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图片超出页面范围

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图片超出页面范围

我在显示图像时遇到问题tabular。这是我实际使用的代码：

\begin{tabular}{p{10.5cm}ll}     Brandschutz: R30, 3-seitiger Abbrand &  \multirow{3}{4pt}{\includegraphics[width=0.32\textwidth]{images/Druckstab/Brandschutz/Brand_rechteck-oxlr.png}}   
\end{tabular}

在某些情况下，图像位于页面的最末端，因此图像无法完全显示，也不会将页面拆分为新页面。它看起来像这样（黑线处是新页面的开始）：

在此处输入图片描述

是否有可能强制分页，在这种情况下图像不会完全显示在页面上？非常感谢您的任何提示！

这是完整的示例 TeX 代码（您需要替换值graphicspath，并且需要在此处有一张图片：images/Druckstab/Brandschutz/Brand_rechteck-oxlr.png。

\documentclass[fleqn]{report}    
\usepackage[utf8]{inputenc}  
\usepackage[a4paper]{geometry}  
\geometry{    left=2.5cm,    right=1.5cm,    top=2cm,    bottom=2.5cm,    bindingoffset=0.5mm  }  
\usepackage{fancyhdr}  
\pagestyle{fancy}    
\usepackage{tabu}  
\usepackage{lastpage}  
\usepackage[document]{ragged2e}  
\usepackage{multicol}  
\usepackage{graphicx}  
\usepackage[space]{grffile}  
\usepackage{marginnote}  
\usepackage{multirow}  
\usepackage{amsmath}  
\usepackage[sfdefault]{ClearSans}  
\usepackage[T1]{fontenc}  
\usepackage[usenames, dvipsnames]{color}  
\graphicspath{ {D:/Private/some-domain.de/public/} }  
\setlength{\mathindent}{0pt}    
\definecolor{reportGrey}{rgb}{0.4, 0.4, 0.4}     
\setlength{\parindent}{2em}    
\setlength{\parskip}{1em}    
\renewcommand{\baselinestretch}{1.5}    
\headheight = 39pt   
\pagestyle{fancy}  
\fancyhf{}   
\rhead{ \textcolor{reportGrey}{22.04.21\\Seite \thepage\hspace{1pt}/\pageref{LastPage}}}   
\lhead{ \textcolor{reportGrey}{ {\Large{Meine Einzelunternehmung}} \linebreak Straße 1a \linebreak 012345 Stadt} }      
\lfoot{\vspace{-16,8cm}  \hspace{-1cm}   \rotatebox{90}{\textcolor{reportGrey}{Bemessungsmodul: T12 - Druckstab \textbar Version 1.0.0:607 | www.some-domain.de}}   
\\\vspace{.24cm}}       
\begin{document}    
\section*{brandschutztechnische Bemessung}  Bemessungsverfahren nach DIN 1995-1-2:2010-12 nach der Methode mit reduziertem Querschnitt    
\subsection*{Schnittgrößen}  Schnittgrößen am Auflager A (\(x\) = \(L\) = 3.00 m)  
\begin{equation*}       
N_{d,fi} = -F_{c,d,fi} - 1.00 * G_k = -37.45 - 1.00 * 0.45 = -37.90  \text{ kN}  
\end{equation*}  
\begin{equation*}       
V_{z,d,fi} = \dfrac{-F_{c,d,fi} * e_z}{L} = \dfrac{-37.45 * 60}{3.00} * 10^{-3} = -0.56 \text{ kN}  
\end{equation*}  
\begin{equation*}       
V_{y,d,fi} = \dfrac{-F_{c,d,fi} * e_y}{L} = \dfrac{-37.45 * 50}{3.00} * 10^{-3} = -0.62 \text{ kN}  
\end{equation*}  
\begin{equation*}  
M_{y,d,fi} = -F_{c,d,fi} * e_z = -37.45 * 60 * 10^{-3} = -1.67 \text{ kNm}  
\end{equation*}  
\begin{equation*}       
M_{z,d,fi} = -F_{c,d,fi} * e_y = -37.45 * 50 * 10^{-3} = -1.87 \text{ kNm}  
\end{equation*}    
\subsection*{Festigkeits- und Steifigkeitswerte}  
