Implement torsion check according to art. 6.3 EN 1992-1-1

[teknovizier]

Description

This is a draft PR that adds a torsion check according to art. 6.3 EN 1992-1-1.
There are many things that need to be changed, improved, and refactored. Some of these are marked in the code with "FIXME" comments. The calculations need to be verified, and tests should be written.
Please evaluate and help to improve this.

Type of change

• New feature (non-breaking change which adds functionality)
• This change requires a documentation update

Checklist:

• I have added tests that prove my fix is effective or that my feature works
• I have commented my code, particularly in hard-to-understand areas
• I have made corresponding changes to the documentation
• New and existing unit tests pass locally with my changes

[egarciamendez]

Hi @teknovizier! I'm very glad to have spoken to you today and really appreciate your contribution. We'll help you get this PR through and into Blueprints :-)

[teknovizier]

Glad to help. Let's make it happen!

[egarciamendez]

I will get into this one soon, been bit busy lately 😅

[egarciamendez]

Hi @teknovizier , I have restructured the torsion check to make it more traceable and mantainable. It is still a working version (but we are getting there)!

[egarciamendez]

In this file I have separated the concerns of each individual test. So you can use each one of those separately from the main torsion check interface.
I want also to add latex representation to each one of them and unite them at the main interface level. 😄

[egarciamendez]

Working version of the Latex report
Blueprints-3.pdf

\documentclass{article}
\usepackage{amsmath}
\usepackage{booktabs}
\usepackage{geometry}
\geometry{a4paper, margin=1in}

% Increase spacing throughout document
\usepackage{setspace}
\setstretch{1.3}  % Increase line spacing by 30%

% Add space between paragraphs
\setlength{\parskip}{0.5em}

% Add space around equations
\setlength{\abovedisplayskip}{12pt}
\setlength{\belowdisplayskip}{12pt}

\begin{document}

\title{Torsion Check Results}
\date{}
\maketitle

\section{Utilization Summary}

\begin{table}[h]
\centering
\begin{tabular}{lcc}
\toprule
Check & Utilization & Status \\
\midrule
Concrete strut capacity & 0.588 & PASS \\
Torsion moment capacity & 4.825 & FAIL \\
Max longitudinal reinforcement & 0.107 & PASS \\
Min tensile reinforcement & 0.575 & PASS \\
Max shear stirrup spacing & 0.494 & PASS \\
Max torsion stirrup spacing & 0.300 & PASS \\
Shear and torsion stirrup area & 0.894 & PASS \\
Min shear reinforcement ratio & 0.131 & PASS \\
\bottomrule
\end{tabular}
\end{table}

\textbf{Overall Result: FAIL}\\[0.5em]

\section{Individual Checks}

\subsection{Concrete Strut Capacity}

Concrete strut capacity check EN 1992-1-1:2004 art. 6.3.2(4)

\vspace{0.3em}
\textbf{Maximum shear resistance}

(6.9) 
\[
V_{Rd,max} = \alpha_{cw} \cdot b_{w} \cdot z \cdot \nu_{1} \cdot \frac{f_{cd}}{\cot(\theta) + \tan(\theta)} = 1.00 \cdot 400.00 \cdot 486.00 \cdot 0.52 \cdot \frac{23.33}{\cot(45.00) + \tan(45.00)} = 1170288.00 \ \text{N}
\]

\vspace{0.3em}
\textbf{Design torsional resistance moment}

(6.30) 
\[
T_{Rd,max} = 2 \cdot \nu \cdot \alpha_{cw} \cdot f_{cd} \cdot A_{k} \cdot t_{ef,i} \cdot \sin(\theta) \cdot \cos(\theta) = 2 \cdot 0.52 \cdot 1.00 \cdot 23.33 \cdot 134400.00 \cdot 120.00 \cdot \sin(45.00) \cdot \cos(45.00) = 194181120.00 \ \text{Nmm}
\]

\vspace{0.3em}
\textbf{Combined interaction check}

(6.29) 
\[
\frac{T_{Ed}}{T_{Rd,max}} + \frac{V_{Ed}}{V_{Rd,max}} = 0.309 + 0.279 = 0.588 \leq 1.0
\]

\vspace{0.5em}
\textbf{Result: PASS (Utilization: 58.8\%)}

\vspace{1em}

\subsection{Torsion Moment Capacity}

Torsion Moment Capacity Check EN 1992-1-1:2004 art. 6.3.2(5)

