Free AI Moment of Inertia Calculator

Calculate area moment of inertia (second moment of area) and mass moment of inertia for all standard shapes. Live annotated diagrams update as you type. Full step-by-step solutions, parallel axis theorem, and all unit systems.

13 Area Shapes 9 Mass Shapes Live Diagrams Step-by-Step

Moment of Inertia Calculator

Unit system:

Select a cross-section shape:

Parallel Axis Theorem - I = Ic + A·d²

mm
I_new = Enter d above

Theory & Concepts

Area Moment of Inertia (Second Moment of Area)

Units: length⁴ - mm⁴, cm⁴, in⁴, m⁴

The area moment of inertia $I$ of a plane figure about an axis measures how the area is distributed relative to that axis. It is a purely geometric property - it depends only on the shape of the cross-section and the position of the reference axis, not on any material property. A larger $I$ means the cross-section is more resistant to bending for the same material and load.

It appears directly in the flexure formula for bending stress, in beam deflection calculations via flexural rigidity $EI$, and in the Euler column buckling formula.

$$I_{xx} = \int_A y^2\,dA \qquad I_{yy} = \int_A x^2\,dA$$ $y$ = distance from x-axis; $x$ = distance from y-axis; units: length⁴
$$\sigma = \frac{My}{I} \qquad \delta_{max} = \frac{WL^3}{kEI} \qquad P_{cr} = \frac{\pi^2 EI}{L^2}$$ Key engineering formulas where area MOI appears: bending stress, beam deflection, Euler buckling

Why does shape matter so much? Because $I = \int y^2\,dA$, area placed far from the neutral axis contributes $y^2$ times more to $I$. Moving material from the centre to the flanges of an I-beam can increase $I$ by 5 to 10 times with the same total area - this is the fundamental reason I-beams are the dominant structural section.

Polar Moment of Area (J)

Units: length⁴ - same as area MOI

The polar moment of area $J$ is the second moment of area about an axis perpendicular to the cross-section (the z-axis, pointing out of the page). It governs torsional stiffness and torsional shear stress.

$$J = I_{xx} + I_{yy} = \int_A r^2\,dA \qquad \tau_{max} = \frac{Tr}{J}$$ Perpendicular axis theorem: $J = I_x + I_y$ (valid for any planar figure). Torsion formula: $T$ = torque, $r$ = radial distance

For a solid circle: $J = \pi R^4/2 = 2\,I_{xx}$. For a hollow circle: $J = \pi(R^4-r^4)/2$. The perpendicular axis theorem $J = I_x + I_y$ is valid for any planar shape lying in the xy-plane.

Section Modulus (Z)

Units: length³ - mm³, cm³, in³, m³

The elastic section modulus $Z$ combines the moment of inertia and the distance to the extreme fibre into a single property that directly gives the maximum bending stress for a given moment. It is the most convenient section property for beam design.

$$Z = \frac{I}{y_{max}} \qquad \sigma_{max} = \frac{M}{Z}$$ $y_{max}$ = distance from centroidal axis to outermost fibre. A larger Z → lower stress → stronger section.
  • Rectangle: $Z_{xx} = bh^2/6$ - the factor $1/6$ comes from $I_{xx}/(h/2) = (bh^3/12)/(h/2)$
  • Solid circle: $Z = \pi R^3/4$ - symmetric, so $y_{max} = R$
  • Hollow circle: $Z = \pi(R^4-r^4)/(4R)$
  • For asymmetric sections (T-section, L-angle, C-channel), $y_{max}$ is the larger of the two distances from the centroid to the top and bottom fibres, giving a lower $Z$ than a symmetric section of the same area.

Radius of Gyration (k)

Units: length - mm, cm, in, m

The radius of gyration $k$ (also written $r$ in some codes) is the distance from the reference axis at which the entire cross-sectional area (or mass) could be concentrated to produce the same moment of inertia. It converts a geometric property ($I$ and $A$) into a single length that captures the spread of material from the axis.

$$k = \sqrt{\frac{I}{A}} \quad\text{(area)} \qquad k = \sqrt{\frac{I}{m}} \quad\text{(mass)}$$ Units: length (mm, in). A larger $k$ means material is distributed further from the axis.

In column design, the slenderness ratio $\lambda = L/k$ is the key parameter in Euler buckling. A column with large $k$ (material far from the axis) is more resistant to buckling for the same length. For a rectangle about its centroid: $k_{xx} = h/\sqrt{12} \approx 0.289h$.

In dynamics, the mass radius of gyration $k = \sqrt{I/m}$ appears in the equation of motion for rotating bodies. For a solid cylinder spinning about its axis: $k_z = R/\sqrt{2} \approx 0.707R$.

Common confusion: The radius of gyration is NOT the distance to the centroid. It is the RMS (root mean square) distance of all area or mass elements from the axis, weighted by their areas or masses. It is always larger than the centroidal distance for non-uniform distributions.

