Calculate the moment of inertia of an I-beam

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CAS Workspace

The CAS keeps the dimensions as named variables, so both methods use identical inputs and their agreement is a computed result, not a claim. The Moment of Inertia Calculator returns the same value from its I-beam shape and is the quick check once the method is clear.

Problem

An I-beam has flange width $b = 150$ mm, overall depth $H = 250$ mm, flange thickness $t_{f} = 10$ mm, and web thickness $t_{w} = 8$ mm. Find $I_{x}$ , the second moment of area about the horizontal centroidal axis, which is the axis the beam bends about under vertical load.

The web height between the flanges is $h_{w} = H - 2 t_{f} = 230$ mm. Every formula below uses it.

Method 1: subtract the pockets

Treat the section as a solid $b \times H$ rectangle with two rectangular pockets removed, one on each side of the web. Each pocket is $(b - t_{w}) /2$ wide and $h_{w}$ tall, and both are centered on the neutral axis, so their own second moments add with no transfer term.

1Outer rectangle: $\frac{b H ^{3}}{12} = \frac{150 \cdot 25 0 ^{3}}{12} = 195, 312, 500 mm^{4}$ .
2Both pockets together: $\frac{( b - t _{w} ) h _{w}^{3}}{12} = \frac{142 \cdot 23 0 ^{3}}{12} = 143, 976, 167 mm^{4}$ .
3Subtract: $I_{x} = \frac{b H ^{3} - ( b - t _{w} ) h _{w}^{3}}{12} = 195, 312, 500 - 143, 976, 167 = 51, 336, 333 mm^{4}$ .

Method 2: add the parts with the parallel axis theorem

Split the section into two flanges and one web. The web is centered on the neutral axis, so only its own term counts. Each flange sits away from the axis, so it needs the transfer term $A d^{2}$ .

1Distance from the neutral axis to a flange centroid: $d = \frac{H - t _{f}}{2} = 120$ mm.
2One flange: $\frac{b t _{f}^{3}}{12} + (b t_{f}) d^{2} = 12, 500 + 1500 \cdot 14, 400 = 21, 612, 500 mm^{4}$ .
3Web: $\frac{t _{w} h _{w}^{3}}{12} = \frac{8 \cdot 23 0 ^{3}}{12} = 8, 111, 333 mm^{4}$ .
4Total: $I_{x} = 2 (21, 612, 500) + 8, 111, 333 = 51, 336, 333 mm^{4}$ .

Why the two methods agree, and what each is good for

Both methods evaluate the same integral $\int y^{2} d A$ . Subtraction is shorter and works whenever the section is a rectangle with rectangular holes. The parallel axis method is the one to keep for sections that are not symmetric, such as a T or an angle, where the centroid has to be located first.

In the flange term, the own-axis part $b t_{f}^{3} /12 = 12, 500$ is under $0.1%$ of the flange total. Nearly all of the stiffness comes from placing flange area far from the neutral axis. That is the reason I-beams exist.

Check with the moment of inertia calculator

In the Moment of Inertia Calculator, choose the I-beam shape and enter flange width $0.15$ m, flange thickness $0.01$ m, web height $0.23$ m, and web thickness $0.008$ m. The calculator adds the two flange thicknesses to the web height to get the overall depth, so enter $0.23$ m there and not $0.25$ m. It returns $I_{x} = 5.134 \times 1 0^{- 5} m^{4}$ , the value found above.

The result feeds the next calculations directly. With $E = 200$ GPa, a 4 m simply supported span, and a 10 kN center load, the Beam Deflection Calculator gives $δ = P L^{3} / (48 E I) = 1.30$ mm. The Column Buckling Calculator takes the same $I$ for the Euler load.

Common Pitfall

The pocket term uses the web height $h_{w} = 230$ mm, not the overall depth $H = 250$ mm. Using $H$ in both terms gives $(150 - 142) \cdot 25 0^{3} /12 = 1.04 \times 1 0^{7}$ mm $^{4}$ , about one fifth of the right answer. A quick sanity check: $I_{x}$ must be less than the outer rectangle's $1.95 \times 1 0^{8}$ mm $^{4}$ and well above the web alone, $8.1 \times 1 0^{6}$ mm $^{4}$ .

Try a Variation

Increase the web thickness from $8$ mm to $12$ mm, a $50%$ increase, and recompute. Then restore $t_{w} = 8$ mm and increase the flange thickness from $10$ mm to $14$ mm. The web change raises $I_{x}$ by about $8%$ and the flange change by about $28%$ . Which one does the extra steel buy more of, and why?

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