The torsional shear stress $\tau_{t,i}$, can be set to be equal to $f_{ctd} = 1.498$ MPa

\textbf{Torsional cracking moment capacity}

$T_{Rd,c} = 2 \cdot A_k \cdot \tau_{t,i} \cdot t_{ef,i} = 2 \cdot 134400 \cdot 1.498 \cdot 120 = 48318922.64$ Nmm

\textbf{Shear resistance without stirrups}

(6.2) $V_{Rd,c} = \max(C_{Rd,c} \cdot k \cdot \left(100 \cdot \rho_l \cdot f_{ck}\right)^{1/3} + k_1 \cdot \sigma_{cp}, v_{min} + k_1 \cdot \sigma_{cp}) \cdot b_w \cdot d$

$= \max(0.12 \cdot 1.61 \cdot \left(100 \cdot 0.00 \cdot 35.00\right)^{1/3} + 0.15 \cdot 0.00, 0.42 + 0.15 \cdot 0.00) \cdot 400.00 \cdot 540.00 = 91247.31 \ \text{N}$

\textbf{Combined utilization check}

(6.31) $\frac{T_{Ed}}{T_{Rd,c}} + \frac{V_{Ed}}{V_{Rd,c}} = 1.242 + 3.584 = 4.825 > 1.0$

\textbf{Result: FAIL (Utilization: 482.5\%)}

\subsection{Maximum Longitudinal Reinforcement}

Maximum longitudinal reinforcement check EN 1992-1-1:2004 art. 9.2.1.1(3)

\textbf{Maximum allowed longitudinal reinforcement area}

$A_{s,max} = 0.04 \cdot A_c = 0.04 \cdot 240000 = 9600$ mm²

\textbf{Provided longitudinal reinforcement area}

$A_s = 1029$ mm²

\textbf{Check:} $A_s \leq A_{s,max}$ $\Longrightarrow$ $1029 \leq 9600$ mm²

\textbf{Result: PASS (Utilization: 10.7\%)}

\subsection{Minimum Tensile Reinforcement}

Minimum tensile reinforcement check EN 1992-1-1:2004 art. 9.2.1.1(1)

\textbf{Minimum tensile reinforcement area}

(9.1) $A_{s,min} = \max \left\{0.26 \cdot \frac{f_{ctm}}{f_{yk}} \cdot b_t \cdot d; 0.0013 \cdot b_t \cdot d\right\}$

$= \max \left\{0.26 \cdot \frac{3.21}{500.00} \cdot 400.00 \cdot 540.00; 0.0013 \cdot 400.00 \cdot 540.00\right\} = 360.54$ mm²

\textbf{Provided tensile reinforcement area (bottom bars)}

$A_{st} = 627$ mm²

\textbf{Check:} $A_{st} \geq A_{st,min}$ $\Longrightarrow$ $627 \geq 361$ mm²

\textbf{Result: PASS (Utilization: 57.5\%)}

\subsection{Maximum Shear Stirrup Spacing}

Maximum shear stirrup spacing check EN 1992-1-1:2004 art. 9.2.2(6)

\textbf{Maximum allowable stirrup spacing for shear}

(9.6) $s_{l,max} = 0.75 \cdot d \cdot \left( 1 + \cot(\alpha) \right) = 0.75 \cdot 540.00 \cdot \left( 1 + \cot(90.00) \right) = 405.00$ mm

\textbf{Actual stirrup spacing (maximum distance between shear stirrups)}

$s = 200$ mm

\textbf{Check:} $s \leq s_{l,max}$ $\Longrightarrow$ $200 \leq 405$ mm

\textbf{Result: PASS (Utilization: 49.4\%)}

\subsection{Maximum Torsion Stirrup Spacing}

Maximum torsion stirrup spacing check EN 1992-1-1:2004 art. 9.2.3(3)

\textbf{Maximum allowable stirrup spacing for shear (formula 9.6)}

$s_{l,max} = 0.75 \cdot d \cdot \left( 1 + \cot(\alpha) \right) = 0.75 \cdot 540.00 \cdot \left( 1 + \cot(90.00) \right) = 405.00$ mm

\textbf{Maximum allowable stirrup spacing for torsion}

$s_{max} = \min\left(\frac{u}{8}, s_{l,max}, b, h\right) = \min\left(\frac{2000}{8}, 405, 400, 600\right) = 250$ mm

