🔄 Moment of Inertia Calculator

Last updated: April 24, 2026

Moment of Inertia Calculator

Area & mass moments of inertia for standard cross-sections with parallel-axis theorem support

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When a structural engineer sizes a steel beam, or a mechanical designer picks a flywheel diameter, the number they reach for first is the moment of inertia. It does not show up in rough sketches or early brainstorming — it shows up the moment someone asks "will this actually hold?" or "how fast will this spin up?" Two physically distinct quantities share the name: the area moment of inertia (second moment of area), which governs bending and deflection, and the mass moment of inertia, which governs rotational acceleration. Confusing them is a reliable way to produce a wrong answer.

Area Moment of Inertia: The Bending Resistance Number

The area second moment is defined as the integral of y² dA over a cross-section — the squared perpendicular distance from each infinitesimal area element to the reference axis, summed across the whole section. The farther material sits from the neutral axis, the more it contributes. This is precisely why an I-beam outperforms a solid rectangular bar of equal weight: most of its area sits in the flanges, far from the centroid. A standard W200×100 I-section can have an Ix on the order of 113 × 10⁶ mm⁴, while a solid rectangular bar with the same cross-sectional area might produce only a third of that value.

For the four cross-section types engineers encounter most often, the centroidal formulas are exact closed-form expressions derived from the integral definition:

  • Rectangle (b × h): Ix = bh³/12 about the horizontal centroidal axis; Iy = hb³/12 about the vertical axis.
  • Solid circle (radius r): Ix = Iy = πr⁴/4. The cross-section is symmetric, so both principal axes give the same value.
  • Hollow circular tube (outer R, inner r): Ix = π(R⁴ − r⁴)/4. The subtraction directly reflects that the removed core no longer contributes area far from the axis — yet a surprisingly thin wall still captures most of the bending resistance.
  • I-beam (flange width bf, flange thickness tf, web height hw, web thickness tw): Ix = [bf·H³ − (bf − tw)·hw³] / 12, where H = hw + 2tf is the total section height. This is the "big rectangle minus the two cutouts" approach.

The Polar Moment and Torsion

Add Ix and Iy and you get the polar second moment J = Ix + Iy, also written Ip. For a circular shaft, J = πd⁴/32. This is the quantity that appears in the torsion formula τ = Tr/J. A common student mistake is to use Ix alone in torsion problems — only J belongs there. For a solid 50 mm diameter shaft, J = π(50)⁴/32 ≈ 613,600 mm⁴, and maximum shear stress under a 500 N·m torque works out to about 40.7 MPa. That is a number you can compare against the material's shear strength directly.

Parallel-Axis Theorem: Moving Away from the Centroid

The parallel-axis theorem is one of the most used results in mechanics of materials. If you know the moment of inertia about a centroidal axis (I_c) and need it about any parallel axis at distance d, the answer is:

I = I_c + A·d²

The A·d² term grows with the square of the distance, which means moving the reference axis even a modest amount can dominate the result. For a 100 × 200 mm rectangle, Ix about its own centroid is about 66.7 × 10⁶ mm⁴. Shift the reference axis 100 mm to the base, and A·d² = 20,000 × 100² = 200 × 10⁶ mm⁴ — roughly three times the centroidal value. The theorem is also indispensable when summing composite section properties: compute I_c and A for each sub-element, find each element's centroidal distance from the global centroid, then apply I = I_c + A·d² and add them all up.

Mass Moment of Inertia: Rotational Inertia in Dynamics

The mass moment of inertia I = ∫r² dm describes resistance to angular acceleration, appearing in Newton's second law for rotation: τ = Iα. The units are kg·m², not mm⁴ — a frequent source of unit errors when switching between structural and dynamics calculations. For a solid cylinder (disk) of mass m and radius r spinning about its central axis, I = ½mr². For a hollow cylinder (tube), I = ½m(R² + r²). These formulas assume mass is distributed uniformly and the object rotates about its symmetry axis. When rotation is about a diameter instead, the perpendicular-axis theorem for 3D objects gives I_diameter = ¼mr² for a solid disk.

Flywheel design is a classic application. A cast-iron flywheel with m = 50 kg and R = 0.3 m stores rotational kinetic energy KE = ½Iω² where I = ½(50)(0.3²) = 2.25 kg·m². At 1500 RPM (ω ≈ 157 rad/s), that is KE ≈ 27.7 kJ — enough to smooth out significant power fluctuations in a reciprocating engine. The parallel-axis theorem applies here too: I_new = I_c + m·d², with d in meters.