Mass Moment of Inertia

Units: kg·m², kg·mm², lb·in², slug·ft²

The mass moment of inertia $I$ of a three-dimensional body about an axis quantifies its resistance to angular acceleration. It is the rotational equivalent of mass in Newton's second law, and determines how much torque is needed to change a body's rate of spin.

$$I = \int_V r^2\,dm = \int_V r^2\,\rho\,dV \qquad \tau = I\alpha$$ $r$ = perpendicular distance from rotation axis; $\rho$ = density; $\tau$ = torque; $\alpha$ = angular acceleration

Unlike area MOI, mass MOI depends on both the geometry and the mass distribution (density × volume). A hollow sphere and a solid sphere of the same outer radius have different mass MOI because the mass is distributed differently.

Area MOI vs Mass MOI: These are completely different physical quantities with different units. Area MOI (mm⁴) is used in structural beam calculations. Mass MOI (kg·m²) is used in dynamics and rotational mechanics. They cannot be converted into each other - they answer different questions about completely different phenomena.

Parallel Axis Theorem (Steiner's Theorem)

Works for both area MOI and mass MOI

The parallel axis theorem allows the MOI about any axis to be found from the centroidal MOI, provided the new axis is parallel to the centroidal axis. It is essential for calculating MOI of composite sections where each sub-component's centroid does not coincide with the composite centroid.

$$I = I_c + A\,d^2 \quad\text{(area)} \qquad I = I_c + m\,d^2 \quad\text{(mass)}$$ $I_c$ = MOI about centroidal axis; $d$ = perpendicular distance between the two parallel axes

Key property: The centroidal axis always gives the minimum I among all parallel axes. Every transfer away from the centroid adds $Ad^2 > 0$.

For composite sections (n components):

$$I_{total} = \sum_{i=1}^{n}\left(I_{c,i} + A_i\,d_i^2\right)$$ $d_i$ = distance from centroid of component $i$ to the composite centroidal axis. Subtract for holes/cutouts.

Area MOI - All Shapes Reference

All $I_{xx}$ and $I_{yy}$ values are about centroidal axes unless stated otherwise. Units: length⁴.

Rectangle b × h

bh
I_xx = bh³/12I_yy = hb³/12J = bh(b²+h²)/12Z_xx = bh²/6k_xx = h/√12 ≈ 0.289h

Solid circle radius R

R
I_xx = I_yy = πR⁴/4J = πR⁴/2Z = πR³/4k = R/2

Hollow circle outer R, inner r

Rr
I_xx = I_yy = π(R⁴−r⁴)/4J = π(R⁴−r⁴)/2Z = π(R⁴−r⁴)/(4R)

Right triangle base b, height h

bh
I_xx = bh³/36 (centroidal)I_yy = hb³/36I_x (about base) = bh³/12Centroid at h/3 from base

Ellipse semi-major a, semi-minor b

ab
I_xx = πab³/4I_yy = πa³b/4Area = πab

Semicircle radius R

RG
I_xx = R⁴(π/8 − 8/9π) centroidalI_xx (about diameter) = πR⁴/8I_yy = πR⁴/8ȳ = 4R/3π from flat edge

Quarter circle radius R

R
I_xx = I_yy = R⁴(π/16 − 4/9π)Centroid at (4R/3π, 4R/3π) from corner

Regular polygon n sides, circumradius R

R
I_xx = I_yy = nR⁴sin(2π/n)(3+cos(2π/n))/96J = nR⁴sin(2π/n)/24

I-section B, H, tf, tw

I_xx = (BH³ − (B−t_w)h_w³)/12h_w = H − 2t_f (web height)I_yy = (2t_f B³ + h_w t_w³)/12

T-section B, H, tf, tw

Composite method: y̅ = ΣAᵢyᵢ/ΣAᵢI_xx = Σ(I_c,i + Aᵢdᵢ²)I_yy = (t_f B³ + h_w t_w³)/12

C-channel B, H, tf, tw

I_xx = (BH³−(B−t_w)h_w³)/12I_yy by parallel axis (asymmetric)x̅ = centroid from web back face

L-angle B, H, t

Both I_xx and I_yy by parallel axisFind centroid first: y̅ and x̅I_xx = Σ(I_c,i + Aᵢdᵢ²)

Hollow rectangle B, H, wall t

I_xx = (BH³ − bᵢhᵢ³)/12I_yy = (HB³ − hᵢbᵢ³)/12bᵢ = B−2t, hᵢ = H−2t

Mass MOI - All Shapes Reference

Uniform density assumed. $I_z$ = spin axis, $I_x = I_y$ = transverse. Units: kg·m² (or chosen mass × length²).

