Moment of inertia calculator (second moment of area)
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Result
| Part | A (mm²) | y (mm) | A·y (mm³) | I₀ (mm⁴) | d (mm) | A·d² (mm⁴) |
|---|---|---|---|---|---|---|
| flange | 3000 | 190 | 570000 | 100000 | 54.55 | 8925620 |
| web | 3600 | 90 | 324000 | 9720000 | −45.45 | 7438017 |
Calculate the second moment of area (area moment of inertia) of common cross-sections: solid and hollow rectangles, solid and hollow circles, symmetric I-beams, T-sections and C-channels. You get the area, the position of the centroid, Ix and Iy about the centroidal axes, the elastic section moduli and the radii of gyration. For composite shapes the steps apply the parallel axis theorem part by part, exactly as you would by hand.
Step-by-step solution
Split the section into rectangles
Centroid
Ix: parallel axis theorem I = I₀ + A·d²
Iy: parallel axis theorem I = I₀ + A·d²
Section modulus and radius of gyration
Key formulas
- Centroid
- ȳ = ΣAiyi/ΣAi
- Rectangle
- I = bh³/12
- Parallel-axis theorem
- I = Ic + A·d²
- Composite section
- I = Σ(Ic,i + Aidi²)
How to find the moment of inertia of a composite section
- Split the section into rectangles and treat holes as negative areas. Note the area A and the centroid height y of each part, measured from a reference line such as the bottom edge.
- Find the centroid of the whole section: ȳ = ΣA·y/ΣA.
- For each part, take its own moment of inertia about its centroid, b·h³/12 for a rectangle, and add A·d², where d is the distance from the centroid of the part to the centroid of the section.
- Add up the parts (subtract the holes) to get Ix. Repeat with the horizontal distances for Iy.
- Section modulus S = I/y_max, where y_max is the distance to the furthest fibre; radius of gyration r = √(I/A).
Questions and answers
Is this the same as the mass moment of inertia?
No. This is the second moment of area, in mm⁴ or in⁴, which describes how the shape of a cross-section resists bending. The mass moment of inertia, in kg·m², describes how a body resists rotation and depends on its mass.
Why does a T-section have two section moduli?
Its centroid is not at mid-height, so the distance to the top fibre differs from the distance to the bottom fibre. Each gives its own S = I/y, and the smaller one gives the larger bending stress.
Why is it b·h³/12 and not b³·h/12?
The axis matters. For bending about the horizontal x-axis, the height h is the dimension perpendicular to the axis and it appears cubed: Ix = b·h³/12. About the vertical axis the roles swap: Iy = h·b³/12.
What is the parallel axis theorem?
I = I₀ + A·d²: the moment of inertia about any axis equals the moment of inertia about a parallel axis through the centroid plus the area times the squared distance between the axes. It moves the parts of a composite section to a common axis.
Are the root fillets of rolled steel sections included?
No. Flanges and webs are treated as plain rectangles, so the results are slightly lower than the values in steel tables, which include the fillets between web and flange.
Which units can I use?
Millimetres, with results in mm², mm³ and mm⁴, or inches, with results in in², in³ and in⁴. Switch with the units selector.
Practise it yourself
Practise Centroid & second moment of area
A calculator gives the answer. In the exam you need the method: practise it with new problems, hints and worked solutions.