The moment of inertia for a rectangular or square shape can be calculated using the moment of inertia formulation of I = bh3/12. This formulation can be used alone to determine the moments of inertia with respect to the x and y axes for a symmetric section. A section that is symmetric with respect to the x and y axes will have its centroids located in the geometric center of the shape. But what does one do when the cross section is not symmetric about one or both of its axes and one or both of the centroids move? In this case the use of the moment of inertia formula by itself is not sufficient and so we employ the use of the parallel axis theorem.
By definition, the moment of inertia for a cross section with respect to an axis is its measure of resistance to bending about that axis. And since we know that the moment of inertia is a measure of the area distributed on both sides of a centroid, the moment of inertia formula can be rewritten to reflect this change for an unsymmetric section. The standard form of the moment of inertia for an unsymmetric or composite section about any axis can be written as, IC = I + Ad2. IC is the total moment of inertia for the composite section, I is the moment of inertia for each individual section, A is the area of each section that's centroid does not coincide with the centroid of the composite section and d is the distance of each sections centroid to the centroid of the composite section.
Let's see this concept further clarified with an example.
We have an actual dimension 4" x 10" wooden beam, where the 4" face is horizontal and the 10" face is vertical. It was found that this beam underwent too much bending stress about its x-axis and so it had to be reinforced at its base with an actual dimension 2" x 6" board. This 2" x 6" board was positioned such that the 2" face is vertical and the 6" face is horizontal. This 2" x 6" is placed such that the composite section is symmetric (vertically) about the y-axis. So in this case we are not concerned about the y-axis but we must determine the moment of inertia for the composite section. Since the cross section is not symmetric about the x-axis, we must determine the location of its centroid with respect to the x-axis.
y = [(A1)(d1)+ (A2)( d2)] / (A1 + A2)
A1 = (4in.)(10in.) = 40 in.2
A2 = (2in.)(6in.) = 12 in.2
d1 = 5in. + 2in. = 7in. This is the distance from the base of the 2" x 6" to the centroid of the 4" x 10."
d2 = 1in. This is the distance from the base of the 2" x 6" to the centroid of the 2" x 6."
y = [(40 in.2)( 7 in.)+ ( 12 in.2)( 1 in.)] / ( 40 in.2 + 12 in.2)
y = [280 in.3 + 12 in.3] / (52 in.2)
y = (296 in.3) / (52 in.2)
y = 5.62 in.
So the centroid for this composite section with respect to the x-axis is found to be at 5.62 inches from the base.
Now we can calculate the moment of inertia for the section with respect to the x-axis using the parallel axis theorem. We will let the 4" x 10" be A1 and the 2" x 6" be A2.
IC,X = I1 + (A1 )( d12) + I2 + (A2 )( d22)
I1 = (4in.)(10in.) 3/ 12 = 333.3 in.4
(A1 )( d12) = ( 40 in.2)(7in. â 5.62in.)2 = 76.2 in.4
I2 = (6in.)(2in.) 3/ 12 = 4.0 in.4
(A2 )( d22) = ( 12 in.2)(5.62in â 1in..)2 = 256.1 in.4
IC,X = 333.3 in.4 + 76.2 in.4 + 4.0 in.4 + 256.1 in.4
IC,X = 669.6 in.4
So we can see that with the addition of the 2" x 6" wood board to the base of the 4" x 10" we doubled the moment of inertia from 333.3 in.4 to 669.6 in.4. This represents a doubling of the strength of the cross section to resist bending stresses.
Gere, James M. (2004). Mechanics of Materials (6th Ed.). Belmont, CA: Brooks/Cole â