Polar Moment of Inertia Calculator
Calculate the polar moment of inertia (J) and polar section modulus (Zp) for four common cross-section shapes: solid circle, hollow circle, solid square, and rectangle. Switch between metric and imperial units. Results update instantly as you type, and the step-by-step panel shows the full formula with your numbers substituted in.
What is the polar moment of inertia?
The polar moment of inertia (symbol J, sometimes called the second polar moment of area) is a geometric property that quantifies how the area of a cross-section is distributed around a central (polar) axis perpendicular to its plane. A large J means the section is far from twisting easily: more area located farther from the center contributes disproportionately because the formula integrates r² dA over the entire section. J is the key ingredient in the classical torsion formula τ = T×c/J, where T is the applied torque, c is the distance from the center to the outermost fiber, and τ is the resulting shear stress.
How to use this calculator
Select your unit system (metric or imperial), then choose the cross-section shape that matches your shaft or beam. Enter the required dimensions: diameter for a solid circle, outer and inner diameters for a hollow tube, side length for a square, or width and height for a rectangle. J and the polar section modulus Zp update instantly. The step-by-step panel below the result shows the exact formula with your numbers substituted in. To find maximum torsional shear stress once you have J, divide the applied torque T by the polar section modulus Zp, or equivalently compute T×c/J.
Polar moment of inertia vs. second moment of area
Engineers sometimes conflate the polar moment of inertia with the area moments of inertia Ix and Iy (second moments of area about the x- and y-axes). They are related by the perpendicular axis theorem: J = Ix + Iy. For a circle, Ix = Iy = πR⁴/4, so J = πR⁴/2, which equals πD⁴/32 in terms of diameter. In structural analysis, Ix and Iy govern bending stiffness, while J governs torsional stiffness. Both are measured in length⁴ units (mm⁴ or in⁴).
Why hollow circular shafts are efficient in torsion
Material near the center of a solid circular shaft contributes very little to J because of the r² weighting. Removing that low-contribution core to create a hollow shaft reduces weight without much loss of J. This is why driveshafts, drive axles, and structural tubing are typically hollow: a hollow shaft with the same outer diameter as a solid one retains most of its torsional stiffness at a fraction of the mass. The exact saving depends on the wall thickness: the formula J = π(D⁴ - d⁴)/32 shows that reducing d has a diminishing return as d approaches D.
Polar moment of inertia formulas by cross-section
| Shape | J formula | c (outermost fiber) | Zp formula |
|---|---|---|---|
| Solid circle | J = πD⁴/32 | R = D/2 | Zp = πD³/16 |
| Hollow circle | J = π(D⁴ - d⁴)/32 | R = D/2 | Zp = π(D⁴ - d⁴)/(16D) |
| Solid square | J = a⁴/6 | c = a√2/2 | Zp = J/c |
| Rectangle | J = (b⁴ + d⁴)/12 | c = √(b²+d²)/2 | Zp = J/c |
Standard formulas for common cross-section shapes. D = outer diameter, d = inner diameter, a = square side, b/d = rectangle width/height.
Frequently asked questions
What units does J use?
The polar moment of inertia has units of length to the fourth power: mm⁴ in the metric system and in⁴ in the U.S. customary system. The polar section modulus Zp has units of length cubed (mm³ or in³). Always check that your torque is in consistent units (N·mm or lbf·in) before computing shear stress.
What is the difference between J and the area moment of inertia?
The area moments of inertia Ix and Iy describe resistance to bending about the x- and y-axes. The polar moment of inertia J = Ix + Iy describes resistance to twisting about the axis perpendicular to the cross-section. All three are in length⁴ units, but J is used with the torsion formula while Ix/Iy are used with the bending formula σ = M×y/I.
How do I use J to find shear stress?
Apply the torsion formula: τ = T × c / J, where T is the applied torque, c is the outer radius (the distance from the center to the farthest point of the cross-section), and J is the polar moment of inertia. Equivalently, τ = T / Zp where Zp = J/c is the polar section modulus. The result is the maximum shear stress at the outer fiber.
Why is J for a hollow circle almost as large as for a solid one?
Because J weights area by the square of its distance from the center (r² dA). Material near the center contributes very little. Removing the low-r core barely reduces J, so a hollow section retains most of its torsional stiffness while saving significant weight and material cost.
Can I use this for non-circular sections like I-beams?
Circular sections are the most common application and the formulas here are exact. For non-circular sections such as I-beams, channels, or T-sections, the torsional behavior is more complex: open thin-walled sections have a much lower effective torsional stiffness than their J alone would suggest because of warping. Those sections require a separate St. Venant torsion constant (also called J or K) computed by a different method.