Moment of Inertia Calculator

The Moment of Inertia Calculator covers both meanings of the term in one place — the area moment of inertia (second moment of area) used by structural engineers to predict how much a beam bends under load, and the mass moment of inertia used by mechanical and aerospace engineers to predict how a body responds to torque. Pick one of 15 ready-made shapes, type the dimensions in any familiar unit, watch the diagram redraw live, and read the moment of inertia together with the cross-section area, polar moment J, section modulus S, radius of gyration k, and a full step-by-step derivation. A parallel-axis-theorem field lets you shift the result to any axis parallel to the centroidal one with a single number.

How to Use This Moment of Inertia Calculator

1. Click Area Moment of Inertia if you are sizing a beam, or Mass Moment of Inertia if you are studying rotation. The shape gallery filters itself to show only the shapes that apply.
2. Tap a shape card — rectangle, circle, hollow tube, triangle, hollow box, I-beam, semicircle, thin rod, solid or hollow cylinder, solid or hollow sphere, rectangular plate. The required dimension fields show up and the diagram on the right adjusts.
3. Type the dimensions in mm, cm, m, in, or ft. For mass-mode shapes also type the total mass in kg, g, lb, t, or oz.
4. Pick the output unit — mm⁴ / cm⁴ / m⁴ / in⁴ / ft⁴ for area moment of inertia, or kg·m² / kg·cm² / g·cm² / lb·ft² / lb·in² for mass moment of inertia.
5. Optionally enter a parallel-axis offset distance. The calculator applies \(I' = I + A d^2\) (area) or \(I' = I + m d^2\) (mass) automatically.
6. Press Calculate to see the moment of inertia, the polar moment, section modulus, radius of gyration, a SVG diagram of the cross-section showing the centroid and axes, and the LaTeX derivation step by step.

What Makes This Calculator Different

Area vs Mass Moment of Inertia

The two quantities sound similar and share a symbol \(I\), but they live in different worlds. The area moment of inertia \(I_x = \int_A y^2 \,dA\) depends only on the shape of a cross-section — material does not enter. Its units are length to the fourth power, so mm⁴, cm⁴, m⁴, or in⁴. You use it in beam bending: a higher \(I_x\) means more resistance to a bending moment about the same axis. The mass moment of inertia \(I = \int r^2 \,dm\) depends on both how much mass there is and how that mass is distributed away from the rotation axis. Its units are mass × length², so kg·m², g·cm², lb·ft², or lb·in². You use it in rotational dynamics: \(\tau = I\alpha\) is the rotational form of Newton's second law.

Formulas for Common Shapes

Every shape supported by this calculator follows one of the formulas below. They are all about the centroidal axis indicated by the diagram; the parallel-axis theorem extends them to any parallel axis.

The Parallel-Axis Theorem

The formulas above all assume the axis passes through the centroid of the shape. To shift to any axis parallel to the centroidal one, add a single correction term:

\[ I_{x'} \;=\; I_x \;+\; A\,d^{2} \qquad \text{(area)} \qquad I' \;=\; I \;+\; m\,d^{2} \qquad \text{(mass)} \]

where \(d\) is the distance between the two parallel axes, \(A\) is the cross-section area, and \(m\) is the total mass. The calculator applies this automatically when you fill in the optional offset field.

Worked Example: I-Beam Section

A W12×40 wide-flange I-beam has total height H = 12 in, flange width B = 8 in, flange thickness t_f = 0.515 in, and web thickness t_w = 0.295 in. The web height is \(h_w = H - 2 t_f = 10.97\) in.

• \( I_x = B H^{3}/12 - (B - t_w)\,h_w^{3}/12 = 8 \cdot 12^{3}/12 - (8 - 0.295) \cdot 10.97^{3}/12 \approx 1152 - 847 \approx 305 \) in⁴.
• That matches the AISC-table value of 307 in⁴ within engineering tolerance.
• For a bending moment \(M = 50000\) lb·in, the maximum bending stress is \( \sigma = M c / I = 50000 \cdot 6 / 307 \approx 977 \) psi.