\textbf{Actual stirrup spacing (maximum distance between torsion stirrups)}

$s = 75$ mm

\textbf{Check:} $s \leq s_{max}$ $\Longrightarrow$ $75 \leq 250$ mm

\textbf{Result: PASS (Utilization: 30.0\%)}

\subsection{Shear and Torsion Stirrup Area}

Combined shear and torsion stirrup area check EN 1992-1-1:2004

\textbf{Required stirrup area for shear force $V_{Ed}$}

(6.8) $\frac{A_{sw}}{s} = \frac{V_{Ed}}{z \cdot \cot\theta \cdot f_{ywd}} = \frac{327000.0}{486 \cdot \cot 45.0 \cdot 434.8} = 1.5475$ mm²/mm

\textbf{Shear stress in wall due to torsion $T_{Ed}$}

(6.26) $\tau_{t,i}t_{ef,i} = \frac{T_{Ed}}{2 \cdot A_{k}} = \frac{60000000.000}{2 \cdot 134400.000} = 223.214$ N/mm

\textbf{Shear force due to torsion in wall}

(6.27) $V_{Ed,i} = \tau_{t,i} t_{ef,i} \cdot z_{i} = 223.2 \cdot 600.0 = 133928.6$ N

\textbf{Required stirrup area for torsion moment $T_{Ed}$}

(6.8) $\frac{A_{sw,t}}{s} = \frac{V_{Ed,i}}{z_i \cdot \cot\theta \cdot f_{ywd}} = \frac{133928.6}{600 \cdot \cot 45.0 \cdot 434.8} = 0.5134$ mm²/mm

\textbf{Total required stirrup area (shear + 2 $\times$ torsion for closed stirrups)}

$\frac{A_{sw,total}}{s} = \frac{A_{sw,V}}{s} + 2 \cdot \frac{A_{sw,T}}{s} = 1.5475 + 2 \cdot 0.5134 = 2.5743$ mm²/mm

\textbf{Provided stirrup area}

$\frac{A_{sw,prov}}{s} = 2.8798$ mm²/mm

\textbf{Check:} $\frac{A_{sw,prov}}{s} \geq \frac{A_{sw,total}}{s}$ $\Longrightarrow$ $2.8798 \geq 2.5743$ mm²/mm

\textbf{Result: PASS (Utilization: 89.4\%)}

\subsection{Minimum Shear Reinforcement Ratio}

Minimum shear reinforcement ratio check EN 1992-1-1:2004 art. 9.2.2(5)

\textbf{Minimum shear reinforcement ratio}

(9.5N) $\rho_{w,min} = \frac{0.08 \cdot \sqrt{f_{ck}}}{f_{yk}} = \frac{0.08 \cdot \sqrt{35.0000}}{500.0000} = 0.000947$

\textbf{Provided shear reinforcement ratio}

(9.4) $\rho_w = \frac{A_{sw}}{s \cdot b_w \cdot \sin(\alpha)} = \frac{2.8798}{1.0000 \cdot 400.0000 \cdot \sin(90.0000)} = 0.007199$

\textbf{Check:} $\rho_w \geq \rho_{w,min}$ $\Longrightarrow$ $0.0072 \geq 0.0009$

\textbf{Result: PASS (Utilization: 13.1\%)}

\section{Additional Reinforcement Requirements}

\textbf{Shear:}

Additional longitudinal reinforcement due to shear: $A_{sl,shear} = 376.05$ mm²

\textbf{Torsion:}

Additional longitudinal reinforcement due to torsion: $A_{sl,torsion} = 780.36$ mm²

(To be distributed along beam edges)

\section{Conclusions}

\textbf{Warning:} One or more checks have failed.

\textbf{Note:} Torsion moment capacity is insufficient with minimum reinforcement. Additional torsion-specific reinforcement is required as specified above.

