A moment of inertia calculator for beam sections
The area moment of inertia, also called the second moment of area, tells you how well a cross-section resists bending. This moment of inertia calculator computes it for the shapes engineers meet most: rectangle, hollow box, solid circle, circular tube, I-beam or H-section, T-section and C-channel. It also gives the area, centroid, section modulus, radius of gyration, polar moment and the same values converted to mm⁴, cm⁴ and in⁴.
For anything else there is a composite mode. List rectangles and circles with their centres, mark holes with a trailing minus sign, and the calculator applies the parallel-axis theorem to the whole assembly, reporting the product of inertia as a check for symmetry.
The results are exact formulas for the geometry you enter, with sharp corners. Rolled beams have root radii and tapered flanges, so for catalogue sections use the manufacturer’s tabulated properties, and always have structural designs checked by a qualified engineer.
How to use the Moment of Inertia (Section Properties) Calculator
- Choose a shape and the length unit.
- Enter the dimensions that appear for that shape: width, height, flange and web thickness, diameter or wall thickness.
- For a composite shape, type one part per line, for example
r 100 20 50 10for a 100 × 20 rectangle centred at x = 50, y = 10. - Read Ixx, Iyy, section modulus and radius of gyration, and open the working to see each part’s contribution.
Moment of Inertia (Section Properties) Calculator features
Eight shapes
Rectangle, box, circle, tube, I, T and C sections, plus composite assemblies.
Parallel-axis working
Each part’s own inertia and its A d² term are listed.
All the properties
Area, centroid, Ixx, Iyy, Sx, Sy, r, J and unit conversions.
Holes supported
Composite parts ending in a minus sign remove material.
Formulas used
For a rectangle of width b and height h the moment of inertia about its own centroidal axis is I = b h³ ÷ 12, and for a circle of diameter d it is I = π d⁴ ÷ 64. For a section made of several parts, the parallel-axis theorem gives I = Σ (Iown + A d²), where d is the distance from each part’s centroid to the centroid of the whole section, found from ȳ = Σ A y ÷ Σ A.
The section modulus is S = I ÷ c, with c the distance from the neutral axis to the outermost fibre; the smaller of the top and bottom values governs. The radius of gyration is r = √(I ÷ A). Example: a 100 × 200 mm rectangle has I = 100 × 200³ ÷ 12 = 66.67 × 10⁶ mm⁴ and S = 667 × 10³ mm³. A T-section 100 mm wide and deep with 20 mm flange and stem has its centroid 67.8 mm above the base and Ixx = 3.14 × 10⁶ mm⁴.
Reading the answer
- Ixx controls bending about the horizontal axis (loads acting vertically); Iyy controls bending sideways.
- Bending stress is σ = M ÷ S and deflection falls in proportion to 1 ÷ I, so doubling the depth of a rectangle makes it eight times stiffer.
- Torsion of non-circular sections needs the torsion constant, which differs from the polar moment J and is not calculated here.
- Local buckling, shear and connection details are outside this calculator.
Moment of Inertia (Section Properties) Calculator FAQ
How do I calculate the moment of inertia of an I-beam?
Take the outer rectangle b × h and subtract the two side voids: I = (b h³ − (b − tw)(h − 2 tf)³) ÷ 12. For a 100 × 200 mm beam with 10 mm flanges and a 6 mm web that is 20.98 × 10⁶ mm⁴.
What is the moment of inertia of a rectangle?
I = b h³ ÷ 12 about the axis through its centroid parallel to the width b, with h the dimension perpendicular to that axis.
How do I find the moment of inertia of a tube?
Subtract the inner circle from the outer: I = π (D⁴ − d⁴) ÷ 64, with d = D − 2t.
Is this the same as mass moment of inertia?
No. This is the area moment of inertia of a cross-section, in length to the fourth power. Mass moment of inertia, in kg·m², is used for rotating bodies.
Which unit does the answer use?
The length unit you chose, to the fourth power. The table also converts Ixx to mm⁴, cm⁴ and in⁴.
Why does a composite result show Ixy?
If the product of inertia is not zero the section is unsymmetrical about the x and y axes, so those are not the principal axes.