Moment of Inertia of an I-Beam: Worked Example

The moment of inertia I (the second moment of area) measures how a cross section's area is spread away from its bending axis. It sets how much a beam deflects and, through the section modulus, how much stress it takes. An I-beam is efficient because it puts most of its area in the flanges, far from the middle.

This example works out the strong axis moment of inertia I_x of a W12x26 steel beam by hand, then compares it with the value in the AISC tables.

The formula

Split the I-beam into three rectangles: a top flange, a web and a bottom flange. For each one:

• Its own moment of inertia about its own center is bh³/12.
• The parallel axis theorem moves that to the beam's centroid: add A·y², where A is the rectangle's area and y the distance from its center to the beam's centroid.

I_x = Σ (bh³/12 + A·y²)

For a symmetric I-beam this comes out as one formula, with d the depth, b_f and t_f the flange width and thickness, t_w the web thickness and h = d − 2t_f the clear web height:

I_x = t_wh³/12 + 2[b_ft_f³/12 + b_ft_f((d − t_f)/2)²]

or, as the full outer rectangle less the two empty spaces beside the web:

I_x = b_fd³/12 − (b_f − t_w)h³/12

Worked example: W12x26

From the AISC shape tables:

Step 1: the web

Clear web height h = 12.2 − 2(0.380) = 11.44 in. The web is centered on the beam's centroid, so it has no A·y² term:

I_web = 0.230 × 11.44³ / 12 = 28.7 in⁴

Step 2: the flanges

Each flange's center is y = (12.2 − 0.380) / 2 = 5.91 in from the centroid, and its area is 6.49 × 0.380 = 2.466 in².

• Own moment of inertia: 6.49 × 0.380³ / 12 = 0.030 in⁴, which is tiny.
• Parallel axis term: 2.466 × 5.91² = 86.1 in⁴.

I_flanges = 2 × (0.030 + 86.1) = 172.3 in⁴

Step 3: add them up

I_x = 28.7 + 172.3 = 201.0 in⁴

The flanges give 86% of the total, almost all of it from the A·y² term. That's the reason for the I shape: area far from the axis counts with the square of its distance.

Step 4: compare with the AISC table

The AISC table gives I_x = 204 in⁴ for a W12x26. The hand calculation is 1.5% low because a rolled shape has curved fillets where the web meets the flanges, and three rectangles leave them out. For a rolled shape, design with the tabulated value; the steel beam sizes page lists it for every W shape, along with the W12x26 itself. Work it out by hand for a built-up or welded plate girder, which has no fillets.

Section modulus and weak axis

• Elastic section modulus: S_x = I_x / (d/2) = 201.0 / 6.1 = 33.0 in³ (AISC table: 33.4 in³). Bending stress is M/S_x.
• Weak axis: here the flanges are the tall rectangles. I_y = 2 × t_fb_f³/12 + h·t_w³/12 = 2 × 0.380 × 6.49³ / 12 + 11.44 × 0.230³ / 12 = 17.3 in⁴, matching the table. That's 12 times less than I_x, which is why I-beams are set with the web vertical, and why they buckle about the weak axis as columns.

Check it with the calculator

The cross section property calculator has an I shape: enter the flange width and thickness, web height and thickness, and it gives the area, centroid and both moments of inertia, with the formulas. To see what the moment of inertia does to a beam, put it in the beam calculator or try the beam deflection example.

Frequently asked questions

What is the formula for the moment of inertia of an I-beam?

I_x = b_fd³/12 − (b_f − t_w)h³/12, where h = d − 2t_f. It's the outer rectangle less the two empty spaces beside the web.

Why is my hand calculation lower than the steel table?

Rolled shapes have fillets between the web and flanges. They add a little area near the flanges, so tabulated values run 1 to 3% above three-rectangle calculations.

What units is moment of inertia in?

Length to the fourth power: in⁴ in US units, mm⁴ or cm⁴ in metric. 1 in⁴ = 416,231 mm⁴.