Worked Example: Flywheel

A solid steel flywheel of mass 20 kg and outer radius 0.30 m, rotating about its own central axis:

• \( I = m r^{2}/2 = 20 \cdot 0.30^{2} / 2 = 0.9\) kg·m².
• The torque needed to spin it from rest to 60 RPM (\(\omega = 6.28\) rad/s) in 5 seconds (\(\alpha = 1.26\) rad/s²) is \( \tau = I \alpha = 0.9 \cdot 1.26 \approx 1.13\) N·m.
• The rotational kinetic energy at 60 RPM is \( K = \tfrac{1}{2} I \omega^{2} = 0.5 \cdot 0.9 \cdot 6.28^{2} \approx 17.7\) J.

Section Modulus, Radius of Gyration, Polar Moment

For every area-mode shape the calculator also reports three companion quantities every engineering student eventually needs:

• Section modulus \(S = I_x / c\), where \(c\) is the distance from the centroid to the most stressed fiber. Used directly in the bending-stress formula \( \sigma = M / S \).
• Radius of gyration \(k = \sqrt{I / A}\) (area) or \(k = \sqrt{I / m}\) (mass). It is the radius at which all the area or mass could be concentrated at a single point and still produce the same I. Appears in Euler's column-buckling formula and in the rotational-equivalent of \(KE = \tfrac{1}{2} m v^{2}\) when written as \(KE = \tfrac{1}{2} m (k\omega)^{2}\).
• Polar moment of inertia \(J = I_x + I_y\), the area moment about the centroidal axis perpendicular to the cross-section. Drives torsional shear stress in a circular shaft: \(\tau = T r / J\).

Frequently Asked Questions

What is the difference between area and mass moment of inertia?
Area moment of inertia depends only on the cross-section shape and is used for beam bending — units are length⁴ (mm⁴, in⁴). Mass moment of inertia depends on both mass and how it is distributed around the rotation axis and is used for rotational dynamics — units are mass × length² (kg·m², lb·ft²). They share the symbol I but answer different physical questions.

How do I compute I for a rectangle?
About the centroidal x-axis, \(I_x = b h^{3}/12\). About the perpendicular centroidal y-axis it is \(I_y = h b^{3}/12\). The polar moment about the centroidal axis perpendicular to the plane is \(J = I_x + I_y\).

How do I compute I for a circle?
For a solid circle of diameter d, \(I = \pi d^{4}/64\) about any diameter and \(J = \pi d^{4}/32\) about the central perpendicular axis. For a hollow tube subtract the inner from the outer: \(I = \pi (D^{4} - d^{4})/64\).

What is the parallel-axis theorem?
It says \(I_{parallel} = I_{centroidal} + A d^{2}\) for area moments and \(I_{parallel} = I_{centroidal} + m d^{2}\) for mass moments, where d is the distance between the two parallel axes. This calculator applies it automatically when you fill in the offset field.

What is the moment of inertia of a solid sphere?
\(I = \tfrac{2}{5} m r^{2}\) about any diameter. A thin hollow sphere of the same mass and radius is \(\tfrac{2}{3} m r^{2}\) — bigger because more mass sits at the outer edge.

What is section modulus and how do I use it?
\(S = I_x / c\) where c is the distance from the centroid to the extreme fiber. Maximum bending stress is \(\sigma = M / S\). A larger S means the beam can carry a larger moment at the same allowable stress.

Why does the I-beam shape outperform a solid rectangle of the same area?
Because the area moment of inertia weights each piece of material by the square of its distance from the centroid. An I-beam puts most of its material in the flanges, far from the centroid, so each kg contributes much more to I than the same kg sitting near the centroid in a solid bar. That is why steel beams are nearly always shaped like an I.