User:Eas4200c.f08.aero6.villalba/circle second moment of area

Derivation of second moment of area of a circle


Given a circle with $$r=a\,\!$$ as shown, we start from the definition of second moment area:

$$I_y=\iint\limits_A \,z^2 dy\,dz$$

which can also be written as,

$$I_y=\int\,z^2\,dA$$

For a circular cross section, a transformation to polar coordinates is needed. Therefore,

$$z=r\sin \theta\,\!$$

$$y=r\cos \theta\,\!$$

Next, we look at our differential area in polar coordinates. This is obtained from the area of a sector.

The area of the sector is $$A=\frac{1}{2} r^2 d\theta$$ and differentiating we obtain $$dA=r dr d\theta\,\!$$, where this differential area is shown below.



Thus, our integral becomes

$$I_y=\int\limits_0^{2\pi}\int\limits_0^a (r\sin \theta)^2 r dr d\theta$$

Next, we solve the integral

$$I_y=\int\limits_0^{2\pi}\int\limits_0^a r^3 \sin^2 \theta dr d\theta = \int\limits_0^{2\pi}\sin^2 \theta \left [ \int\limits_0^a r^3 dr \right ] d\theta = \int\limits_0^{2\pi}\sin^2 \theta \left [ \frac{r^4}{4} \right ]_0^a d\theta$$

$$= \frac {a^4}{4} \int\limits_0^{2\pi}\sin^2 \theta d\theta = \frac {a^4}{4} \int\limits_0^{2\pi} \frac{1-\cos (2\theta)}{2} d\theta = \frac {a^4}{4} \int\limits_0^{2\pi} \frac{1}{2} - \frac{1}{2} \cos(2\theta) d\theta$$

$$= \frac {a^4}{4} \left [ \frac{1}{2}\theta - \frac{1}{4} \sin(2\theta) \right ]_0^{2\pi} = \frac {a^4}{4} \left [ \frac{1}{2}(2\pi) - \frac{1}{4}\sin(4\pi) - \frac{1}{2}(0) + \frac{1}{4}\sin(0) \right ]$$

$$= \frac {a^4}{4} \left ( \pi - 0 - 0 + 0 \right ) = \frac{\pi a^4}{4}$$

Therefore, for any circular cross-section with a radius of $$r\,\!$$, the second moment of area about the y-axis, $$I_y\,\!$$ is equals to $$\frac{\pi r^4}{4}$$.

Note: Through the same means and a similar derivation, it can also be shown the same results would be obtained for $$I_z\,\!$$.