Moments of inertia for plane figures – example calculation

Second moments of area of plane figures in brief:

• Calculating the second moment of area of an I-section requires dividing the cross-section into three rectangles: two flanges (top and bottom) and one web.
• The first step is to determine the centroid of the entire cross-section, which for symmetric I-sections lies exactly at mid-height.
• The second moments of area of the individual rectangular components are calculated using: Ix = (b · h³) / 12, where h is the dimension perpendicular to the axis being considered.
• Steiner’s theorem (parallel-axis theorem) is necessary for the flanges, because their own centroidal axes do not coincide with the neutral axis of the full cross-section.
• The total second moment of area of the I-section is the sum of the second moments of area of all three rectangles, each transferred to the common centroidal axis.
• For I-sections symmetric about both axes, the flanges have identical second moments of area, so the calculation can be simplified by computing one flange and multiplying the result by two.

You mostly want to know…

The moments of inertia of flat figures (cross-sections) are key parameters in structural strength analysis, particularly when analyzing the bending of beams.

In this article, we will focus on practical examples of calculating moments of inertia for various types of cross-sections, We will cover cross-sections based on rectangles (I-beams) and more complex shapes made up of several different simple figures. Each example will be broken down step by step to show how to correctly determine moments of inertia and how these results affect the structural behavior under load.

With these calculations and exercises in EquiBeam, you will gain confidence in strength analysis for different cross-sections.

I-Beam

In this example, we will calculate the moment of inertia for the cross-section shown in the following figure:

This cross-section, called an I-beam, consists of two horizontal flanges (elements at the top and bottom of the cross-section) and a vertical web. It is one of the most commonly used structural elements due to its excellent strength properties related to bending resistance.

Additionally, in this case, we are dealing with an asymmetric I-beam because the lengths of the top and bottom flanges are not the same. Why is it so strong, and why should we be cautious about this strength? You’ll learn from the calculations below.

Before we start the calculations, it’s worth taking a moment to perform an initial classification of the cross-section.

Our I-beam is a figure symmetric with respect to the vertical axis, which means that before performing any calculations, we know that the centroid of the cross-section is located on this axis.

Furthermore, we know that with respect to a coordinate system aligned with this axis, the product of inertia will be zero, meaning that the vertical axis will be one of the principal axes of inertia. The location of the second axis must, however, be determined through calculations since the figure does not have horizontal symmetry.

Let’s move on to the calculations.

Dividing into simple figures

To calculate the moments of inertia for a complex flat figure, you first need to divide it into simple shapes such as rectangles, triangles, or circles.

The advantage of simple shapes is that we already know the locations of their centroids and the formulas for their central moments of inertia. Both can be determined analytically, as I explained in the theoretical articles ????.

For an asymmetric I-beam, the easiest way to divide it is into three rectangles. The division of the composite figure, along with the assumed (deliberately not aligned with the axis of symmetry) division, is shown below:

In the figure, the centroids of the simple figures and their local coordinate systems are marked, aligned with the global coordinate system. Immediately, we can observe that the centroids of all simple figures lie on the axis of symmetry ????. Knowing the locations of these points allows us to calculate the overall centroid.

Determining the centroid

Let’s determine the locations of the centroids of the flat figures:

• starting with the bottom flange:
  • the z-coordinate is related to the symmetry axis: c_1z=3.5a
  • the y-coordinate is half the height of the rectangle: c_1y=0.5a
• next, the web:
  • c_3z=3.5a
  • the vertical coordinate is the sum of the bottom flange’s height and half the height of the web: c_3y= a+0.5 4a = 3a
• finally, the top flange:
  • c_2z=3.5a
  • the vertical coordinate is the sum of the heights of the bottom flange and web, plus half the height of the top flange: c_2y= a+ 4a 0.5 a = 5.5a

Now we can calculate the first moments of area relative to the global coordinate system, which are the products of the centroids’ coordinates and the areas of the figures.