\end{document}

[egarciamendez]

Working version of the Latex report Blueprints-3.pdf

\documentclass{article}
\usepackage{amsmath}
\usepackage{booktabs}
\usepackage{geometry}
\geometry{a4paper, margin=1in}

% Increase spacing throughout document
\usepackage{setspace}
\setstretch{1.3}  % Increase line spacing by 30%

% Add space between paragraphs
\setlength{\parskip}{0.5em}

% Add space around equations
\setlength{\abovedisplayskip}{12pt}
\setlength{\belowdisplayskip}{12pt}

\begin{document}

\title{Torsion Check Results}
\date{}
\maketitle

\section{Utilization Summary}

\begin{table}[h]
\centering
\begin{tabular}{lcc}
\toprule
Check & Utilization & Status \\
\midrule
Concrete strut capacity & 0.588 & PASS \\
Torsion moment capacity & 4.825 & FAIL \\
Max longitudinal reinforcement & 0.107 & PASS \\
Min tensile reinforcement & 0.575 & PASS \\
Max shear stirrup spacing & 0.494 & PASS \\
Max torsion stirrup spacing & 0.300 & PASS \\
Shear and torsion stirrup area & 0.894 & PASS \\
Min shear reinforcement ratio & 0.131 & PASS \\
\bottomrule
\end{tabular}
\end{table}

\textbf{Overall Result: FAIL}\\[0.5em]

\section{Individual Checks}

\subsection{Concrete Strut Capacity}

Concrete strut capacity check EN 1992-1-1:2004 art. 6.3.2(4)

\vspace{0.3em}
\textbf{Maximum shear resistance}

(6.9) 
\[
V_{Rd,max} = \alpha_{cw} \cdot b_{w} \cdot z \cdot \nu_{1} \cdot \frac{f_{cd}}{\cot(\theta) + \tan(\theta)} = 1.00 \cdot 400.00 \cdot 486.00 \cdot 0.52 \cdot \frac{23.33}{\cot(45.00) + \tan(45.00)} = 1170288.00 \ \text{N}
\]

\vspace{0.3em}
\textbf{Design torsional resistance moment}

(6.30) 
\[
T_{Rd,max} = 2 \cdot \nu \cdot \alpha_{cw} \cdot f_{cd} \cdot A_{k} \cdot t_{ef,i} \cdot \sin(\theta) \cdot \cos(\theta) = 2 \cdot 0.52 \cdot 1.00 \cdot 23.33 \cdot 134400.00 \cdot 120.00 \cdot \sin(45.00) \cdot \cos(45.00) = 194181120.00 \ \text{Nmm}
\]

\vspace{0.3em}
\textbf{Combined interaction check}

(6.29) 
\[
\frac{T_{Ed}}{T_{Rd,max}} + \frac{V_{Ed}}{V_{Rd,max}} = 0.309 + 0.279 = 0.588 \leq 1.0
\]

\vspace{0.5em}
\textbf{Result: PASS (Utilization: 58.8\%)}

\vspace{1em}

\subsection{Torsion Moment Capacity}

Torsion Moment Capacity Check EN 1992-1-1:2004 art. 6.3.2(5)

The torsional shear stress $\tau_{t,i}$, can be set to be equal to $f_{ctd} = 1.498$ MPa

\textbf{Torsional cracking moment capacity}

$T_{Rd,c} = 2 \cdot A_k \cdot \tau_{t,i} \cdot t_{ef,i} = 2 \cdot 134400 \cdot 1.498 \cdot 120 = 48318922.64$ Nmm

\textbf{Shear resistance without stirrups}

(6.2) $V_{Rd,c} = \max(C_{Rd,c} \cdot k \cdot \left(100 \cdot \rho_l \cdot f_{ck}\right)^{1/3} + k_1 \cdot \sigma_{cp}, v_{min} + k_1 \cdot \sigma_{cp}) \cdot b_w \cdot d$

$= \max(0.12 \cdot 1.61 \cdot \left(100 \cdot 0.00 \cdot 35.00\right)^{1/3} + 0.15 \cdot 0.00, 0.42 + 0.15 \cdot 0.00) \cdot 400.00 \cdot 540.00 = 91247.31 \ \text{N}$

\textbf{Combined utilization check}

(6.31) $\frac{T_{Ed}}{T_{Rd,c}} + \frac{V_{Ed}}{V_{Rd,c}} = 1.242 + 3.584 = 4.825 > 1.0$

\textbf{Result: FAIL (Utilization: 482.5\%)}

\subsection{Maximum Longitudinal Reinforcement}

Maximum longitudinal reinforcement check EN 1992-1-1:2004 art. 9.2.1.1(3)

\textbf{Maximum allowed longitudinal reinforcement area}

$A_{s,max} = 0.04 \cdot A_c = 0.04 \cdot 240000 = 9600$ mm²

\textbf{Provided longitudinal reinforcement area}

$A_s = 1029$ mm²

\textbf{Check:} $A_s \leq A_{s,max}$ $\Longrightarrow$ $1029 \leq 9600$ mm²

\textbf{Result: PASS (Utilization: 10.7\%)}

\subsection{Minimum Tensile Reinforcement}

Minimum tensile reinforcement check EN 1992-1-1:2004 art. 9.2.1.1(1)

