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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) 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":

${I}_{total}&space;=&space;\sum{(\bar{I}_{i}&space;+&space;{A}_{i}&space;{{d}_{i}}^{2})}&space;\textup{&space;where:}\\&space;\bar{I}_{i}&space;=&space;\textup{The&space;moment&space;of&space;inertia&space;of&space;the&space;individual&space;segment&space;about&space;its&space;own&space;centroid&space;axis}\\&space;{A}_{i}&space;=&space;\textup{The&space;area&space;of&space;the&space;individual&space;segment}\\&space;{d}_{i}&space;=&space;\textup{The&space;vertical&space;distance&space;from&space;the&space;centroid&space;of&space;the&space;segment&space;to&space;the&space;Netrual&space;Axis&space;(NA)}$

It is widely known that the moment of inertia equation of a rectangle about its centroid axis is simply:
$\bar{I}=\frac{1}{12}b{h}^{3}&space;\textup{&space;where:}\\\\&space;b&space;=&space;\textup{The&space;base&space;or&space;width&space;of&space;the&space;rectangle}\\&space;h&space;=&space;\textup{The&space;height&space;of&space;the&space;rectangle}$

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:

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

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.

• Shaun Murrin

This is a tutorial question I’m struggling a bit with axis to base the i value on..
“Based on calculations it has been determined that a universal steel column must have a minimum value of I of 38748 cm4. Select a column suitable for this scenario showing clearly why you have selected that particular column. “

• Glenn

Used your calculator to find for simple ‘i’ beam. Whilst we both find identical Zy, my Zx is out compared to yours. I looked at your formula to find the centroid and it shows the addition of web thickness to the vertical centroid. Beam is: TFw & BFw = 100; TFt & BFt = 10; Wh = 80, Wt = 6. Following your tute on centroid for simple I beam, Cx would be 56, however your app shows (correctly in my mind) Cx and Cy both as 50. Yet my Ix = 4.566 x 106
mm4 ; calculator shows 4.323 x 106
mm4 (Iy is same as mine 1.668 x 106
mm4 ).
Been over calcs left and right and can’t find error. Unless Cx as 56 is culprit. Any ideas?

• Hi Glenn. As your section is symmetric about both horizontal and vertical and your section depth and width are both 100mm, then the centroid is definitely 50mm in both directions. Recalculate your Cx following the tutorial (see the image attached below).

• Cristian

There is an error in the last picture, it shows the subscripts of some answers for the “x” axis, while there is the axis “z”, maybe it will lead to some misunderstandings. Anyways nice tutorial.

• Thank you. This has been amended.