\begin{tabular}{p{1.6cm}p{3.5cm}p{1.9cm}p{5.5cm}ll}  
\(k_{mod,fi}\) : &  1.00 & \(k_{fi}\)  : & 1.25 & \(\gamma_{M,fi}\) : & 1.00i\\  
\end{tabular}  
\begin{equation*}  
f_{c,0,d,fi} = k_{mod,fi} * k_{fi} * \dfrac{f_{c,0,k}}{\gamma_{M,fi}} = 1.00 * 1.25 * \dfrac{21.00}{1.00} = 26.25 \text{ N/mm$^2$}  
\end{equation*}  
\begin{equation*}  
\begin{split}  
k_{h,y} &= min  
\begin{cases}  
\left( \dfrac{150}{b} \right)^{0.2} = \left( \dfrac{150}{140} \right)^{0.2} = 1.01 \\  1.3  
\end{cases}  
\end{split}  
\end{equation*}        
\begin{equation*}       
f_{m,y,d_fi} = k_{mod,fi} * k_{fi} * \dfrac{f_{m,k}}{\gamma_M,fi} = 1.00 * 1.25 * \dfrac{24.00}{1.00} = 30.00 \text{ N/mm$^2$}  
\end{equation*}  
\begin{equation*}       
f_{m,z,d,fi} = k_{mod,fi} * k_{fi} * \dfrac{k_{h,y} * f_{m,k}}{\gamma_M,fi} = 1.00 * 1.25 * \dfrac{1.01 * 24.00}{1.00} = 30.30 \text{ N/mm$^2$}  
\end{equation*}    
\subsection*{Querschnittswerte}  
\begin{tabular}{p{2.6cm}p{2.5cm}p{1.9cm}p{4.0cm}ll}  
Abbrandrate \(\beta_n\): &  0.80 mm/min & \(d_{0}\)  : & 7.00 mm & Branddauer \(t\) : & 30.00 min\\  
\end{tabular}  
\begin{equation*}  
d_{char,n} = \beta_n * t = 0.80 * 30.00  = 24.00 \text{ mm}  \tag{\text{DIN EN 1995-1-2:2010-12, Gl. 3.2}}  
\end{equation*}  
\begin{equation*}       
d_{ef} = d_{char,n} + k_0 * d_0 = 24.00 + 1.00 * 7.00  = 31.00 \text{ mm}  \tag{\text{DIN EN 1995-1-2:2010-12, Gl. 4.1}}  
\end{equation*}  
\begin{tabular}{p{10.5cm}ll}     
Brandschutz: R30, 3-seitiger Abbrand &  \multirow{3}{4pt}{\includegraphics[width=0.32\textwidth]{images/Druckstab/Brandschutz/Brand_rechteck-oxlr.png}}   
\end{tabular}  
\begin{equation*}       
b_t = b - 2 * d_{ef} = 140 - 2 * 31 = 78 \text{ mm}  
\end{equation*}  
\begin{equation*}       
h_t = h - 2 * d_{ef} = 180 - 2 * 31 = 118 \text{ mm}  
\end{equation*}  
\begin{equation*}       
A_r = b_t * h_t = 78 * 118 * 10^{-2} = 92.04 \text{ cm$^2$}  
\end{equation*}  
\begin{equation*}       
W_{y,r} = \dfrac{b_t*h_t^2}{6} = \dfrac{78*118^2}{6} * 10^{-3} = 181.01 \text{ cm$^3$}  
\end{equation*}  
\begin{equation*}       
W_{z,r} = \dfrac{h_t*b_t^2}{6} = \dfrac{118*78^2}{6} * 10^{-3} = 119.65 \text{ cm$^3$}  
\end{equation*}  
\begin{equation*}       
i_{y,r} = \dfrac{h_t}{\sqrt{12}} = \dfrac{118}{\sqrt{12}} = 34.06 \text{ mm}  
\end{equation*}  
\begin{equation*}       
i_{z,r} = \dfrac{b_t}{\sqrt{12}} = \dfrac{78}{\sqrt{12}} = 22.52 \text{ mm}  
\end{equation*}    
\subsubsection{verwendete Normen}  
\begin{tabular}{p{5cm}ll}  
DIN EN 338:2016-07 & Bauholz für tragende Zwecke   \\  
DIN EN 1995-1-1:2010-12 & Eurocode 5: Bemessung und Konstruktion von Holzbauteilen, Teil 1-1 \\  
DIN EN 1995-1-1/A2:2014-07 & Änderung  A2 zu EC5 \\    
DIN EN 1995-1-1/NA:2013-08 & Nationaler Anhang (EC5) \\   \\  
\end{tabular}        
\end{document}

