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Second moment of area

The second moment of area, or second area moment, or quadratic moment of area and also known as the area moment of inertia, 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 or with a . In both cases, it is calculated with a multiple integral over the object in question. Its dimension is L (length) to the fourth power. Its unit of dimension, when working with the International System of Units, is meters to the fourth power, m4, or inches to the fourth power, in4, when working in the Imperial System of Units or the US customary system.

Definition
The second moment of area for an arbitrary shape  with respect to an arbitrary axis BB' (BB' axis is not drawn in the adjacent image; is an axis coplanar with x and y axes and is perpendicular to the line segment \rho) is defined as J_{BB'} = \iint_{R} {\rho}^2 \, dA where • dA is the infinitesimal area element, and • \rho is the distance from the BB' axis. 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_{R} y^2\, dx\, dy 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 I_{xy} = \iint_{R} yx\, dx\, dy == Parallel axis theorem ==
Parallel axis theorem
axis x. The parallel axis theorem can be used to obtain the second moment of area with respect to the x' axis. It is sometimes necessary to calculate the second moment of area of a shape with respect to an 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, x, and use the parallel axis theorem to derive the second moment of area with respect to the x' axis. The parallel axis theorem states I_{x'} = I_x + A d^2 where • A is the area of the shape, and • d is the perpendicular distance between the x and x' axes. A similar statement can be made about a y' axis and the parallel centroidal y axis. Or, in general, any centroidal B axis and a parallel B' axis. == Perpendicular axis theorem ==
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 J_z to I_x and I_y. J_z = \iint_{R} \rho^2\, dA = \iint_{R} \left(x^2 + y^2\right) dA = \iint_{R} x^2 \, dA + \iint_{R} y^2 \, 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 ==
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==
Examples
See list of second moments of area 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. \begin{align} 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{b h^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}} h x^2\, dx = \frac{b^3 h}{12} \end{align} Using the perpendicular axis theorem we get the value of J_z. J_z = I_x + I_y = \frac{b h^3}{12} + \frac{h b^3}{12} = \frac{b h}{12}\left(b^2 + h^2\right) Annulus centered at origin Consider an annulus whose center is at the origin, outside radius is r_2, and inside radius is r_1. 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_2 minus the polar moment of inertia of a circle with radius r_1, 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. \begin{align} 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{align} 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_2 and a circle with radius r_1. 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 dr integral the first time around to reflect the fact that there is a hole. This would be done like this. \begin{align} 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{align} Any polygon 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 n vertices, numbered in counter-clockwise fashion. If polygon vertices are numbered clockwise, returned values will be negative, but absolute values will be correct. \begin{align} 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} + 2 x_i y_i + 2 x_{i+1} y_{i+1} + x_{i+1} y_i \right) \end{align} where x_i,y_i are the coordinates of the i-th polygon vertex, for 1 \le i \le n. Also, x_{n+1}, y_{n+1} are assumed to be equal to the coordinates of the first vertex, i.e., x_{n+1} = x_1 and y_{n+1} = y_1. ==See also==
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