MechSimulator

Moment of Inertia of a Hollow Rectangle

Box & RHS Sections • Iₓ, Iₖ & Section Modulus

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Moment of Inertia of a Hollow Rectangle — Formula and Worked Example

A rectangular hollow section is simply the outer rectangle minus the inner void. Unlike an I-beam it is closed, which gives it far greater torsional stiffness — the usual reason for choosing RHS over an open section.

Formulas

QuantityExpression
AreaA = BH − bh
Second moment about xIₓ = (BH³ − bh³) / 12
Second moment about yIₖ = (HB³ − hb³) / 12
Section modulusSₓ = 2Iₓ / H
Radius of gyrationrₓ = √(Iₓ/A)

Subtraction only works because both rectangles share the same centroid. If the void is offset, you must use the parallel-axis theorem on each part separately rather than subtracting directly — a frequent mistake with unsymmetric box sections.

Worked Example — An 80 × 120 mm box section with a 10 mm wall

QuantityValue
B × H (outer)80 × 120 mm
b × h (inner)60 × 100 mm
wall thickness10 mm
A3,600 mm²
Iₓ6.520 × 10⁶ mm⁴
Iₖ3.320 × 10⁶ mm⁴
Sₓ108.7 × 10³ mm³
rₓ42.6 mm

Values computed by the calculator above. Change any dimension and every property updates live.

What is the moment of inertia of a box section?

Ix = (BH³ − bh³)/12, where B and H are the outer width and depth and b and h the inner ones. This works because the outer and inner rectangles are concentric, so their centroidal axes coincide and the second moments can simply be subtracted.

Why choose RHS over an I-beam?

A closed box has vastly higher torsional stiffness than an open section of similar size, and it has no weak axis in the way an I-beam does, since Iy is comparatively large. It is also easier to protect and looks cleaner in exposed steelwork. The trade-off is that the inside is hard to inspect and connect to.

Does the subtraction method always work?

Only when the void shares the centroid of the outer shape. For an offset or unsymmetric void, compute each rectangle's contribution about the true combined centroid using the parallel axis theorem, subtracting the void's area and its transfer term.

Common Square and Rectangular Hollow Sections

Properties for typical SHS and RHS sizes, quoted as depth × width × wall. Iₓ is about the horizontal axis (the strong axis when the section stands upright):

Section (mm)A (mm²)Iₓ (10⁶ mm⁴)Iₖ (10⁶ mm⁴)Sₓ (10³ mm³)
50 × 50 × 35640.2080.2088.3
60 × 60 × 48960.4710.47115.7
80 × 80 × 51,5001.4131.41335.3
100 × 50 × 41,1361.4410.47428.8
120 × 80 × 51,9003.7561.97662.6
150 × 100 × 62,8568.8524.663118.0

For a square section Iₓ = Iₖ, which is exactly why SHS is favoured for columns — there is no weak axis to buckle about. A rectangular section deliberately trades that away for more depth, and therefore more Iₓ, when it is used as a beam.

Why RHS Beats an I-Beam in Torsion

An I-beam is an open section: under torque its thin elements twist almost independently, and its torsion constant is only about Σbt³/3 — very small. A box is closed, so shear flows continuously around the perimeter and the Bredt–Batho relation applies, J ≈ 4Aᶜ²t/p, where Aᶜ is the area enclosed by the wall centreline. The result is routinely one to two orders of magnitude stiffer in torsion for a comparable section. That, rather than bending, is usually the reason to specify RHS.

When does the subtraction method fail?

Only when the void is not concentric with the outer shape. Subtracting (BH³ − bh³)/12 quietly assumes both rectangles share a centroidal axis. If the bore is offset — an eccentric box, or a section with an off-centre slot — you must compute the combined centroid first, then apply Iᶜ + Ad² to the outer shape and subtract the void's own Iᶜ + Ad² about that same axis.

Other Cross-Sections

The same calculator handles all eight standard sections:

For the full section-property reference and the parallel-axis theorem, see the Moment of Inertia Calculator hub.