# Second moment of area

(Redirected from Polar moment)
This article is about the geometrical property of an area, termed the second moment of area. For the moment of inertia dealing with the rotation of an object with mass, see mass moment of inertia.
For a list, see list of second moments of area.

The second moment of area, also known as moment of inertia of plane area, area moment of inertia, or second area moment, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. The second moment of area is typically denoted with either an ${\displaystyle I}$ for an axis that lies in the plane or with a ${\displaystyle J}$ for an axis perpendicular to the plane. In both cases, it is calculated with a multiple integral over the object in question. Its unit of dimension is length to fourth power, L4.

In the field of structural engineering, the second moment of area of the cross-section of a beam is an important property used in the calculation of the beam's deflection and the calculation of stress caused by a moment applied to the beam.

Note: Different disciplines use "Moment of Inertia" (MOI) to refer to either or both of the planar second moment of area, ${\displaystyle I=\textstyle \iint _{R}x^{2}\,\mathrm {d} A}$, where x is the distance to some reference plane, or the polar second moment of area, ${\displaystyle I=\textstyle \iint _{R}r^{2}\,\mathrm {d} A}$, where r is the distance to some reference axis. In each case the integral is over all the infinitesimal elements of area, dA, in some two-dimensional cross-section. In math and physics, "Moment of Inertia" is strictly the second moment of mass with respect to distance from an axis: ${\displaystyle I=\textstyle \int _{Q}r^{2}\mathrm {d} m}$, where r is the distance to some potential rotation axis, and the integral is over all the infinitesimal elements of mass, dm, in a three-dimensional space occupied by an object Q. The MOI, in this sense, is the analog of mass for rotational problems. In engineering (especially mechanical and civil), "Moment of Inertia" commonly refers to the second moment of the area.[1]

## Definition

An arbitrary shape. ρ is the radial distance to the element dA, with projections x and y on the axes.

The second moment of area for an arbitrary shape R with respect to an arbitrary axis ${\displaystyle BB'}$ is defined as

${\displaystyle J_{BB'}=\iint \limits _{R}{\rho }^{2}\,\mathrm {d} A}$

where

${\displaystyle \mathrm {d} A}$ is the differential area of the arbitrary shape, and
${\displaystyle \rho }$ is the distance from the axis ${\displaystyle BB'}$ to ${\displaystyle \mathrm {d} A}$.[2]

For example, when the desired reference axis is the x-axis the second moment of area, ${\displaystyle I_{xx}}$ (often denoted as ${\displaystyle I_{x}}$) can be computed in Cartesian coordinates as

${\displaystyle I_{x}=\iint \limits _{R}y^{2}\,\mathrm {d} x\,\mathrm {d} y}$

The second moment of the area is crucial in Euler–Bernoulli theory of slender beams.

### Product moment of area

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

${\displaystyle I_{xy}=\iint \limits _{R}xy\,\mathrm {d} x\,\mathrm {d} y}$

## Parallel axis theorem

A shape with centroidal axis x. The parallel axis theorem can be used to obtain the second moment of area with respect to the x' axis.
Main article: Parallel axis theorem

It is sometimes necessary to calculate the second moment of area of a shape with respect to an ${\displaystyle x'}$ axis different to the centroidal axis of the shape. However, it is often easier to derive the second moment of area with respect to its centroidal axis, ${\displaystyle x}$, and use the parallel axis theorem to derive the second moment of area with respect to the ${\displaystyle x'}$ axis. The parallel axis theorem states

${\displaystyle I_{x'}=I_{x}+Ad^{2}}$

where

${\displaystyle A}$ is the area of the shape, and
${\displaystyle d}$ is the perpendicular distance between the ${\displaystyle x}$ and ${\displaystyle x'}$ axes.[4][5]

A similar statement can be made about a ${\displaystyle y'}$ axis and the parallel centroidal ${\displaystyle y}$ axis. Or, in general, any centroidal ${\displaystyle B}$ axis and a parallel ${\displaystyle B'}$ axis.

