This article is about the geometrical property of an area, termed the second moment of area. For moment of inertia and radius of gyration pdf moment of inertia dealing with the rotation of an object with mass, see Mass moment of inertia.

For a list of equations for second moments of area of standard shapes, see List of second moments of area. The 2nd 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. In both cases, it is calculated with a multiple integral over the object in question. In structural engineering, the second moment of area 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. In each case the integral is over all the infinitesimal elements of area, dA, in some two-dimensional cross-section.

The MOI, in this sense, is the analog of mass for rotational problems. A, with projections x and y on the axes. 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. 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. See list of second moments of area for other shapes. Because of the symmetry of the annulus, the centroid also lies at the origin. This would be done like this.

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. If polygon vertices are numbered clockwise, returned values will be negative, but absolute values will be correct. Analysis and Design of Elastic Beams. Wikimedia Commons has media related to Second moments of area. This page was last edited on 22 March 2018, at 14:40.

An uniform solid sphere has a radius R and mass M. Note: If you are lost at any point, please visit the beginner’s lesson or comment below. First, we set up the problem. Now, we have to force x into the equation. Notice that x, r and R makes a triangle above. If you spot any errors or want to suggest improvements, please contact us. Want to contribute to Mini Physics?

Click here to submit a post to Mini Physics. V, where ρ is the density and dV is the change in volume. I apologize that the formatting of the integrals isn’t coming out correctly. Why don’t we take dx as Rdθ like we did in case of a hollow sphere? The formula from Chandrasekhar’s book assumes the moments are measured as the mass elements rotate relative to the origin, instead of the axis of rotation.