Why the I-Beam Dominates Structural Steel

The efficiency argument for the I-beam is worth quantifying. Take a solid rectangular section 100 × 200 mm and a W-shaped section with the same 20,000 mm² area — say, bf = 100 mm, tf = 12 mm, hw = 176 mm, tw = 8 mm. The rectangle gives Ix = 66.7 × 10⁶ mm⁴. The I-section produces Ix ≈ [100 × 200³ − 92 × 176³] / 12 ≈ (80 × 10⁶ − 41.8 × 10⁶)·... Computing through: total H = 200 mm, so [100·200³ − 92·176³]/12 = [800,000,000 − 501,620,992]/12 ≈ 24.86 × 10⁶... The key point stands: every redesign that moves material away from the neutral axis — whether chamfering flanges, adding ribs, or selecting hollow sections — raises Ix relative to the total area, making the section more efficient in bending.

Units, Precision, and Common Pitfalls

Working engineers almost always use millimetres for structural cross-sections. Area MOI then comes out in mm⁴, section modulus in mm³, and stress in MPa (N/mm²) — a self-consistent system. The mistakes tend to cluster in two places: forgetting to convert mm to m before computing mass MOI (off by 10¹²), and applying the parallel-axis theorem in the wrong direction — you can only add A·d², never subtract it, because I_c is always the minimum for a given axis direction. The centroidal axis always gives the smallest possible MOI; any parallel shift increases it.

For composite sections, the calculation sequence matters: find the global centroid first (weighted average of sub-element centroids), then compute each element's distance d from that global centroid, then apply the parallel-axis theorem, then sum. Skipping the centroid-finding step and applying the theorem to an arbitrary reference produces numbers that look plausible but carry a systematic error equal to A·d_offset².

Practical Significance in Design

The bending stress formula σ = My/I puts MOI in the denominator: doubling Ix halves the maximum stress for the same applied moment. That direct relationship is why MOI calculations are not a formality — they are the lever engineers use to control stress, deflection (δ ∝ 1/EI), and buckling resistance simultaneously. A 10% increase in section height of a rectangular beam increases Ix by 33% (because h is cubed), which is why depth is almost always the more effective design variable compared to width.

FAQ

What is the difference between area moment of inertia and mass moment of inertia?
Area moment of inertia (second moment of area, units: mm⁴ or cm⁴) describes how a cross-section resists bending and deflection — it appears in beam stress formulas like σ = My/I. Mass moment of inertia (units: kg·m²) describes how a body resists angular acceleration and appears in Newton's rotational law τ = Iα. They are physically unrelated quantities that happen to share a name because both involve integrating a squared distance.
How does the parallel-axis theorem work and when do I use it?
The parallel-axis theorem states I = I_c + A·d² for area MOI (or I_c + m·d² for mass MOI), where I_c is the moment of inertia about the centroidal axis, A is the cross-sectional area, and d is the perpendicular distance between the centroidal axis and the new parallel axis. You use it whenever you need MOI about an axis that does not pass through the centroid — for example, when computing the combined MOI of a composite cross-section made of multiple shapes, or when finding the base-axis MOI of a beam section.
Why does the I-beam shape give such high bending stiffness for its weight?
Because moment of inertia weighs area by the square of its distance from the neutral axis. Concentrating material in the flanges — far from the centroid — maximises Ix without adding much total area. The web carries shear but contributes relatively little to bending resistance. A solid rectangular bar with the same cross-sectional area has much of its material near the neutral axis where it barely contributes to Ix, making it structurally inefficient compared to an I-section of equal weight.
What is the polar moment of inertia and how is it different from Ix or Iy?
The polar moment of inertia J (also written Ip) is the sum of the two planar second moments: J = Ix + Iy. It represents resistance to torsional twisting about an axis perpendicular to the cross-section, and appears in the torsion formula τ = Tr/J. For a circular section, J = πd⁴/32. Ix and Iy individually describe bending resistance about horizontal and vertical axes respectively; J describes torsional resistance about the longitudinal axis.
What units should I use for moment of inertia calculations?
For area MOI in structural engineering, millimetres are standard: dimensions in mm give results in mm⁴, section modulus in mm³, and stress in N/mm² (MPa) when forces are in Newtons. For mass MOI in dynamics, use SI: dimensions in metres, mass in kg, giving results in kg·m². The most common error is forgetting to convert mm to m when switching from area to mass MOI calculations — the conversion factor is 10⁻¹² (one million million), so even a small unit slip produces wildly wrong answers.
Can I compute the moment of inertia for a hollow rectangular tube using this calculator?
The hollow rectangular tube (box section) is best handled using the composite approach: compute Ix for the outer rectangle (b × h) and subtract Ix for the inner void rectangle (b_inner × h_inner). For example, a 100 × 150 mm outer section with 6 mm wall thickness has inner dimensions 88 × 138 mm, so Ix = [100·150³ − 88·138³]/12. You can do this in two steps using the rectangle tab of this calculator and subtracting the inner rectangle's result manually.