The moment of inertia of an object represents its resistance to bending. The moment of inertia of an object is determined by the relationship of its cross sectional area to where its centroid is located with respect to the sections x and y axes. This article will show you how to calculate the moment of inertia about these axes for a standard rectangular shape.

Since the moment of inertia is a property that is determined by the geometry of an object and considered with respect to how much bending it can resist, we will look at a real world example. But first, we must recognize that a standard object such as a rectangle or square will have two primary axes of bending, its x and y axis. By convention, we will say that the x-axis is the horizontal axis and the y-axis is the vertical axis. We will consider that any standard shape based on its geometry will have some resistance to bending about its x and y axes that can be calculated determining its moment of inertia with respect to each axis, I_{X} and I_{Y}.

Let's consider the example of a wooden floor beam in a house. The actual dimensions of the beam are 2" x 8" and it is oriented so that the 8" is vertical and the 2" is horizontal. This floor beam will resist vertical loads acting downward due to the law of gravity; this creates bending about the x-axis that must be resisted by I_{X}. However, the floor beam has to resist no loads or forces acting horizontally and so there is no bending that occurs about the y-axis, but there is still a resisting moment of inertia, I_{Y.} Intuition tells us that this beam is oriented such that it will have the greatest moment of inertia about its x-axis, in other words, I_{X} > I_{Y}, but just to be sure we will calculate the values of each to see.

Before we calculate any moment of inertia we must consider where the centroid of the section is for both the x and y axes. The centroid of the section with respect to the x-axis is the line (parallel to the x-axis) about which the section bends. Basically the location where there is an even amount above and below the centroid of cross sectional area distributed according to its distance from the centroid. This applies the same to the centroid of the y-axis. In this case the section is symmetric about both axes and the centroids of each axis are located in the center of the section and so the use of the parallel axis theorem is not needed, so we can calculate the moment of inertia by simply using: I = bh^{3}/12. Where b is the width of the section and h is the resisting height.

So the moment of inertia about the x-axis can be calculated as I_{X }= [(2 in.)(8 in.)^{3}]/12,

where I_{X} = 85.33 in.^{4}. The moment of about the y-axis is calculated from,

I_{X }= [(8 in.)(2 in.)^{3}]/12, where I_{X} = 5.33 in.^{4}. So from the numbers we can see that I_{X} > I_{Y} and so the beam is in fact oriented such that the greatest resistance to bending is about the x-axis and it resists the floor loads.

RESOURCE:

Gere, James M. (2004). Mechanics of Materials (6^{th} Ed.). Belmont, CA: Brooks/Cole â€“

Thomson Learning

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