\textbf{Minimum tensile reinforcement area}

(9.1) $A_{s,min} = \max \left\{0.26 \cdot \frac{f_{ctm}}{f_{yk}} \cdot b_t \cdot d; 0.0013 \cdot b_t \cdot d\right\}$

$= \max \left\{0.26 \cdot \frac{3.21}{500.00} \cdot 400.00 \cdot 540.00; 0.0013 \cdot 400.00 \cdot 540.00\right\} = 360.54$ mm²

\textbf{Provided tensile reinforcement area (bottom bars)}

$A_{st} = 627$ mm²

\textbf{Check:} $A_{st} \geq A_{st,min}$ $\Longrightarrow$ $627 \geq 361$ mm²

\textbf{Result: PASS (Utilization: 57.5\%)}

\subsection{Maximum Shear Stirrup Spacing}

Maximum shear stirrup spacing check EN 1992-1-1:2004 art. 9.2.2(6)

\textbf{Maximum allowable stirrup spacing for shear}

(9.6) $s_{l,max} = 0.75 \cdot d \cdot \left( 1 + \cot(\alpha) \right) = 0.75 \cdot 540.00 \cdot \left( 1 + \cot(90.00) \right) = 405.00$ mm

\textbf{Actual stirrup spacing (maximum distance between shear stirrups)}

$s = 200$ mm

\textbf{Check:} $s \leq s_{l,max}$ $\Longrightarrow$ $200 \leq 405$ mm

\textbf{Result: PASS (Utilization: 49.4\%)}

\subsection{Maximum Torsion Stirrup Spacing}

Maximum torsion stirrup spacing check EN 1992-1-1:2004 art. 9.2.3(3)

\textbf{Maximum allowable stirrup spacing for shear (formula 9.6)}

$s_{l,max} = 0.75 \cdot d \cdot \left( 1 + \cot(\alpha) \right) = 0.75 \cdot 540.00 \cdot \left( 1 + \cot(90.00) \right) = 405.00$ mm

\textbf{Maximum allowable stirrup spacing for torsion}

$s_{max} = \min\left(\frac{u}{8}, s_{l,max}, b, h\right) = \min\left(\frac{2000}{8}, 405, 400, 600\right) = 250$ mm

\textbf{Actual stirrup spacing (maximum distance between torsion stirrups)}

$s = 75$ mm

\textbf{Check:} $s \leq s_{max}$ $\Longrightarrow$ $75 \leq 250$ mm

\textbf{Result: PASS (Utilization: 30.0\%)}

\subsection{Shear and Torsion Stirrup Area}

Combined shear and torsion stirrup area check EN 1992-1-1:2004

\textbf{Required stirrup area for shear force $V_{Ed}$}

(6.8) $\frac{A_{sw}}{s} = \frac{V_{Ed}}{z \cdot \cot\theta \cdot f_{ywd}} = \frac{327000.0}{486 \cdot \cot 45.0 \cdot 434.8} = 1.5475$ mm²/mm

\textbf{Shear stress in wall due to torsion $T_{Ed}$}

(6.26) $\tau_{t,i}t_{ef,i} = \frac{T_{Ed}}{2 \cdot A_{k}} = \frac{60000000.000}{2 \cdot 134400.000} = 223.214$ N/mm

\textbf{Shear force due to torsion in wall}

(6.27) $V_{Ed,i} = \tau_{t,i} t_{ef,i} \cdot z_{i} = 223.2 \cdot 600.0 = 133928.6$ N

\textbf{Required stirrup area for torsion moment $T_{Ed}$}

(6.8) $\frac{A_{sw,t}}{s} = \frac{V_{Ed,i}}{z_i \cdot \cot\theta \cdot f_{ywd}} = \frac{133928.6}{600 \cdot \cot 45.0 \cdot 434.8} = 0.5134$ mm²/mm

\textbf{Total required stirrup area (shear + 2 $\times$ torsion for closed stirrups)}

$\frac{A_{sw,total}}{s} = \frac{A_{sw,V}}{s} + 2 \cdot \frac{A_{sw,T}}{s} = 1.5475 + 2 \cdot 0.5134 = 2.5743$ mm²/mm