The first moment relative to the z-axis is calculated as the distance along the y-axis:

S_z=A_1 c_{y1}+A_2 c_{y2}+A_3 c_{y3}=(5a⋅a)⋅0.5a+(7a⋅a)⋅5.5a+(4a⋅a)⋅3a=

=2.5a^3+38.5a^3+12a^3=53a^3

The first moment relative to the y-axis is calculated as the distance along the z-axis:

S_y=A_1 c_{z1}+A_2 c_{z2}+A_3 c_{z3}=(5a⋅a)⋅3.5a+(7a⋅a)⋅3.5a+(4a⋅a)⋅3.5a=

=17.5a^3+24.5a^3+14a^3=56a^3

Meanwhile, we can calculate the total area of the figure:

A=5a^2+7a^2+4a^2=16a^2

Finally, we calculate the location of the centroid:

c_y={\frac {S_z} {A}}={\frac{53a^3} {16a^2}}=3.3125 a,c_z={\frac{S_y}{A}}={\frac{56a^3}{16a^2}}=3.5a

The calculations confirmed that the centroid is located on the symmetry axis– coordinate c_z = 3.5a (as expected)—and determined that the horizontal principal axis is slightly above the midpoint of the cross-section. We could have suspected this from the beginning since the top flange is slightly longer than the bottom one ????.

The diagram showing the position of the centroid and its principal axes z_c and  y_c is below:

Central moments of inertia – calculations

Once we know the location of the principal axes, we can calculate the moments of inertia for the flat figure. To recap, the calculations are done by summing the central moments of inertia of the simple shapes relative to the respective axes and accounting for the distances of their centroids from the principal axes of the entire figure (the Steiner theorem).

We’ll start with the vertical axis y_c – this will be a bit easier:

I_{yc}=\Big ({\frac{(5a)^3⋅a}{12}}\Big)+\Big ({\frac{(7a)^3⋅a}{12}}\Big)+\Big ({\frac{a^3⋅4a}{12}}\Big)=\frac{125a^4+343a^4+4a^4} {12}=39.333 a^4

in the above calculations, we see formulas for the moments of inertia of a rectangle relative to the vertical axis (since we’re calculating the moment of inertia relative to the y_c axis), meaning the third power involves the dimensions perpendicular to this axis.

Additionally, the Steiner theorem components don’t appear in the equation because all points lie on the y_c and their distances from this axis are zero.

Let’s move on to the calculations for the z_c axis:

I_{zc}=\Big(\frac{5a⋅a^3}{12}+5a^2⋅(0.5a-3.3125a)^2\Big)+\Big(\frac{7a⋅a^3}{12}+7a^2⋅\Big(5.5a-3.3125a)^2 \Big)+\Big(\frac{a⋅(4a)^3}{12}+4a^2⋅(3a-3.3125a)^2 )=

=(0.41667+39.55078) a^4+(0.58333+33.49609) a^4+(5.33333+0.390625) a^4

39.96745a^4+34.07942a^4+5.72396a^4=79.77083 a^4

In the above calculations, we again see formulas for the moments of inertia of a rectangle—this time relative to the horizontal axis—so the third power involves the dimensions perpendicular to the horizontal axis.

The Steiner theorem components are present in the equation since none of the points pass through the z_c axis.

In the Steiner moment calculations, the distance from the axis can easily be computed by subtracting the centroid coordinates in the initial system from the global centroid coordinates.

Remember to square the distance, a step that students often forget. What’s more, squaring ensures that the order of subtraction doesn’t matter ????.

In summary, we have calculated the central moments of inertia for our I-beam. There is no need to calculate the product of inertia as it will be zero (the figure is symmetric, and the principal system is aligned with the symmetry axis!)

Determining principal central moments of inertia

One thing we should notice is that the geometric moment of inertia relative to the z_c axis is significantly larger (about twice) than the moment relative to the y_caxis, even though the figure is “wider” than it is “tall.”

As you know from the theoretical articles, in the analysis of the moment of inertia, not just the dimension of the cross-section matters, but more importantly, the distribution of its area. In this example, the area is significantly more spread out from the z_c axisaxis due to the flanges than from the y_c axis.

As you can guess, this is precisely the intention behind I-beams—to maximize the area distribution away from the preferred bending axis.

And here’s the caution: the I-beam (in this orientation) is primarily resistant to bending relative to the horizontal axis, and that’s how it should be used. Its bending parameters relative to the vertical axis are not as favorable.

In this context, we can recall the principal central moments of inertia. These are the moments of inertia that reach extreme values and are related to the principal system, where the product of inertia is zero.

Thus, the first (maximum) principal moment is associated with the z_c axis and equals:

I_1=I_{max}=I_{zc}=79.77083 a^4

The second (minimum) moment of inertia is associated with the y_c axis and equals:

I_2=I_{min}=I_{yc}=39.333 a^4

In this example, I performed the moment of inertia calculations for a fairly simple figure. You can practice the next examples yourself: that is, not by yourself, but with the EquiBeam app.

You’re invited to read the related article on cross-sections composed of more complex shapes. These types of cross-sections are often assigned by instructors to “thoroughly” test students’ knowledge ????.