----- | ------------- | ------------ | ----------- | | W920x382 | 3,260,000,000 | 404,000,000 | 382 | | W610x101 | 774,000,000 | 46.3,000,000 | 101 | | W460x52 | 213,000,000 | 7,080,000 | 52.0 | | W310x45 | 104,000,000 | 6,320,000 | 45.2 | | W250x18 | 22.2,000,000 | 1,620,000 | 17.9 | | W150x13 | 6.77,000,000 | 739,000 | 13.0 |

HSS (Hollow Structural Section) shapes

Section Ix (mm^4) Iy (mm^4) J (mm^4)
HSS 305x305x13 161,000,000 161,000,000 215,000,000
HSS 254x152x6.4 50.2,000,000 21.9,000,000 33.8,000,000
HSS 203x203x9.5 53.5,000,000 53.5,000,000 74.0,000,000
HSS 89x89x4.8 1,310,000 1,310,000 2,070,000

Note: HSS sections have Ix approximately equal to Iy, and J is much larger than for open sections. This makes HSS ideal for columns and members subject to torsion.

Channel (C) shapes

Section Ix (mm^4) Iy (mm^4)
C380x74 168,000,000 4,570,000
C310x45 99.2,000,000 2,090,000
C250x45 61.2,000,000 1,500,000
C150x19 14.5,000,000 421,000

Channels have much lower Iy than Ix, making them weak about the minor axis. They need lateral bracing when used as beams.

How moment of inertia drives beam design

Deflection check

For a simply supported beam with uniform load:

delta = 5 * w * L^4 / (384 * E * I)

Where w is the load per unit length, L is the span, E is the elastic modulus, and I is the moment of inertia.

For a W460x52 spanning 8 m with a service load of 25 kN/m:

delta = 5 * 25 * 8000^4 / (384 * 200000 * 213000000)
delta = 5 * 25 * 4.096e15 / 1.635e16
delta = 31.3 mm

L/360 limit = 8000/360 = 22.2 mm. This beam fails the deflection check. You need a larger I.

A W530x82 (Ix = 475,000,000 mm^4) would give:

delta = 5 * 25 * 8000^4 / (384 * 200000 * 475000000)
delta = 14.0 mm  < 22.2 mm  -- OK

Bending stress

sigma = M * y / I

Where M is the bending moment, y is the distance from the neutral axis to the extreme fiber (h/2 for symmetric sections), and I is the moment of inertia.

The section modulus S = I / (h/2) combines these into a single property:

sigma = M / S

This is why beam tables list both I and S.

Lateral-torsional buckling

The critical moment for lateral-torsional buckling depends on Iy, J, and the warping constant Cw. A section with low Iy (like a W-shape loaded about its strong axis) is susceptible to LTB unless properly braced.

Using the moment of inertia calculator

Our moment of inertia calculator handles:

  1. Standard sections: Select from 500+ W, HSS, C, L, WT, and custom shapes across AISC, AS 4100, EN 1993, and CSA S16 databases.
  2. Built-up sections: Combine plates and standard shapes using the parallel axis theorem.
  3. Custom shapes: Enter dimensions for non-standard cross-sections.
  4. Unit support: Toggle between metric (mm) and imperial (in) units.

The calculator returns Ix, Iy, J, Sx, Sy, rx, ry, and the centroid location instantly.

Common mistakes

  1. Using the wrong axis: Ix and Iy are not interchangeable. A W310x45 has Ix = 104,000,000 mm^4 but Iy = 6,320,000 mm^4. Using Ix when the beam bends about the weak axis gives results 16x too stiff.

  2. Forgetting the parallel axis theorem: When adding a cover plate to a W-shape, you must include the Ad^2 term. Just adding the plate's own I underestimates the total.

  3. Confusing I and S: I is the second moment of area (mm^4). S is the section modulus (mm^3). S = I / c, where c is the distance to the extreme fiber. S is for stress checks, I is for deflection checks.

  4. Ignoring units: I values in mm^4 differ from in^4 by a factor of 2.54^4 = 41.6. Always verify units before comparing or using in formulas.

Related calculators

References