Second moment of area

More generally, the product moment of area is defined as[3]

$${\displaystyle I_{xy}=\iint _{R}yx\,dx\,dy}$$

Rectangle with base b and height h

Consider a rectangle with base ${\displaystyle b}$ and height ${\displaystyle h}$ whose centroid is located at the origin. ${\displaystyle I_{x}}$ represents the second moment of area with respect to the x-axis; ${\displaystyle I_{y}}$ represents the second moment of area with respect to the y-axis; ${\displaystyle J_{z}}$ represents the polar moment of inertia with respect to the z-axis.

$${\displaystyle {\begin{aligned}I_{x}&=\iint _{R}y^{2}\,dA=\int _{-{\frac {b}{2}}}^{\frac {b}{2}}\int _{-{\frac {h}{2}}}^{\frac {h}{2}}y^{2}\,dy\,dx=\int _{-{\frac {b}{2}}}^{\frac {b}{2}}{\frac {1}{3}}{\frac {h^{3}}{4}}\,dx={\frac {bh^{3}}{12}}\\I_{y}&=\iint _{R}x^{2}\,dA=\int _{-{\frac {b}{2}}}^{\frac {b}{2}}\int _{-{\frac {h}{2}}}^{\frac {h}{2}}x^{2}\,dy\,dx=\int _{-{\frac {b}{2}}}^{\frac {b}{2}}hx^{2}\,dx={\frac {b^{3}h}{12}}\end{aligned}}}$$

Using the perpendicular axis theorem we get the value of ${\displaystyle J_{z}}$.

$${\displaystyle J_{z}=I_{x}+I_{y}={\frac {bh^{3}}{12}}+{\frac {hb^{3}}{12}}={\frac {bh}{12}}\left(b^{2}+h^{2}\right)}$$

Annulus with inner radius r₁ and outer radius r₂

Consider an annulus whose center is at the origin, outside radius is ${\displaystyle r_{2}}$, and inside radius is ${\displaystyle r_{1}}$. Because of the symmetry of the annulus, the centroid also lies at the origin. We can determine the polar moment of inertia, ${\displaystyle J_{z}}$, about the ${\displaystyle z}$ axis by the method of composite shapes. This polar moment of inertia is equivalent to the polar moment of inertia of a circle with radius ${\displaystyle r_{2}}$ minus the polar moment of inertia of a circle with radius ${\displaystyle r_{1}}$, both centered at the origin. First, let us derive the polar moment of inertia of a circle with radius ${\displaystyle r}$ with respect to the origin. In this case, it is easier to directly calculate ${\displaystyle J_{z}}$ as we already have ${\displaystyle r^{2}}$, which has both an ${\displaystyle x}$ and ${\displaystyle y}$ component. Instead of obtaining the second moment of area from Cartesian coordinates as done in the previous section, we shall calculate ${\displaystyle I_{x}}$ and ${\displaystyle J_{z}}$ directly using polar coordinates.

$${\displaystyle {\begin{aligned}I_{x,{\text{circle}}}&=\iint _{R}y^{2}\,dA=\iint _{R}\left(r\sin {\theta }\right)^{2}\,dA=\int _{0}^{2\pi }\int _{0}^{r}\left(r\sin {\theta }\right)^{2}\left(r\,dr\,d\theta \right)\\&=\int _{0}^{2\pi }\int _{0}^{r}r^{3}\sin ^{2}{\theta }\,dr\,d\theta =\int _{0}^{2\pi }{\frac {r^{4}\sin ^{2}{\theta }}{4}}\,d\theta ={\frac {\pi }{4}}r^{4}\\J_{z,{\text{circle}}}&=\iint _{R}r^{2}\,dA=\int _{0}^{2\pi }\int _{0}^{r}r^{2}\left(r\,dr\,d\theta \right)=\int _{0}^{2\pi }\int _{0}^{r}r^{3}\,dr\,d\theta \\&=\int _{0}^{2\pi }{\frac {r^{4}}{4}}\,d\theta ={\frac {\pi }{2}}r^{4}\end{aligned}}}$$

Now, the polar moment of inertia about the ${\displaystyle z}$ axis for an annulus is simply, as stated above, the difference of the second moments of area of a circle with radius ${\displaystyle r_{2}}$ and a circle with radius ${\displaystyle r_{1}}$.

$${\displaystyle J_{z}=J_{z,r_{2}}-J_{z,r_{1}}={\frac {\pi }{2}}r_{2}^{4}-{\frac {\pi }{2}}r_{1}^{4}={\frac {\pi }{2}}\left(r_{2}^{4}-r_{1}^{4}\right)}$$

Alternatively, we could change the limits on the ${\displaystyle dr}$ integral the first time around to reflect the fact that there is a hole. This would be done like this.

$${\displaystyle {\begin{aligned}J_{z}&=\iint _{R}r^{2}\,dA=\int _{0}^{2\pi }\int _{r_{1}}^{r_{2}}r^{2}\left(r\,dr\,d\theta \right)=\int _{0}^{2\pi }\int _{r_{1}}^{r_{2}}r^{3}\,dr\,d\theta \\&=\int _{0}^{2\pi }\left[{\frac {r_{2}^{4}}{4}}-{\frac {r_{1}^{4}}{4}}\right]\,d\theta ={\frac {\pi }{2}}\left(r_{2}^{4}-r_{1}^{4}\right)\end{aligned}}}$$

A simple polygon. Here, ${\displaystyle n=6}$, notice point "7" is identical to point 1.

The second moment of area about the origin for any simple polygon on the XY-plane can be computed in general by summing contributions from each segment of the polygon after dividing the area into a set of triangles. This formula is related to the shoelace formula and can be considered a special case of Green's theorem.

A polygon is assumed to have ${\displaystyle n}$ vertices, numbered in counter-clockwise fashion. If polygon vertices are numbered clockwise, returned values will be negative, but absolute values will be correct.

$${\displaystyle {\begin{aligned}I_{y}&={\frac {1}{12}}\sum _{i=1}^{n}\left(x_{i}y_{i+1}-x_{i+1}y_{i}\right)\left(x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}\right)\\I_{x}&={\frac {1}{12}}\sum _{i=1}^{n}\left(x_{i}y_{i+1}-x_{i+1}y_{i}\right)\left(y_{i}^{2}+y_{i}y_{i+1}+y_{i+1}^{2}\right)\\I_{xy}&={\frac {1}{24}}\sum _{i=1}^{n}\left(x_{i}y_{i+1}-x_{i+1}y_{i}\right)\left(x_{i}y_{i+1}+2x_{i}y_{i}+2x_{i+1}y_{i+1}+x_{i+1}y_{i}\right)\end{aligned}}}$$

where ${\displaystyle x_{i},y_{i}}$ are the coordinates of the ${\displaystyle i}$-th polygon vertex, for ${\displaystyle 1\leq i\leq n}$. Also, ${\displaystyle x_{n+1},y_{n+1}}$ are assumed to be equal to the coordinates of the first vertex, i.e., ${\displaystyle x_{n+1}=x_{1}}$ and ${\displaystyle y_{n+1}=y_{1}}$.[6][7] [8] [9]