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Added calculations of moment of inertia of arbitrary sub-triangle

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Karma Riuk 2023-06-07 14:20:22 +02:00
parent c6f845dc83
commit 3de6a66721
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sections/appendix.tex
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@ -46,11 +46,19 @@ it will be useful to simplify the result of the integral.
\end{equation}
\newpage
\paragraph{Moment of inertia of sub-triangle of arbitrary polygon}
\paragraph{Moment of inertia of sub-triangle of arbitrary polygon} Recall
equation \ref{eq:r} defines
$$ \vec r = \alpha \vv{CA} + \beta \alpha \vv{AB} $$
\begin{equation}
\label{eq:subtriangle_arbitrary_moment_long}
\begin{split}
I_{T_i} = \rho \int_0^1 \int_0^1 \vec r^2 hb \alpha \diff \alpha \diff \beta\\
&=
I_{T_i} &= \rho \int_0^1 \int_0^1 \vec r^2 hb \alpha \diff \alpha \diff \beta\\
&=\rho hb\int_0^1 \int_0^1 \left(\alpha \vv{CA} + \beta \alpha \vv{AB}\right)^2 \alpha \diff \alpha \diff \beta\\
&=\rho hb\int_0^1 \int_0^1 \left(\alpha^2 \vv{CA}^2 + 2 \alpha^2 \beta \vv{AB} \cdot \vv{CA} + \alpha^2 \beta^2 \vv{AB}^2\right) \alpha \diff \alpha \diff \beta\\
&=\rho hb\int_0^1 \int_0^1 \alpha^3 \left(\vv{CA}^2 + 2 \beta \vv{AB} \cdot \vv{CA} + \beta^2 \vv{AB}^2\right) \diff \alpha \diff \beta\\
&=\rho hb\int_0^1 \Biggl[\frac{1}{4} \alpha^4 \left(\vv{CA}^2 + 2 \beta \vv{AB} \cdot \vv{CA} + \beta^2 \vv{AB}^2\right) \Biggr]_0^1\diff \beta\\
&= \frac{\rho hb}{4}\int_0^1 \beta^2 \vv{AB}^2 + 2 \beta \vv{AB} \cdot \vv{CA} + \vv{CA}^2 \diff \beta\\
&= \frac{\rho hb}{4} \Biggl[\frac{1}{3} \beta^3 \vv{AB}^2 + \beta^2 \vv{AB} \cdot \vv{CA} + \beta\vv{CA}^2 \Biggr]_0^1\\
&= \frac{\rho hb}{4} \left(\frac{1}{3}\vv{AB}^2 + \vv{AB} \cdot \vv{CA} + \vv{CA}^2\right) \\
\end{split}
\end{equation}