Polar Moment of Inertia: Formulas, Polar Section Modulus and a Worked Example

The polar moment of inertia, \( J \), is the second moment of area taken about an axis perpendicular to the section rather than one lying in it. It measures how far the material sits from the centroid in every direction at once, which makes it the property that governs twisting of a circular shaft: shear stress is \( \tau = T r / J \) and the angle of twist is \( \theta = T L / G J \). This page gives the formulas for the common shapes, the polar section modulus that goes with them, a checked worked example, and the one caveat that trips people up: for anything other than a circle or circular tube, \( J \) is not the torsion constant.

What the polar moment of inertia is

Take an element of area \( dA \) at radius \( \rho \) from the centroid. The polar moment of inertia sums \( \rho^2 \) over the whole section:

$$ J = \int \rho^2 \, dA $$

Since \( \rho^2 = x^2 + y^2 \), this splits into the two familiar second moments of area about the centroidal axes:

$$ J = I_x + I_y $$

That identity is the practical way to find \( J \) for any shape whose \( I_x \) and \( I_y \) you already know, and it is why \( J \) is sometimes written \( I_p \) or \( I_z \). It also means \( J \) is independent of how the x and y axes are rotated in the plane, because the sum \( I_x + I_y \) is invariant. The formulas for \( I_x \) and \( I_y \) are on our moment of inertia guide. Units are \( \text{mm}^4 \), the same as \( I \).

Polar section modulus

Just as the elastic section modulus \( Z = I / y_{max} \) converts bending moment to extreme fibre stress, the polar section modulus converts torque to maximum shear stress:

$$ Z_p = \frac{J}{r_{max}}, \qquad \tau_{max} = \frac{T}{Z_p} = \frac{T \, r_{max}}{J} $$

where
\( T \) is the applied torque
\( r_{max} \) is the distance from the centroid to the furthest point on the section
\( Z_p \) is the polar section modulus, in \( \text{mm}^3 \), also written \( W_p \) or \( W_t \)

These two relations are exact for solid and hollow circular sections, where plane sections stay plane under torsion and the shear stress grows linearly from zero at the centre to \( \tau_{max} \) at the surface.

Polar moment of inertia formulas

All about the centroid. The torsion caveat after this list applies to every non-circular shape.

Solid circle, diameter d

$$ J = \frac{\pi d^4}{32} = \frac{\pi r^4}{2}, \qquad Z_p = \frac{\pi d^3}{16} $$

Circular tube, outer D and inner d

$$ J = \frac{\pi \left( D^4 - d^4 \right)}{32}, \qquad Z_p = \frac{\pi \left( D^4 - d^4 \right)}{16 D} $$

For a thin tube of mean radius \( r_m \) and wall \( t \), \( J \approx 2 \pi r_m^3 t \).

Square, side a

$$ J = \frac{a^4}{6}, \qquad Z_p = \frac{J}{a / \sqrt{2}} = \frac{a^3 \sqrt{2}}{6} $$

Rectangle, b wide and h deep

$$ J = \frac{b h^3 + h b^3}{12} = \frac{b h \left( b^2 + h^2 \right)}{12} $$

\( r_{max} \) is half the diagonal, \( \sqrt{b^2 + h^2} / 2 \).

Hollow rectangle, B by H outside, b by h inside

$$ J = \frac{B H^3 - b h^3}{12} + \frac{H B^3 - h b^3}{12} $$

I or H section

Add the strong and weak axis values: \( J = I_x + I_y \). For the 300 mm deep welded section on our I beam moment of inertia page, \( I_x = 77.7 \times 10^6 \) and \( I_y = 5.64 \times 10^6 \), so \( J = 83.4 \times 10^6 \text{ mm}^4 \). Read the next section before using that number for anything.

Regular hexagon, side a

$$ J = \frac{5 \sqrt{3}}{8} a^4 \approx 1.083 \, a^4 $$

Equilateral triangle, side a

$$ J = \frac{\sqrt{3}}{48} a^4 \approx 0.0361 \, a^4 $$

For a general triangle use \( I_x + I_y \) about the centroid, with \( I = b h^3 / 36 \) about the centroidal axis parallel to each base.

Circle with a cut-out

A keyway, a square socket in a bolt head, or any hole is handled by subtraction: \( J = J_{outer} - J_{hole} \), provided both are centred on the same point. If the hole is off-centre, shift its own \( J \) with the parallel axis theorem first, \( J_{hole,c} = J_{hole} + A_{hole} \, e^2 \), where \( e \) is the offset.

