Second moment of area

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 area moments of inertia.

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 $I$ for an axis that lies in the plane or with a $J$ for an axis perpendicular to the plane. 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.

Definition

A schematic showing how the polar moment of inertia is calculated for an arbitrary shape with respect to the z axis. 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 with respect to an arbitrary axis $BB$ is defined as

$J_{BB}=\int _{A}{\rho }^{2}\,\mathrm {d} A$

$\mathrm {d} A$ = Differential area of the arbitrary shape

$\rho$ = Distance from the axis BB to dA

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

$I_{x}=\iint _{A}y^{2}\,\mathrm {d} x\,\mathrm {d} y$

Parallel axis theorem

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It is often easier to derive the second moment of area with respect to its centroidal axis, $x'$. However, it may be necessary to calculate the second moment of area with respect to a different, parallel axis, say the $x$ axis. The parallel axis theorem states

$I_{x}=I_{x'}+Ad_{y}^{2}$

where

$A$ = Area of the shape

$d_{y}$ = Perpendicular distance between the $x'$ and $x$ axes

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

Perpendicular axis theorem

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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 $J_{z}$ to $I_{x}$ and $I_{y}$.

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

This relationship relies on the Pythagorean theorem which relates $x$ and $y$ to $\rho$ 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 area moments of inertia for other shapes.

Rectangle with centroid at the origin

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

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

$I_{y}=\int _{A}x^{2}\,\mathrm {d} A=\int _{-b/2}^{b/2}\int _{-h/2}^{h/2}x^{2}\,\mathrm {d} y\,\mathrm {d} x=\int _{-b/2}^{b/2}hx^{2}\,\mathrm {d} x={\frac {b^{3}h}{12}}$

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

Annulus centered at origin

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

$I_{x,circle}=\iint y^{2}\,dA=\iint (r\sin {\theta )}^{2}\,dA=\int _{0}^{2\pi }\int _{0}^{r}(r\sin {\theta })^{2}\left(r\,dr\,d\theta \right)=\int _{0}^{2\pi }\int _{0}^{r}r^{3}\sin ^{2}{\theta }\,dr\,d\theta =\int _{0}^{2\pi }{\frac {r^{4}\sin ^{2}{\theta }}{4}}\,d\theta ={\frac {\pi }{4}}r^{4}$

$J_{z,circle}=\iint r^{2}\,dA=\int _{0}^{2\pi }\int _{0}^{r}r^{2}\left(r\,dr\,d\theta \right)=\int _{0}^{2\pi }\int _{0}^{r}r^{3}\,dr\,d\theta =\int _{0}^{2\pi }{\frac {r^{4}}{4}}\,d\theta ={\frac {\pi }{2}}r^{4}$

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

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

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

$J_{z}=\iint r^{2}\,dA=\int _{0}^{2\pi }\int _{r_{i}}^{r_{o}}r^{2}\left(r\,dr\,d\theta \right)=\int _{0}^{2\pi }\int _{r_{i}}^{r_{o}}r^{3}\,dr\,d\theta =\int _{0}^{2\pi }\left[{\frac {r_{o}^{4}}{4}}-{\frac {r_{i}^{4}}{4}}\right]\,d\theta ={\frac {\pi }{2}}\left(r_{o}^{4}-r_{i}^{4}\right)$

Any polygon

Any 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.. polygon is assumed to be counter clock wise (for clockwise polygon all values will be negative with same absolute value)

$I_{x}={\frac {1}{12}}\sum _{i=1}^{i=N}(y_{i}^{2}+y_{i}y_{i+1}+y_{i+1}^{2})(x_{i}y_{i+1}-x_{i+1}y_{i})$

$I_{y}={\frac {1}{12}}\sum _{i=1}^{i=N}(x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2})(x_{i}y_{i+1}-x_{i+1}y_{i})$

$I_{xy}={\frac {1}{24}}\sum _{i=1}^{i=N}(x_{i}y_{i+1}+2x_{i}y_{i}+2x_{i+1}y_{i+1}+x_{i+1}y_{i})(x_{i}y_{i+1}-x_{i+1}y_{i})$

where $x_{i},y_{i}$ (with $x_{n+1}=x_{1},y_{n+1}=y_{1}$) are the coordinates of any polygon vertex.

See also

• List of area moments of inertia
• List of moments of inertia
• Moment of inertia
• Parallel axis theorem
• Perpendicular axis theorem
• Radius of gyration

References

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External links

• Calculator for Second Moment of Area