Dissemination of IT for the Promotion of Materials Science (DoITPoMS)

# Calculating I for simple shapes

The second moment of area for a rectangular section beam of width w and thickness h is given by

$I = \int\limits_0^{h/2} {{y^2}dA} = 2\int\limits_0^{h/2} {{y^2}w {\rm{d}}y} = 2w\left[ {\frac{{{y^3}}}{3}} \right]_0^{h/2} = \frac{{w{h^3}}}{{12}}$

The corresponding operation for a circular cross-section of diameter D gives

$\begin{array}{l} I = \int\limits_A^{} {{y^2}{\rm{d}}A} = \int\limits_{\theta = 0}^{2\pi } {\int\limits_{r = 0}^{D/2} {{{\left( {r\sin \theta } \right)}^2}} } \;\left[ {\left( {r{\rm{d}}\theta } \right){\rm{d}}r} \right] = \int\limits_{\theta = 0}^{2\pi } {\left\{ {\int\limits_{r = 0}^{D/2} {{r^3}{\rm{d}}r} } \right\}} {\sin ^2}\theta {\rm{d}}\theta \\ = \frac{{{D^4}}}{{64}}\int\limits_{\theta = 0}^{2\pi } {{{\sin }^2}\theta {\rm{d}}\theta } = \frac{{{D^4}}}{{64}}\int\limits_{\theta = 0}^{2\pi } {\left( {\frac{{1 - \cos 2\theta }}{2}} \right){\rm{d}}\theta } = \frac{{{D^4}}}{{64}}\left[ {\frac{\theta }{2} - \frac{{\sin 2\theta }}{4}} \right]_0^{2\pi } = \frac{{\pi {D^4}}}{{64}} \end{array}$