Polar Moment of Inertia Calculator

Calculate the polar second moment of area for solid circular, hollow circular, and rectangular cross-sections.

Cross-section calculator
Choose a shape and enter its dimensions in metres.

About polar moment of inertia

The polar second moment of area, commonly written J, describes how cross-sectional area is distributed about an axis perpendicular to the section. It has dimensions of length to the fourth power. A larger value means more area lies farther from the polar axis, which generally improves a circular shaft's resistance to torsional stress and angular twist. This calculator accepts dimensions in metres and returns J in metres to the fourth power. For a solid circle of diameter d, J = πd⁴ / 32. For a hollow circle with outer diameter D and inner diameter d, J = π(D⁴ - d⁴) / 32. The fourth-power dependence explains why moving material outward is so effective: a hollow shaft can retain substantial torsional performance while using less material than a solid shaft of similar outside diameter. The inner diameter must remain smaller than the outer diameter, and all dimensions must use the same unit. For any planar area, the polar area moment about a point equals the sum of the two perpendicular area moments through that point: J = Ix + Iy. For a rectangle centered on the axis, this gives J = bh(b² + h²) / 12. However, this rectangular polar area moment is not generally the same as the Saint-Venant torsional constant used to calculate twist and torsional shear in a noncircular bar. The distinction matters because noncircular sections warp under torsion. The calculator labels the rectangular result as a polar area moment and should not be substituted blindly into a shaft-twist formula. For circular shafts, elastic torsion relationships include maximum shear stress τmax = Tr / J and angle of twist θ = TL / (GJ), where T is torque, r is outside radius, L is shaft length, and G is shear modulus. These relations assume homogeneous linear-elastic material, small deformation, uniform circular geometry, and loading about the centroidal longitudinal axis. Keyways, splines, cracks, abrupt diameter changes, and stress concentrations require additional analysis. Choose dimensions from the actual load-carrying section and preserve unit consistency. If dimensions are entered in millimetres, the numerical result is in millimetres to the fourth power rather than metres to the fourth power; convert before comparing with an SI calculation. Use this result for coursework, preliminary sizing, and formula checks. Detailed machine design should also evaluate strength, fatigue, deflection, manufacturing tolerances, stress concentrations, and the correct torsion theory for the chosen shape.

Polar moment examples

The same formulas apply at any consistent length scale.

Cross-sectionPolar momentUse
Solid circle, diameter 0.10 m9.817 × 10⁻⁶ m⁴Centroidal polar moment for a solid round shaft.
Hollow circle, outer 0.10 m, inner 0.05 m9.204 × 10⁻⁶ m⁴Removing central material causes a relatively small reduction.
Rectangle, width 0.10 m, height 0.20 m8.333 × 10⁻⁵ m⁴This is Ix plus Iy, not the rectangle's torsional constant.

How to use the calculator

  1. Select Solid Circle, Hollow Circle, or Rectangle.
  2. Measure the required cross-section dimensions and convert them to metres.
  3. Enter diameter values for circles or width and height for a rectangle.
  4. Select Calculate J and interpret the result according to the selected shape.

Polar moment of inertia FAQ

What units does polar moment of inertia use?

It uses length to the fourth power, such as m⁴ or mm⁴. The calculator expects metres and therefore displays m⁴.

Is polar moment of inertia the same as mass moment of inertia?

No, polar area moment depends only on cross-sectional geometry and uses length-to-the-fourth units. Mass moment of inertia describes rotational mass distribution and includes mass units.

Why are hollow shafts efficient?

Material far from the axis contributes strongly because diameter appears to the fourth power. Removing lightly contributing central material can reduce mass without a proportional loss of circular-shaft torsional stiffness.

Can I use the rectangle result in the circular shaft twist formula?

Not directly in general, because a rectangle warps and its Saint-Venant torsional constant differs from its polar area moment. Use a noncircular torsion formula for twist and stress.

Which axis does this calculator use?

It uses the centroidal axis perpendicular to the cross-section. Results about another point require the parallel-axis theorem before summing the planar moments.