Polar moment of inertia is not the torsion constant

This is the point that matters most. The relations \( \tau = T r / J \) and \( \theta = T L / G J \) assume the section does not warp, which is only true for circular sections. Every other shape warps out of plane when twisted, and its real torsional stiffness is set by a separate property, the torsion constant \( J_t \) (written \( I_T \) in Eurocode 3, \( J \) in Australian section tables, which is a naming collision to watch for). \( J_t \) is always less than \( I_x + I_y \), and for open thin-walled sections it is less by orders of magnitude:

• Solid rectangle, 100 mm by 50 mm. Polar: \( J = 100 \times 50 \times (100^2 + 50^2) / 12 = 5.21 \times 10^6 \text{ mm}^4 \). Torsion constant from the standard closed-form approximation \( J_t \approx b h^3 \left[ 1/3 - 0.21 (h/b)(1 - h^4 / 12 b^4) \right] \), with \( b = 100 \) and \( h = 50 \): \( 2.86 \times 10^6 \text{ mm}^4 \). The polar value overstates stiffness by 82 percent.
• The 300 mm welded I section above. Polar: \( 83.4 \times 10^6 \text{ mm}^4 \). Torsion constant from \( J_t = \sum b t^3 / 3 = (2 \times 150 \times 10^3 + 280 \times 8^3) / 3 = 148 \times 10^3 \text{ mm}^4 \). The polar value is about 560 times too high.
• Circular tube. \( J_t = J \) exactly. Closed sections with thick walls are the only non-circular shapes where the two are even in the same range, through the Bredt formula \( J_t = 4 A_m^2 / \oint (ds/t) \).

So use \( J = I_x + I_y \) when you need the polar second moment of area itself: for the polar radius of gyration \( r_p = \sqrt{J/A} \) in torsional buckling checks, for the stress at a bolt in a bolt group under torsion (the bolt group calculator is this calculation), and for shafts. Use the torsion constant for the twist and shear stress of any non-circular member.

Worked example: solid shaft and hollow shaft

The circular shaft torsion calculator runs this example live: enter the diameters, torque, length and shear modulus and read \( J \), \( Z_p \), the shear stress and the twist.

A 60 mm diameter steel shaft carries a torque of 2.0 kNm over a 1.5 m length. Shear modulus \( G = 80\,000 \) MPa.

$$ J = \frac{\pi \times 60^4}{32} = 1.272 \times 10^6 \text{ mm}^4, \qquad Z_p = \frac{\pi \times 60^3}{16} = 42.4 \times 10^3 \text{ mm}^3 $$

$$ \tau_{max} = \frac{T}{Z_p} = \frac{2.0 \times 10^6}{42.4 \times 10^3} = 47.2 \text{ MPa} $$

$$ \theta = \frac{T L}{G J} = \frac{2.0 \times 10^6 \times 1500}{80\,000 \times 1.272 \times 10^6} = 0.0295 \text{ rad} = 1.69^\circ $$

Now a 100 mm outside diameter tube with a 5 mm wall, so \( d = 90 \) mm:

$$ J = \frac{\pi \left( 100^4 - 90^4 \right)}{32} = 3.376 \times 10^6 \text{ mm}^4, \qquad \tau_{max} = \frac{2.0 \times 10^6 \times 50}{3.376 \times 10^6} = 29.6 \text{ MPa} $$

The tube has a cross-sectional area of \( 1492 \text{ mm}^2 \) against \( 2827 \text{ mm}^2 \) for the solid bar, so it uses 53 percent of the material and carries 37 percent less peak stress. Material near the centre of a solid shaft is at low stress and contributes little to \( J \), which is the same logic that puts an I beam's steel in its flanges. The same relations can be rearranged to size a shaft for an allowable stress or a twist limit; \( G \) for steel follows from \( E \) and Poisson's ratio, as in our shear modulus relations page.

Codes and where J appears

• Steel. EN 1993-1-1 Clause 6.2.7 covers torsion, using the torsion constant \( I_T \) and, for open sections, the warping constant \( I_w \). AS 4100 has no dedicated torsion design clause; Australian practice takes \( J \) (the torsion constant) and \( I_w \) from the section tables and designs by elastic analysis, with the warping terms entering the lateral torsional buckling moment \( M_o \) in AS 4100 Clause 5.6.1.
• Concrete. AS 3600 and EN 1992-1-1 Clause 6.3 both design for torsion with a thin-walled tube model rather than the polar moment of inertia. The uncracked torsional stiffness in AS 3600 uses a torsion constant of the form \( 0.4 x^2 y \) for rectangles, which is a torsion constant, not \( J \). Our EC2 combined bending, shear and torsion template works the Eurocode side.
• Mechanical. Shafts, axles and couplings are the one place \( J \) and the torsion constant coincide, and the formulas above apply directly.

Run it in CalcTree

The circular shaft torsion calculator does exactly this for a solid or hollow shaft. Shaft sizing is a handful of formulas with units, so it suits a live, checkable CalcTree page: enter the diameters, torque, length and material, and have \( J \), \( Z_p \), stress and twist update together with the working on show. The circular hollow section properties template is a starting point for the hollow shaft, and the template library has the torsion design calculators for concrete to ACI 318 and Eurocode 2.