Area Moment of Inertia Calculator
Second moment of area and section modulus for common beam cross-sections.
Extreme-fibre distance c = 50 mm · section modulus S = I ÷ c = 83330 mm³. This is the area (second) moment of area, not the mass moment of inertia.
The area moment of inertia (second moment of area) of a rectangle is I = b·h³ ÷ 12 and of a solid circle is I = π·d⁴ ÷ 64. For b = 50 mm and h = 100 mm, I ≈ 4.17×10⁶ mm⁴. This is the geometric stiffness of a section, not the mass moment of inertia used for rotation.
What the area moment of inertia is
The area moment of inertia, also called the second moment of area, measures how a cross-section’s material is spread about its centroidal neutral axis. It has units of length to the fourth power (mm⁴ or m⁴) and governs how much a beam resists bending: a larger I means less deflection and lower bending stress for the same load. Because area far from the neutral axis counts most, moving material outward — a deeper section or a tube wall — raises I sharply.
Do not confuse it with the mass moment of inertia (units kg·m²), which describes resistance to angular acceleration in dynamics. This tool computes the purely geometric area quantity that feeds beam bending and deflection.
Rectangle (base b, height h) and solid circle (diameter d), about the horizontal centroidal axis · units mm → mm⁴
Worked example
Take a solid rectangular section 50 mm wide and 100 mm deep, bending about the horizontal axis so the 100 mm dimension is the depth h.
- 1 Identify the cross-section and its bending axis. A rectangle uses I = b·h³ ÷ 12 about the horizontal centroidal axis, where h is the depth in the direction of bending.
- 2 Put all dimensions in the same length unit. Here b = 50 mm and h = 100 mm, so the result comes out in mm⁴.
- 3 Substitute into the formula. I = 50 × 100³ ÷ 12 = 50 × 1 000 000 ÷ 12 = 50 000 000 ÷ 12.
- 4 Read the second moment of area. I = 4 166 667 mm⁴ ≈ 4.17×10⁶ mm⁴.
- 5 Get the section modulus for bending stress. The extreme fibre is c = h ÷ 2 = 50 mm, so S = I ÷ c = 4 166 667 ÷ 50 ≈ 83 300 mm³.
Second moment of area and section modulus by cross-section
All about the horizontal centroidal (neutral) axis. c is the distance to the extreme fibre; the section modulus S = I ÷ c feeds bending stress σ = M ÷ S.
| Cross-section | Moment of inertia I | Extreme fibre c | Section modulus S = I ÷ c |
|---|---|---|---|
| Rectangle (b × h) | b·h³ ÷ 12 | h ÷ 2 | b·h² ÷ 6 |
| Solid circle (d) | π·d⁴ ÷ 64 | d ÷ 2 | π·d³ ÷ 32 |
| Hollow circle / tube (D, d) | π·(D⁴ − d⁴) ÷ 64 | D ÷ 2 | π·(D⁴ − d⁴) ÷ (32·D) |
| Hollow rectangle / box (b, h, bᵢ, hᵢ) | (b·h³ − bᵢ·hᵢ³) ÷ 12 | h ÷ 2 | (b·h³ − bᵢ·hᵢ³) ÷ (6·h) |
How to use I and S
Feed I straight into a deflection calculation — for a simply supported beam under a central load, δ = F·L³ ÷ (48·E·I), so doubling I halves the deflection. Use the section modulus S for strength: the maximum bending stress is σ = M ÷ S, where M is the bending moment. A section survives when that stress stays below the material’s allowable limit.
The strongest lesson is the cube on height. Because I scales with h³, orienting a plank on edge instead of flat, or adding depth, buys far more bending resistance than adding width. Hollow tubes and box sections exploit the same idea: they keep material at the outer fibres where it does the most work while shedding the near-axis material that contributes little.