Second moment of area

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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, polar moment of area 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 deflection.

Definition[edit]

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 dA
\mathrm dA = 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 dx\, \mathrm dy

Parallel axis theorem[edit]

Main article: Parallel axis theorem

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'} + A d_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[edit]

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 dA = \int_A (x^2 + y^2)\,\mathrm dA = \int_A x^2\,\mathrm dA + \int_A y^2\,\mathrm dA = 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[edit]

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[edit]

See list of area moments of inertia for other shapes.

Rectangle with centroid at the origin[edit]

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.

Block centroid axes.svg
I_{x} = \int_A y^2\,\mathrm dA = \int^{b/2}_{-b/2} \int^{h/2}_{-h/2} y^2 \,\mathrm dy \,\mathrm dx = \int^{b/2}_{-b/2} \frac{1}{3}\frac{h^3}{4}\,\mathrm dx = \frac{b h^3}{12}
I_{y} = \int_A x^2\,\mathrm dA = \int^{b/2}_{-b/2} \int^{h/2}_{-h/2} x^2 \,\mathrm dy \,\mathrm dx = \int^{b/2}_{-b/2} h x^2\,\mathrm dx = \frac{b^3 h}{12}
J_{z} = I_x + I_y = \frac{b h^3}{12} + \frac{h b^3}{12} = \frac{b h}{12}(b^2 + h^2) (see Perpendicular axis theorem)

Annulus centered at origin[edit]

Second Moment of Area, Annulus.svg

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[edit]

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} + 2 x_i y_i + 2 x_{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.[1]

See also[edit]

References[edit]

  • Pilkey, Walter D. (2002). Analysis and Design of Elastic Beams. John Wiley & Sons, Inc. ISBN 0-471-38152-7. 
  • Hibbeler, R. C. (2004). Statics and Mechanics of Materials (Second ed.). Pearson Prentice Hall. ISBN 0-13-028127-1. 
  1. ^ [1]

External links[edit]