An I-beam puts most of its area in the flanges, as far from the neutral axis as possible. That gives a high moment of inertia for little steel, which is why it is the standard shape for steel beams.
Formula
Worked example (millimetres)
Given
- Flange width b_f = 200 mm
- Flange thickness t_f = 15 mm
- Web thickness t_w = 10 mm
- Overall depth h = 400 mm
Steps
- Clear web height: h_w = h − 2t_f = 400 − 2 × 15 = 370 mm.
- Closed form: Iₓ = (b_f × h³ − (b_f − t_w) × h_w³) / 12 = (200 × 400³ − 190 × 370³) / 12 = 264,660,833 mm⁴ (26,466.1 cm⁴).
- Check by parallel axis — each flange: b_f × t_f³ / 12 + b_f × t_f × d², with d = (h − t_f) / 2 = 192.5 mm, gives 111,225,000 mm⁴.
- Web: t_w × h_w³ / 12 = 42,210,833 mm⁴. Total: 2 × 111,225,000 + 42,210,833 = 264,660,833 mm⁴, the same answer.
- Section modulus: Zₓ = Iₓ / (h/2) = 1,323,304 mm³; radius of gyration rₓ = √(Iₓ / A) = 165.2 mm.
| Area, A | 9,700 mm² |
|---|---|
| Moment of inertia, Iₓ | 264,660,833 mm⁴ |
| Section modulus, Zₓ | 1,323,304 mm³ |
| Radius of gyration, rₓ | 165.2 mm |
Worked example (inches)
Given
- Flange width b_f = 8 in
- Flange thickness t_f = 0.5 in
- Web thickness t_w = 0.3 in
- Overall depth h = 12 in
Steps
- Clear web height: h_w = h − 2t_f = 12 − 2 × 0.5 = 11 in.
- Closed form: Iₓ = (b_f × h³ − (b_f − t_w) × h_w³) / 12 = (8 × 12³ − 7.7 × 11³) / 12 = 297.9 in⁴.
- Check by parallel axis — each flange: b_f × t_f³ / 12 + b_f × t_f × d², with d = (h − t_f) / 2 = 5.75 in, gives 132.3 in⁴.
- Web: t_w × h_w³ / 12 = 33.28 in⁴. Total: 2 × 132.3 + 33.28 = 297.9 in⁴, the same answer.
- Section modulus: Zₓ = Iₓ / (h/2) = 49.66 in³; radius of gyration rₓ = √(Iₓ / A) = 5.13 in.
| Area, A | 11.3 in² |
|---|---|
| Moment of inertia, Iₓ | 297.9 in⁴ |
| Section modulus, Zₓ | 49.66 in³ |
| Radius of gyration, rₓ | 5.13 in |
Try your own dimensions
The free calculator gives area, centroid, Iₓ, I_y, section modulus and radius of gyration in mm or inches.
Open the Moment of Inertia Calculator →Frequently asked questions
- How do you calculate the moment of inertia of an I-beam?
- Treat it as the outer rectangle b_f × h minus the two empty rectangles beside the web: Iₓ = (b_f × h³ − (b_f − t_w) × h_w³) / 12, where h_w = h − 2t_f. The parallel-axis theorem on the two flanges and the web gives the same result.
- Why does an I-beam have a high moment of inertia?
- Moment of inertia weights each bit of area by the square of its distance from the axis. The flanges sit at the top and bottom, so they contribute most of Iₓ while the thin web adds little weight.
- Does this match the values in steel section tables?
- Closely, but tabulated rolled sections include the fillet radii where web meets flange, so their Iₓ is slightly higher than this idealised plate calculation.