Solid cylinder R, L

I_z = mR²/2 (about spin axis)I_x = I_y = m(3R²+L²)/12k_z = R/√2

Hollow cylinder R, r, L

I_z = m(R²+r²)/2I_x = m(3R²+3r²+L²)/12

Solid sphere radius R

I = 2mR²/5 (all axes equal)k = R√(2/5) = 0.632R

Hollow sphere R, r

I = 2m(R⁵−r⁵)/[5(R³−r³)]Thin shell limit: I = 2mR²/3

Rectangular prism a, b, c

I_z = m(a²+b²)/12I_x = m(b²+c²)/12I_y = m(a²+c²)/12

Thin rod length L

I (centre, transverse) = mL²/12I (end, transverse) = mL²/3I about own axis ≈ 0

Thin disk radius R

I_z = mR²/2 (spin)I_x = I_y = mR²/4

Solid cone R, h

I_z = 3mR²/10I_x = 3m(R²/4+h²)/5Centroid at h/4 from base

Torus R major, r minor

I_z = m(R²+3r²/4)I_x = I_y = m(5R²+4r²)/8

Unit Conversion - Area MOI (length⁴)

FromTo mm⁴To cm⁴To in⁴To m⁴
1 mm⁴11 × 10⁻⁴2.402 × 10⁻⁶1 × 10⁻¹²
1 cm⁴10,00010.024021 × 10⁻⁸
1 in⁴416,23141.62314.162 × 10⁻⁷
1 ft⁴8.631 × 10⁹8.631 × 10⁵20,7368.631 × 10⁻³
1 m⁴1 × 10¹²1 × 10⁸2.403 × 10⁶1

For mass MOI: 1 kg·m² = 10⁶ kg·mm² = 10⁴ kg·cm² = 23.73 slug·ft² = 3417.2 lb·in².

Frequently Asked Questions

1. What is area moment of inertia and what are its units?

Area moment of inertia (second moment of area) is a geometric property of a 2D cross-section that measures how area is distributed relative to an axis. Units are length⁴ - mm⁴, cm⁴, in⁴, or m⁴. It governs bending stiffness via σ = My/I, beam deflection via EI, and column buckling via Pcr = π²EI/L². It has absolutely nothing to do with material density or mass.

2. What is the difference between area MOI and mass MOI?

Area MOI (units: length⁴) is a purely geometric property of a cross-section shape. Mass MOI (units: kg·m²) is a dynamic property of a 3D body. Area MOI tells you how stiff a beam is in bending. Mass MOI tells you how hard a flywheel is to spin up. They cannot be converted into each other - they describe completely different physical phenomena.

3. Why does nothing appear in the output when I first open the calculator?

Make sure you click the Calculate button or type into one of the input fields. The calculator updates live as you type. If the output shows 'Invalid dimensions', check that inner dimensions (like the inner radius of a hollow circle) are smaller than the outer dimensions.

4. How does the parallel axis theorem work?

I = Ic + A·d² (area) or I = Ic + m·d² (mass). Ic is the MOI about the centroidal axis. d is the perpendicular distance between the centroidal axis and the new parallel axis. The centroidal axis always gives the minimum I among all parallel axes. Every transfer away from the centroid adds the positive term A·d². Use the parallel axis calculator in the right panel of the calculator section.

5. How do you calculate the MOI of an I-section?

Method 1 (recommended): treat the I-section as a large outer rectangle minus two side cutouts. Ixx = (BH³ − (B−tw)hw³)/12, where hw = H − 2tf. Method 2: three separate rectangles (top flange + web + bottom flange), apply the parallel axis theorem to each, then sum. The composite centroid of a symmetric I-section is at mid-height.

6. Why do I-beams have such high moment of inertia?

Because I = ∫y²dA, area far from the neutral axis contributes y² times more to I. Moving material from the centre to the flanges - far from the neutral axis - increases I dramatically with the same total area. An I-beam can have 5 to 10 times the I of a solid rectangular bar of the same cross-sectional area.

7. What is the polar moment of area J used for?

J = Ixx + Iyy is the second moment of area about an axis perpendicular to the cross-section. It governs torsional stiffness in the torsion formula τ = Tr/J. For a solid circle J = πR⁴/2 = 2Ixx. The perpendicular axis theorem J = Ix + Iy is valid for any plane figure.

8. What is the section modulus Z and why is it useful?

Z = I/ymax is the elastic section modulus (units: length³). The maximum bending stress is simply σmax = M/Z. It is the most convenient section property for beam design because it combines I and ymax into one number. For a rectangle Z = bh²/6. For a symmetric I-section Z = I/(H/2). For asymmetric sections (T, L, C), take ymax as the larger of the two distances from centroid to extreme fibre.

9. What is the radius of gyration and when is it used?

k = √(I/A) for area MOI (or √(I/m) for mass MOI). It is the distance from the axis at which all area (or mass) could be concentrated to give the same I. In structural engineering, the slenderness ratio L/k is the key parameter for column buckling - a higher k means better buckling resistance for the same length. For a rectangle: k_xx = h/√12 ≈ 0.289h.

10. How do you find the centroid of a composite section?

For a composite section with n components: ȳ = ΣAᵢyᵢ / ΣAᵢ where yᵢ is the distance from a reference line to the centroid of each component. For holes or cutouts, use negative areas. Once the composite centroid is found, apply the parallel axis theorem to each component: I = Σ(Ic,i + Aᵢdᵢ²) where dᵢ = distance from component centroid to composite centroid.

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