更新 @Zarko 的解决方案有效。如果图像不适合页面，则显示在下一页上。但这不是最佳解决方案，因为这样会在左侧出现较大的“间隙”（见下图）：

在此处输入图片描述

有没有办法避免这些“间隙”，并使所有文本和公式直接显示在彼此下方，同时将图像放在右侧？

答案1

反而

\begin{tabular}{p{10.5cm}ll}     
Brandschutz: R30, 3-seitiger Abbrand &  \multirow{3}{4pt}{\includegraphics[width=0.32\textwidth]{images/Druckstab/Brandschutz/Brand_rechteck-oxlr.png}}   
\end{tabular}

写

Brandschutz: R30, 3-seitiger Abbrand\hfill  
    \includegraphics[width=0.32\textwidth, valign=c]{example-image-duck} %{images/Druckstab/Brandschutz/Brand_rechteck-oxlr.png}}

并在文档序言中将其替换usepackage{graphicx}为usepackage[export]{adjustbox}。

笔记：

  • 我会删除*乘法符号，并在\cdot必要时使用它们
  • 对于单位我会使用siunitx包
  • 我会将一系列equations*写成gather*，例如，在您的文档示例中，第一个为：
\begin{gather*}
N_{d,fi} = -F_{c,d,fi} - 1.00 G_k = -37.45 - 1.00 0.45 = \SI{-37.90}{kN} 
\\
V_{z,d,fi} = \dfrac{-F_{c,d,fi} e_z}{L} = \dfrac{-37.45 60}{3.00} 10^{-3} = \SI{-0.56}{kN}
\\
V_{y,d,fi} = \dfrac{-F_{c,d,fi} e_y}{L} = \dfrac{-37.45 50}{3.00} 10^{-3} = \SI{-0.62}{kN}
\\
M_{y,d,fi} = -F_{c,d,fi} e_z = -37.45\cdot 60\cdot10^{-3} = \SI{-1.67}{kNm}
\\
M_{z,d,fi} = -F_{c,d,fi} e_y = -37.45\cdot50\cdot10^{-3} = -1.87 \text{kNm}
\end{gather*}

这使：

在此处输入图片描述

（红线表示文本左边框）

附录： 在最后的评论/投诉之后，您似乎在寻找类似这样的内容：

在此处输入图片描述

为了得到上述结果，您需要在序言中加载wrapfig包。生成（第二页）上述图像的 MWE 是：

\documentclass[fleqn]{report}
\usepackage[a4paper]{geometry}
\geometry{hmargin={2.5cm,1.5cm},    
          vmargin={2.5cm},    
          bindingoffset=0.5mm,
          headheight = 39pt
          }
\usepackage[T1]{fontenc}
\usepackage[sfdefault]{ClearSans}

\usepackage[document]{ragged2e}
\usepackage[export]{adjustbox}  % <---
%\graphicspath{ {D:/Private/some-domain.de/public/} }
\usepackage{siunitx}            % <---
\usepackage{wrapfig}            % <---

\usepackage{amsmath}
\setlength{\mathindent}{0pt}
\usepackage[usenames, dvipsnames]{color}
\definecolor{reportGrey}{rgb}{0.4, 0.4, 0.4}
\setlength{\parindent}{2em}
\setlength{\parskip}{1em}
\renewcommand{\baselinestretch}{1.5}


\usepackage{fancyhdr}
\pagestyle{fancy}
\fancyhf{}
\rhead{ \textcolor{reportGrey}{22.04.21\\Seite \thepage\hspace{1pt}/\pageref{LastPage}}}
\lhead{ \textcolor{reportGrey}{ {\Large{Meine Einzelunternehmung}} \linebreak Straße 1a \linebreak 012345 Stadt} }
\lfoot{\vspace{-16,8cm}  \hspace{-1cm}   \rotatebox{90}{\textcolor{reportGrey}{Bemessungsmodul: T12 - Druckstab \textbar Version 1.0.0:607 | www.some-domain.de}}
\\\vspace{.24cm}}

\begin{document}
\section*{brandschutztechnische Bemessung}  
Bemessungsverfahren nach DIN 1995-1-2:2010-12 nach der Methode mit reduziertem Querschnitt

\subsection*{Schnittgrößen}  
Schnittgrößen am Auflager A (\(x\) = \(L\) = 3.00 m)
\begin{gather*}
N_{d,fi} = -F_{c,d,fi} - 1.00 G_k = -37.45 - 1.00 0.45 = \SI{-37.90}{kN}
\\
V_{z,d,fi} = \dfrac{-F_{c,d,fi} e_z}{L} = \dfrac{-37.45 60}{3.00} 10^{-3} = \SI{-0.56}{kN}
\\
V_{y,d,fi} = \dfrac{-F_{c,d,fi} e_y}{L} = \dfrac{-37.45 50}{3.00} 10^{-3} = \SI{-0.62}{kN}
\\
M_{y,d,fi} = -F_{c,d,fi} e_z = -37.45\cdot 60\cdot10^{-3} = \SI{-1.67}{kNm}
\\
M_{z,d,fi} = -F_{c,d,fi} e_y = -37.45\cdot50\cdot10^{-3} = -1.87 \text{kNm}
\end{gather*}