## Perpendicular axis theorem

For the simplicity of calculation, it is often desired to define the polar moment of area (with respect to a perpendicular axis) in terms of two area moments of inertia (both with respect to in-plane axes). The simplest case relates ${\displaystyle J_{z}}$ to ${\displaystyle I_{x}}$ and ${\displaystyle I_{y}}$.

${\displaystyle J_{z}=\iint \limits _{R}\rho ^{2}\,\mathrm {d} A=\iint \limits _{R}(x^{2}+y^{2})\,\mathrm {d} A=\iint \limits _{R}x^{2}\,\mathrm {d} A+\iint \limits _{R}y^{2}\,\mathrm {d} A=I_{x}+I_{y}}$

This relationship relies on the Pythagorean theorem which relates ${\displaystyle x}$ and ${\displaystyle y}$ to ${\displaystyle \rho }$ and on the linearity of integration.

## Composite shapes

For more complex areas, it is often easier to divide the area into a series of "simpler" shapes. The second moment of area for the entire shape is the sum of the second moment of areas of all of its parts about a common axis. This can include shapes that are "missing" (i.e. holes, hollow shapes, etc.), in which case the second moment of area of the "missing" areas are subtracted, rather than added. In other words, the second moment of area of "missing" parts are considered negative for the method of composite shapes.

## Examples

See list of second moments of area for other shapes.

### Rectangle with centroid at the origin

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.

${\displaystyle I_{x}=\iint \limits _{R}y^{2}\,\mathrm {d} A=\int _{-b/2}^{b/2}\int _{-h/2}^{h/2}y^{2}\,\mathrm {d} y\,\mathrm {d} x=\int _{-b/2}^{b/2}{\frac {1}{3}}{\frac {h^{3}}{4}}\,\mathrm {d} x={\frac {bh^{3}}{12}}}$
${\displaystyle I_{y}=\iint \limits _{R}x^{2}\,\mathrm {d} A=\int _{-b/2}^{b/2}\int _{-h/2}^{h/2}x^{2}\,\mathrm {d} y\,\mathrm {d} x=\int _{-b/2}^{b/2}hx^{2}\,\mathrm {d} x={\frac {b^{3}h}{12}}}$

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

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

### Annulus centered at origin

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.

${\displaystyle I_{x,circle}=\iint \limits _{R}y^{2}\,dA=\iint \limits _{R}(r\sin {\theta )}^{2}\,dA=\int _{0}^{2\pi }\int _{0}^{r}(r\sin {\theta })^{2}\left(r\,dr\,d\theta \right)=}$

${\displaystyle =\int _{0}^{2\pi }\int _{0}^{r}r^{3}\sin ^{2}{\theta }\,dr\,d\theta =\int _{0}^{2\pi }{\frac {r^{4}\sin ^{2}{\theta }}{4}}\,d\theta ={\frac {\pi }{4}}r^{4}}$

${\displaystyle J_{z,circle}=\iint \limits _{R}r^{2}\,dA=\int _{0}^{2\pi }\int _{0}^{r}r^{2}\left(r\,dr\,d\theta \right)=\int _{0}^{2\pi }\int _{0}^{r}r^{3}\,dr\,d\theta =}$

${\displaystyle =\int _{0}^{2\pi }{\frac {r^{4}}{4}}\,d\theta ={\frac {\pi }{2}}r^{4}}$

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 ${\displaystyle r_{1}}$.

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

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.

${\displaystyle J_{z}=\iint \limits _{R}r^{2}\,dA=\int _{0}^{2\pi }\int _{r_{1}}^{r_{2}}r^{2}\left(r\,dr\,d\theta \right)=\int _{0}^{2\pi }\int _{r_{1}}^{r_{2}}r^{3}\,dr\,d\theta =}$

${\displaystyle =\int _{0}^{2\pi }\left[{\frac {r_{2}^{4}}{4}}-{\frac {r_{1}^{4}}{4}}\right]\,d\theta ={\frac {\pi }{2}}\left(r_{2}^{4}-r_{1}^{4}\right)}$

### Any polygon

A simple polygon.

The second moment of area for any simple polygon on the XY-plane can be computed in general by summing contributions from each segment of the polygon. 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.

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

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]