Click here to use the Free Moment of Inertia Calculator

## How to Calculate the Moment of Inertia of a Beam Section

(Second Moment of Area)

Before we find the moment of inertia (or second moment of area) of a beam section, its centroid (or center of mass) must be known. For instance, if the moment of inertia of the section about its horizontal (XX) axis was required then the vertical (y) centroid would be needed first (Please view our Tutorial on how to calculate the Centroid of a Beam Section).

Before we start, if you were looking for our Free Moment of Inertia Calculator please click the link to learn more. Consider the I-beam section below, which was also featured in our Centroid Tutorial. As shown below the section was split into 3 segments:

The Neutral Axis (NA)

The Neutral Axis (NA) or the horizontal XX axis is located at the centroid or center of mass. In our Centroid Tutorial, the centroid of this section was previously found to be 216.29 mm from the bottom of the section. Now to calculate the total moment of ienrtia of the section we need to use the "Parallel Axis Theorem":[math] {I}_{total} = \sum{(\bar{I}_{i} + {A}_{i} {{d}_{i}}^{2})} \text{ where:}\\ \begin{align} \bar{I}_{i} &= \text{The moment of inertia of the individual segment about its own centroid axis}\\ {A}_{i} &= \text{The area of the individual segment}\\ {d}_{i} &= \text{The vertical distance from the centroid of the segment to the Netrual Axis (NA)} \end{align} [math]

It is widely known that the moment of inertia equation of a rectangle about its centroid axis is simply:

[math]
\bar{I}=\frac{1}{12}b{h}^{3} \text{ where:}\\\\
\begin{align}
b &= \text{The base or width of the rectangle}\\
h &= \text{The height of the rectangle}
\end{align}
[math]

The moment of inertia of other shapes are often stated in the front/back of textbooks, however the rectangular shape is very common for beam sections.

Now we have all the information we need to use the "Parallel Axis Theorem" and find the total moment of inertia of the I-beam section:

[math] \text{Segment 1:}\\ \begin{align} \bar{I}_{1} &= \tfrac{1}{12}(250)(38)^{3} = 1,143,166.667 {\text{ mm}}^{4}\\ {A}_{1} &= 250\times38 = 9500 {\text{ mm}}^{2}\\ {d}_{1} &= \left|{y}_{1} - \bar{y} \right| = \left|(38 +300 +\tfrac{38}{2}) - 216.29\right| = 140.71 \text{ mm}\\\\ \end{align} [math] [math] \text{Segment 2:}\\ \begin{align} \bar{I}_{2} &= \tfrac{1}{12}(25)(300)^{3} = 56,250,000 {\text{ mm}}^{4}\\ {A}_{2} &= 300\times25 = 7500 {\text{ mm}}^{2}\\ {d}_{2} &= \left|{y}_{2} - \bar{y}\right| = \left|(38 +\tfrac{300}{2}) - 216.29\right| = 28.29 \text{ mm}\\\\ \end{align} [math] [math] \text{Segment 3:}\\ \begin{align} \bar{I}_{3} &= \tfrac{1}{12}(150)(38)^{3} = 685,900 {\text{ mm}}^{4}\\ {A}_{3} &= 150\times38 = 5700 {\text{ mm}}^{2}\\ {d}_{3} &= \left|{y}_{3} - \bar{y}\right| = \left|\tfrac{38}{2} - 216.29\right| = 197.29 \text{ mm}\\\\ \end{align} [math] [math] \begin{align} \therefore {I}_{total} &= \sum{(\bar{I}_{i} + {A}_{i} {{d}_{i}}^{2})} \\ &= (\bar{I}_{1} + {A}_{1}{{d}_{1}}^{2}) + (\bar{I}_{2} + {A}_{2}{{d}_{2}}^{2}) + (\bar{I}_{3} + {A}_{3}{{d}_{3}}^{2})\\ &= (1,143,166.667 + 9500\times140.71^{2}) + (56,250,000 + 7500\times28.29^{2}) + (685,900 + 5700\times197.29^{2})\\ &= 474,037,947.7 {\text{ mm}}^{4}\\ {I}_{total} &= 4.74 \times 10^{8} {\text{ mm}}^{4} \end{align} [math]

Click here to use the Free Moment of Inertia Calculator

So in summary, you can see the results (calculated from our Free Moment of Inertia Calculator) along with some other section properties to check your work.

Of course you don't need to do all these calculations manually because you can use our fantastic Free Moment of Inertia Calculator to find the important properties and information about beam sections.