\subsection*{Festigkeits- und Steifigkeitswerte}
\begin{tabular}{p{1.6cm}p{3.5cm}p{1.9cm}p{5.5cm}ll}
\(k_{mod,fi}\) : &  1.00 & \(k_{fi}\)  : & 1.25 & \(\gamma_{M,fi}\) : & 1.00i\\
\end{tabular}

\begin{gather*}
f_{c,0,d,fi} = k_{mod,fi} * k_{fi} * \dfrac{f_{c,0,k}}{\gamma_{M,fi}} = 1.00 * 1.25 * \dfrac{21.00}{1.00} = 26.25 \text{ N/mm$^2$}
\\
\begin{split}
k_{h,y} = \min\begin{cases}
    \left( \dfrac{150}{b} \right)^{0.2} = \left( \dfrac{150}{140} \right)^{0.2} = 1.01 \\  
    1.3
    \end{cases}
\end{split}
\\
f_{m,y,d_fi} = k_{mod,fi} * k_{fi} * \dfrac{f_{m,k}}{\gamma_M,fi} = 1.00 * 1.25 * \dfrac{24.00}{1.00} = 30.00 \text{ N/mm$^2$}
\\
f_{m,z,d,fi} = k_{mod,fi} * k_{fi} * \dfrac{k_{h,y} * f_{m,k}}{\gamma_M,fi} = 1.00 * 1.25 * \dfrac{1.01 * 24.00}{1.00} = 30.30 \text{ N/mm$^2$}
\end{gather*}

\subsection*{Querschnittswerte}
\begin{tabular}{p{2.6cm}p{2.5cm}p{1.9cm}p{4.0cm}ll}
Abbrandrate \(\beta_n\): &  0.80 mm/min & \(d_{0}\)  : & 7.00 mm & Branddauer \(t\) : & 30.00 min\\
\end{tabular}
\begin{equation*}
d_{char,n} = \beta_n * t = 0.80 * 30.00  = 24.00 \text{ mm}  \tag{\text{DIN EN 1995-1-2:2010-12, Gl. 3.2}}
\end{equation*}
\begin{equation*}
d_{ef} = d_{char,n} + k_0 * d_0 = 24.00 + 1.00 * 7.00  = 31.00 \text{ mm}  \tag{\text{DIN EN 1995-1-2:2010-12, Gl. 4.1}}
\end{equation*}

\begin{wrapfigure}[12]{r}{0.32\linewidth} % <----
    \includegraphics[width=\linewidth]{example-image-duck} %{images/Druckstab/Brandschutz/Brand_rechteck-oxlr.png}}
\end{wrapfigure}
Brandschutz: R30, 3-seitiger Abbrand   
\begin{gather*}
b_t = b - 2 * d_{ef} = 140 - 2 * 31 = 78 \text{ mm}
\\
h_t = h - 2 * d_{ef} = 180 - 2 * 31 = 118 \text{ mm}
\\
A_r = b_t * h_t = 78 * 118 * 10^{-2} = 92.04 \text{ cm$^2$}
\\
W_{y,r} = \dfrac{b_t*h_t^2}{6} = \dfrac{78*118^2}{6} * 10^{-3} = 181.01 \text{ cm$^3$}
\\
W_{z,r} = \dfrac{h_t*b_t^2}{6} = \dfrac{118*78^2}{6} * 10^{-3} = 119.65 \text{ cm$^3$}
\\
i_{y,r} = \dfrac{h_t}{\sqrt{12}} = \dfrac{118}{\sqrt{12}} = 34.06 \text{ mm}
\\
i_{z,r} = \dfrac{b_t}{\sqrt{12}} = \dfrac{78}{\sqrt{12}} = 22.52 \text{ mm}
\end{gather*}

\subsubsection{Verwendete Normen}
\begin{tabular}{p{5cm}ll}
DIN EN 338:2016-07 & Bauholz für tragende Zwecke   \\
DIN EN 1995-1-1:2010-12 & Eurocode 5: Bemessung und Konstruktion von Holzbauteilen, Teil 1-1 \\
DIN EN 1995-1-1/A2:2014-07 & Änderung  A2 zu EC5 \\
DIN EN 1995-1-1/NA:2013-08 & Nationaler Anhang (EC5) \\   \\
\end{tabular}
\end{document}

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