\textbf{Provided stirrup area}

$\frac{A_{sw,prov}}{s} = 2.8798$ mm²/mm

\textbf{Check:} $\frac{A_{sw,prov}}{s} \geq \frac{A_{sw,total}}{s}$ $\Longrightarrow$ $2.8798 \geq 2.5743$ mm²/mm

\textbf{Result: PASS (Utilization: 89.4\%)}

\subsection{Minimum Shear Reinforcement Ratio}

Minimum shear reinforcement ratio check EN 1992-1-1:2004 art. 9.2.2(5)

\textbf{Minimum shear reinforcement ratio}

(9.5N) $\rho_{w,min} = \frac{0.08 \cdot \sqrt{f_{ck}}}{f_{yk}} = \frac{0.08 \cdot \sqrt{35.0000}}{500.0000} = 0.000947$

\textbf{Provided shear reinforcement ratio}

(9.4) $\rho_w = \frac{A_{sw}}{s \cdot b_w \cdot \sin(\alpha)} = \frac{2.8798}{1.0000 \cdot 400.0000 \cdot \sin(90.0000)} = 0.007199$

\textbf{Check:} $\rho_w \geq \rho_{w,min}$ $\Longrightarrow$ $0.0072 \geq 0.0009$

\textbf{Result: PASS (Utilization: 13.1\%)}

\section{Additional Reinforcement Requirements}

\textbf{Shear:}

Additional longitudinal reinforcement due to shear: $A_{sl,shear} = 376.05$ mm²

\textbf{Torsion:}

Additional longitudinal reinforcement due to torsion: $A_{sl,torsion} = 780.36$ mm²

(To be distributed along beam edges)

\section{Conclusions}

\textbf{Warning:} One or more checks have failed.

\textbf{Note:} Torsion moment capacity is insufficient with minimum reinforcement. Additional torsion-specific reinforcement is required as specified above.

\end{document}

Hi @teknovizier, it's been a while since we last spoke. Have you had time to check this working version of the Torsion check report? I would love to sharpen the rough edges of this PR and hopefully merge somewhere before the holydays.

[teknovizier]

@egarciamendez, sorry, I've been very busy with other tasks recently.

[GerjanDorgelo]

Suggestion for equations: personally im a bigger fan of starting and ending equations like this

\begin{equation} V_{Rd,max} = \alpha_{cw} \cdot b_{w} \cdot z \cdot \nu_{1} \cdot \frac{f_{cd}}{\cot(\theta) + \tan(\theta)} = 1.00 \cdot 400.00 \cdot 486.00 \cdot 0.52 \cdot \frac{23.33}{\cot(45.00) + \tan(45.00)} = 1170288.00 \ \text{N} \tag{6.9} \end{equation}

Makes it easier to use a translate or to_word function.

[GerjanDorgelo]

Suggestion for text: Id love to have all text wrapped in a \text{..} or \textbf{..}. that way we can more easilly do the translation functions. This also requires an explicit \nextline, which makes sure all LaTeX processors handle the next line in a same way.

[GerjanDorgelo]

(a shorter version of) the above would then look like:

\documentclass{article}
\usepackage{amsmath}
\usepackage{booktabs}
\usepackage{geometry}
\geometry{a4paper, margin=1in}

% Increase spacing throughout document
\usepackage{setspace}
\setstretch{1.3}  % Increase line spacing by 30%

% Add space between paragraphs
\setlength{\parskip}{0.5em}

% Add space around equations
\setlength{\abovedisplayskip}{12pt}
\setlength{\belowdisplayskip}{12pt}

\begin{document}

\title{Torsion Check Results}
\date{}
\maketitle

\section{Utilization Summary}

\begin{table}[h]
\centering
\begin{tabular}{lcc}
\toprule
Check & Utilization & Status \\
\midrule
\text{Concrete strut capacity} & 0.588 & \text{PASS} \\
\text{Torsion moment capacity} & 4.825 & \text{FAIL} \\
\text{Max longitudinal reinforcement} & 0.107 & \text{PASS} \\
\text{Min tensile reinforcement} & 0.575 & \text{PASS} \\
\text{Max shear stirrup spacing} & 0.494 & \text{PASS} \\
\text{Max torsion stirrup spacing} & 0.300 & \text{PASS} \\
\text{Shear and torsion stirrup area} & 0.894 & \text{PASS} \\
\text{Min shear reinforcement ratio} & 0.131 & \text{PASS} \\
\bottomrule
\end{tabular}
\end{table}

