Moment of Inertia of Beam Sections

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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. This will calculate the centroid, moi and other results and even show you the step by step calculations! But for now, let's look at a step-by-step guide and example of how to calculate moment of inertia:

Step 1: Segment the beam section into parts

When calculating the area moment of inertia, we must calculate the moment of inertia of smaller segments. Try to break them into simple rectangular sections. For instance, consider the I-beam section below, which was also featured in our Centroid Tutorial. We have chosen to split this section into 3 rectangular segments:

Step 2: Calculate 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.

Step 3: Calculate Moment of Inertia

To calculate the total moment of inertia 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]

Since we have split it into three rectangular parts, we must calculate the moment of inertia of each of these sections. 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 or from this guide of moment of inertia shapes. However the rectangular shape is very common for beam sections, so it is probably worth memorizing.

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. In our moment of inertia example:

[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]

So there you have our guide on calculating the area of moment for beam sections. This result is critical in structural engineering and is an important factor in the deflection of a beam. We hope you enjoyed the tutorial and look forward to any comments you have.

BONUS: Using our Moment of Inertia Calculator

SkyCiv's Account shows full calculations of moment of inertia. This interactive module will show you the step-by-step calculations of how to find moment of inertia:

Alternatively, you can see the results of our Free Moment of Inertia Calculator to check your work. This will calculate all the properties of your cross section and is a useful reference to calculate the Centroid, Area and Moment of Inertia of your beam sections!

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  • Owain

    hi, some guidance on how to calculate Ixx when you have rivets in the bottom flange would be great. Do you remove the X-sectional area of holes, determine NA and Ixx? or, do not remove holes and determine NA, then calculate Ixx with holes removed?

  • Kabila

    when is 1/12 factor used in calculating the moment of inertia? and why?

    • side-fish

      Hi Kabila, I hope I’m not too late. But the 1/12 coefficient from the moment of inertia of a rectangle is derived from calculus by integrating y^2dA. Your upper limit is the extreme fiber and the lower limit is the axis of rotation. In the case of a rectangle, you will integrate the top and the bottom of the rectangle. Note that dA=xdy. The new equation becomes I = ∫xy^2dy with the upper limit as h and the lower limit as h/2 if you want to integrate the top half (or h/2-0 if you want the bottom). Since the neutral axis of the rectangle is exactly in the middle, you simply integrate say… the top half and multiply twice. The coefficient that you should get is 1/12. If the axis of rotation is shifted to say the bottom fiber, integrating will yield a coefficient of 1/3, though you could use the transfer formula I=Io+Ad^2 and still get the same answer.

  • 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.

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