[thick] (0,5) — (1,5) node[pos=0.9,below left]{$e_0$};
\draw [line width=1pt,dashed] (0,0) arc[start angle=0,end angle=30,x radius=-12cm,y radius=9.5cm];
%Vincolo incastro
\node (g1) at (0,0) [ground,anchor=north]{};
\draw (g1.north west) — (g1.north east);
%Forze
\draw[->] (1,6.2) — (1,5.1) node[pos=0.3 ,below right] {$P$};
\draw[->] (-1.2,5) — (-0.1,5) node[pos=0.3 ,above right] {$F$};
\end{tikzpicture}
&
$\displaystyle
a = \frac{Fh^3}{3EI} \frac{3}{\alpha^3 h^3}(\tan \alpha h-\alpha h)
$
\\
\begin{tikzpicture}[>=latex,baseline=(current bounding box.center)]
%%% struttura
%campiture:
\tikzset{
ground/.style={fill,pattern=north east lines,draw=none,
minimum width=1.2cm,minimum height=0.2cm}
}
% disegno della struttura e del telaio
\draw[thick] (0,0) — (0,5) node[pos=0.5,above right]{$l$};
\draw[thick] (0,5) — (1,5) node[pos=0.9,below left]{$e_0$};
\draw [line width=1pt,dashed] (0,0) arc[start angle=0,end angle=30,x radius=-12cm,y radius=9.5cm];
%Vincolo incastro
\node (g1) at (0,0) [ground,anchor=north]{};
\draw (g1.north west) — (g1.north east);
%Momento
\draw[<-] (0.5,5.5) arc[start angle=45,end angle=180,radius=0.5cm];
\path (-0.2,6) node(M) {$M$};
%Carico distribuito
\draw[<-] (0,0.1) -- (-.7,.1);
\draw (-.7,.1) -- (-.7,4.9) node[pos=0.5,above left]{$q$};
\draw[<-] (0,4.9) -- (-.7,4.9);
\draw[<-] (0,0.5) -- (-.7,0.5);
\draw[<-] (0,1) -- (-.7,1);
\draw[<-] (0,1.5) -- (-.7,1.5);
\draw[<-] (0,2) -- (-.7,2);
\draw[<-] (0,2.5) -- (-.7,2.5);
\draw[<-] (0,3) -- (-.7,3);
\draw[<-] (0,3.5) -- (-.7,3.5);
\draw[<-] (0,4) -- (-.7,4);
\draw[<-] (0,4.5) -- (-.7,4.5);
\end{tikzpicture}
&
$\displaystyle
a = \frac{Fh^3}{3EI} \frac{3}{\alpha^3 h^3}(\tan \alpha h-\alpha h)
$
\end{tabular}
\end{center}
\end{document}`
Ciao
Enrico