\textbf{Overall Result: FAIL}
\section{Individual Checks}
\subsection{Concrete Strut Capacity}
\text{Concrete strut capacity check EN 1992-1-1:2004 art. 6.3.2(4)} \newline \newline
\textbf{Maximum shear resistance}
\begin{equation} V_{Rd,max} = \alpha_{cw} \cdot b_{w} \cdot z \cdot \nu_{1} \cdot \frac{f_{cd}}{\cot(\theta) + \tan(\theta)} = 1.00 \cdot 400.00 \cdot 486.00 \cdot 0.52 \cdot \frac{23.33}{\cot(45.00) + \tan(45.00)} = 1170288.00 \ \text{N} \tag{6.9} \end{equation}
\textbf{Design torsional resistance moment}
\begin{equation} T_{Rd,max} = 2 \cdot \nu \cdot \alpha_{cw} \cdot f_{cd} \cdot A_{k} \cdot t_{ef,i} \cdot \sin(\theta) \cdot \cos(\theta) = 2 \cdot 0.52 \cdot 1.00 \cdot 23.33 \cdot 134400.00 \cdot 120.00 \cdot \sin(45.00) \cdot \cos(45.00) = 194181120.00 \ \text{Nmm} \tag{6.30} \end{equation}
\textbf{Combined interaction check}
\begin{equation} \frac{T_{Ed}}{T_{Rd,max}} + \frac{V_{Ed}}{V_{Rd,max}} = 0.309 + 0.279 = 0.588 \leq 1.0 \tag{6.29} \end{equation}
\textbf{Result: PASS (Utilization: 58.8\%)}

\section{Conclusions}

\textbf{Warning:} One or more checks have failed.

\textbf{Note:} Torsion moment capacity is insufficient with minimum reinforcement. Additional torsion-specific reinforcement is required as specified above.

\end{document}

[egarciamendez]

(a shorter version of) the above would then look like:

\documentclass{article}
\usepackage{amsmath}
\usepackage{booktabs}
\usepackage{geometry}
\geometry{a4paper, margin=1in}

% Increase spacing throughout document
\usepackage{setspace}
\setstretch{1.3}  % Increase line spacing by 30%

% Add space between paragraphs
\setlength{\parskip}{0.5em}

% Add space around equations
\setlength{\abovedisplayskip}{12pt}
\setlength{\belowdisplayskip}{12pt}

\begin{document}

\title{Torsion Check Results}
\date{}
\maketitle

\section{Utilization Summary}

\begin{table}[h]
\centering
\begin{tabular}{lcc}
\toprule
Check & Utilization & Status \\
\midrule
\text{Concrete strut capacity} & 0.588 & \text{PASS} \\
\text{Torsion moment capacity} & 4.825 & \text{FAIL} \\
\text{Max longitudinal reinforcement} & 0.107 & \text{PASS} \\
\text{Min tensile reinforcement} & 0.575 & \text{PASS} \\
\text{Max shear stirrup spacing} & 0.494 & \text{PASS} \\
\text{Max torsion stirrup spacing} & 0.300 & \text{PASS} \\
\text{Shear and torsion stirrup area} & 0.894 & \text{PASS} \\
\text{Min shear reinforcement ratio} & 0.131 & \text{PASS} \\
\bottomrule
\end{tabular}
\end{table}

\textbf{Overall Result: FAIL}
\section{Individual Checks}
\subsection{Concrete Strut Capacity}
\text{Concrete strut capacity check EN 1992-1-1:2004 art. 6.3.2(4)} \newline \newline
\textbf{Maximum shear resistance}
\begin{equation} V_{Rd,max} = \alpha_{cw} \cdot b_{w} \cdot z \cdot \nu_{1} \cdot \frac{f_{cd}}{\cot(\theta) + \tan(\theta)} = 1.00 \cdot 400.00 \cdot 486.00 \cdot 0.52 \cdot \frac{23.33}{\cot(45.00) + \tan(45.00)} = 1170288.00 \ \text{N} \tag{6.9} \end{equation}
\textbf{Design torsional resistance moment}
\begin{equation} T_{Rd,max} = 2 \cdot \nu \cdot \alpha_{cw} \cdot f_{cd} \cdot A_{k} \cdot t_{ef,i} \cdot \sin(\theta) \cdot \cos(\theta) = 2 \cdot 0.52 \cdot 1.00 \cdot 23.33 \cdot 134400.00 \cdot 120.00 \cdot \sin(45.00) \cdot \cos(45.00) = 194181120.00 \ \text{Nmm} \tag{6.30} \end{equation}
\textbf{Combined interaction check}
\begin{equation} \frac{T_{Ed}}{T_{Rd,max}} + \frac{V_{Ed}}{V_{Rd,max}} = 0.309 + 0.279 = 0.588 \leq 1.0 \tag{6.29} \end{equation}
\textbf{Result: PASS (Utilization: 58.8\%)}

