How to Calculate the Moment of Inertia for an Area

Areas occupy two dimensions, like the chalk outline shown here.

The area moment of inertia is the property of an areal object that quantifies its tendency to deflect or endure stress when an external force acts on it. A calculation of this quantity follows the same procedure as that of the better-known moment of inertia for a mass that occupies a spatial volume. While not a complicated procedure, determining the moment of inertia for an area does require preliminary knowledge of integration methods in introductory calculus.

Things You'll Need

• Pencil
• Paper

Instructions

1. Orient the areal object whose moment of inertia you wish to determine. In this example, the square will have its centroid located at the origin of the coordinate system, and will therefore span from (-1/2)a to (1/2)a in both the x and y dimensions.

2. Write down the area moment of inertia tensor. This quantity takes the form of a four-element (two by two) matrix. Thus, for individual elements J(nm) of a n by m matrix, you would write each element as follows: J(11) = y^2, J(12) = J(21) = -xy, and J(22) = x^2.

3. Integrate each element of the inertia tensor over the boundary values for the object. In the present example, you would have four double integrals with integration limits in x of (1/2)a and (-1/2)a, and integration limits in y of (1/2)a and (-1/2)a. For the square, the integration of the off-diagonal components of the tensor appears as follows:

   int(int(-xydx, x = (1/2)a . . . (-1/2)a))dy, y = (1/2)a . . . (-1/2)a)

   = int(-1/2x^2y|x = (1/2)a . . . (-1/2)a)dy, y = (1/2)a . . . (-1/2)a)

   = int(0, y = (1/2)a . . . (-1/2)a) = 0,

   where "int" means "integrate," and "|" means "evaluate at the limits."

   Note that the diagonal components J(12) and J(21) are equivalent for a square. For the on-diagonal components, the integration appears as:

   int(int((x^2)dx, x = (1/2)a . . . (-1/2)a))dy, y = (1/2)a . . . (-1/2)a)

   = int(1/3x^3|x = (1/2)a . . . (-1/2)a)dy, y = (1/2)a . . . (-1/2)a)

   = int(1/12a^3dy, y = (1/2)a . . . (-1/2)a)

   = 1/12a^3y|y = (1/2)a . . . (-1/2)a

   = 1/12(a^3)a = 1/12(a^4).

   Note that in the case of a square, the on-diagonal components J(11) and J(22) are also equivalent to one another.

4. Write each integrated element of the inertia tensor in place of the original element relationship to get the moment of inertia for the area. Here, the elements J(nm) appear as J(11) = J(22) = 1/12a^4 and J(12) = J(21) = 0.

References

• University of Nebraska -- Lincoln: Area Moment of inertia