1. Definition

A schematic showing how the polar moment of inertia is calculated for an arbitrary shape about an axis ${\displaystyle O}$. Where ${\displaystyle \rho }$ is the radial distance to the element ${\displaystyle dA}$.

The equation describing the polar moment of area is a multiple integral over the cross-sectional area, ${\displaystyle A}$, of the object.

${\displaystyle J={\iint }_A{r}^{2}dA}$ where ${\displaystyle r}$ is the distance to the element ${\displaystyle dA}$.

Substituting the ${\displaystyle x}$ and ${\displaystyle y}$ components, using the Pythagorean theorem: ${\displaystyle J={\iint }_A(x^2+y^2)dxdy}$ ${\displaystyle J={\iint }_A{x}^{2}dxdy+{\iint }_A{y}^{2}dxdy}$

Given the planar second moments of area equations, where: ${\displaystyle I_x={\iint }_A{y}^{2}dxdy}$ ${\displaystyle I_y={\iint }_A{x}^{2}dxdy}$

It is shown that the polar moment of area can be described as the summation of the ${\displaystyle x}$ and ${\displaystyle y}$ planar moments of area, ${\displaystyle I_x}$ and ${\displaystyle I_y}$ $\text{Undefined control sequence therefore}$

This is also shown in the perpendicular axis theorem.^[1] For objects that have rotational symmetry,^[2] such as a cylinder or hollow tube, the equation can be simplified to: ${\displaystyle J=2I_x}$ or ${\displaystyle J=2I_y}$

For a circular section with radius ${\displaystyle R}$: ${\displaystyle I_z={\int }_0^{2\pi }{\int }_0^R{r}^2(rdrd\varphi )=\frac{\pi R^4}{2}}$

2. Unit

The SI unit for polar moment of inertia, like the area moment of inertia, is meters to the fourth power (m⁴), and inches to the fourth power (in⁴) in U.S. Customary units and imperial units.

3. Limitations

The polar moment of inertia is insufficient for use to analyze beams and shafts with non-circular cross-sections, due their tendency to warp when twisted, causing out-of-plane deformations. In such cases, a torsion constant should be substituted, where an appropriate deformation constant is included to compensate for the warping effect. Within this, there are articles that differentiate between the polar moment of inertia, ${\displaystyle I_z}$, and the torsional constant, ${\displaystyle J_t}$, no longer using ${\displaystyle J}$ to describe the polar moment of inertia.^[3]

In objects with significant cross-sectional variation (along the axis of the applied torque), which cannot be analyzed in segments, a more complex approach may have to be used. See 3-D elasticity.

4. Application

Though the polar moment of inertia is most often used to calculate the angular displacement of an object subjected to a moment (torque) applied parallel to the cross-section, the provided value of rigidity does not have any bearing on the torsional resistance provided to an object as a function of its constituent materials. The rigidity provided by an object's material is a characteristic of its shear modulus, ${\displaystyle G}$. Combining these two features with the length of the shaft, ${\displaystyle L}$, one is able to calculate a shaft's angular deflection, ${\displaystyle \theta }$, due to the applied torque, ${\displaystyle T}$: ${\displaystyle \theta =\frac{TL}{JG}}$

As shown, the larger the material's shear modulus and polar moment of area (i.e. larger cross-sectional area), the greater resistance to torsional deflection.

The polar moment of area appears in the formulae that describe torsional stress and angular displacement.

Torsional stresses: ${\displaystyle \tau =\frac{Tr}{J_z}}$ where ${\displaystyle \tau }$ is the torsional shear stress, ${\displaystyle T}$ is the applied torque, ${\displaystyle r}$ is the distance from the central axis, and ${\displaystyle J_z}$ is the polar moment of area.

Note: In a circular shaft, the shear stress is maximal at the surface of the shaft.

5. Sample Calculation

The rotor of a modern steam turbine. https://handwiki.org/wiki/index.php?curid=1484887

Calculation of the steam turbine shaft radius for a turboset:

Assumptions:

• Power carried by the shaft is 1000 MW; this is typical for a large nuclear power plant.
• Yield stress of the steel used to make the shaft (τ_yield) is: 250 × 10⁶ N/m².
• Electricity has a frequency of 50 Hz; this is the typical frequency in Europe. In North America the frequency is 60 Hz. This is assuming that there is a 1:1 correlation between rotational velocity of turbine and the frequency of mains power.

The angular frequency can be calculated with the following formula: ${\displaystyle \omega =2\pi f}$

The torque carried by the shaft is related to the power by the following equation: ${\displaystyle P=T\omega }$

The angular frequency is therefore 314.16 rad/s and the torque 3.1831 × 10⁶ N·m.

The maximal torque is: ${\displaystyle T_{max}=\frac{{\tau }_{max}{J}_z}{r}}$

After substitution of the polar moment of inertia the following expression is obtained: ${\displaystyle r=\sqrt[3]{\frac{2T_{max}}{\pi {\tau }_{max}}}}$

The radius is r=0.200 m = 200 mm, or a diameter of 400 mm. If one adds a factor of safety of 5 and re-calculates the radius with the admissible stress equal to the τ_adm=τ_yield/5 the result is a radius of 0.343 m, or a diameter of 690 mm, the approximate size of a turboset shaft in a nuclear power plant.

6. Comparing Polar and Mass Moments of Inertia

6.1. Hollow Cylinder

Polar moment of inertia: ${\displaystyle I_z=\frac{\pi (D^4-d^4)}{32}}$

Mass moment of inertia: ${\displaystyle I_c=I_z\rho l=\frac{\pi \rho l(D^4-d^4)}{32}}$

6.2. Solid Cylinder

Polar moment of inertia ${\displaystyle I_z=\frac{\pi D^4}{32}}$

Mass moment of inertia ${\displaystyle I_c=I_z\rho l=\frac{\pi \rho l{D}^4}{32}}$ where:

• ${\displaystyle d}$ is the inner diameter in meters (m)
• ${\displaystyle D}$ is the outer diameter in meters (m)
• ${\displaystyle I_c}$ is the mass moment of inertia in kg·m²
• ${\displaystyle I_z}$ is the polar moment of inertia in meters to the fourth power (m⁴)
• ${\displaystyle l}$ is the length of cylinder in meters (m)
• ${\displaystyle \rho }$ is the specific mass in kg/m³