\section{Conclusions}

\textbf{Warning:} One or more checks have failed.

\textbf{Note:} Torsion moment capacity is insufficient with minimum reinforcement. Additional torsion-specific reinforcement is required as specified above.

\end{document}

Thanks! I will make sure to implement this :-)

[GerjanDorgelo]

(a shorter version of) the above would then look like:

\documentclass{article}
\usepackage{amsmath}
\usepackage{booktabs}
\usepackage{geometry}
\geometry{a4paper, margin=1in}

% Increase spacing throughout document
\usepackage{setspace}
\setstretch{1.3}  % Increase line spacing by 30%

% Add space between paragraphs
\setlength{\parskip}{0.5em}

% Add space around equations
\setlength{\abovedisplayskip}{12pt}
\setlength{\belowdisplayskip}{12pt}

\begin{document}

\title{Torsion Check Results}
\date{}
\maketitle

\section{Utilization Summary}

\begin{table}[h]
\centering
\begin{tabular}{lcc}
\toprule
Check & Utilization & Status \\
\midrule
\text{Concrete strut capacity} & 0.588 & \text{PASS} \\
\text{Torsion moment capacity} & 4.825 & \text{FAIL} \\
\text{Max longitudinal reinforcement} & 0.107 & \text{PASS} \\
\text{Min tensile reinforcement} & 0.575 & \text{PASS} \\
\text{Max shear stirrup spacing} & 0.494 & \text{PASS} \\
\text{Max torsion stirrup spacing} & 0.300 & \text{PASS} \\
\text{Shear and torsion stirrup area} & 0.894 & \text{PASS} \\
\text{Min shear reinforcement ratio} & 0.131 & \text{PASS} \\
\bottomrule
\end{tabular}
\end{table}

\textbf{Overall Result: FAIL}
\section{Individual Checks}
\subsection{Concrete Strut Capacity}
\text{Concrete strut capacity check EN 1992-1-1:2004 art. 6.3.2(4)} \newline \newline
\textbf{Maximum shear resistance}
\begin{equation} V_{Rd,max} = \alpha_{cw} \cdot b_{w} \cdot z \cdot \nu_{1} \cdot \frac{f_{cd}}{\cot(\theta) + \tan(\theta)} = 1.00 \cdot 400.00 \cdot 486.00 \cdot 0.52 \cdot \frac{23.33}{\cot(45.00) + \tan(45.00)} = 1170288.00 \ \text{N} \tag{6.9} \end{equation}
\textbf{Design torsional resistance moment}
\begin{equation} T_{Rd,max} = 2 \cdot \nu \cdot \alpha_{cw} \cdot f_{cd} \cdot A_{k} \cdot t_{ef,i} \cdot \sin(\theta) \cdot \cos(\theta) = 2 \cdot 0.52 \cdot 1.00 \cdot 23.33 \cdot 134400.00 \cdot 120.00 \cdot \sin(45.00) \cdot \cos(45.00) = 194181120.00 \ \text{Nmm} \tag{6.30} \end{equation}
\textbf{Combined interaction check}
\begin{equation} \frac{T_{Ed}}{T_{Rd,max}} + \frac{V_{Ed}}{V_{Rd,max}} = 0.309 + 0.279 = 0.588 \leq 1.0 \tag{6.29} \end{equation}
\textbf{Result: PASS (Utilization: 58.8\%)}

\section{Conclusions}

\textbf{Warning:} One or more checks have failed.

\textbf{Note:} Torsion moment capacity is insufficient with minimum reinforcement. Additional torsion-specific reinforcement is required as specified above.

\end{document}

Thanks! I will make sure to implement this :-)

yay. i spend a while today making sure the to_docx also is able to handle all that stuff and ill use exactly that as testcase
test_complex_document.docx

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