Polar moment of inertia

Polar moment of inertia is a quantity used to predict an object's ability to resist torsion, in objects (or segments of objects) with an invariant circular cross section and no significant warping or out-of-plane deformation.^[1] It is used to calculate the angular displacement of an object subjected to a torque. It is analogous to the area moment of inertia, which characterizes an object's ability to resist bending and is required to calculate displacement.

The larger the polar moment of inertia, the less the beam will twist, when subjected to a given torque.

Polar moment of inertia should not be confused with moment of inertia, which characterizes an object's angular acceleration due to a torque. See moment (physics).

1 Limitations
2 Definition
3 Unit
4 Conversion from Area Moment of Inertia
5 Application
6 Sample calculation
7 Comparing various moments of inertias for a solid cylinder
8 See also
9 References
10 External links

Limitations

The polar moment of inertia cannot be used to analyze shafts with non-circular cross-sections. In such cases, the torsion constant can be substituted instead.

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.

However the polar moment of inertia can be used to calculate the moment of inertia of an object with arbitrary cross-section.

Definition

$J_z = \int_A \rho ^2\, dA$

J_z = the polar moment of inertia about the axis z
dA = an elemental area
ρ = the radial distance to the element dA from the axis z

For a circular section with radius r:

$J_z = \int_0^{2\pi} \int_0^r \rho^2 \rho\, d\rho\, d\phi = \frac{\pi r^4}{2}$

Unit

The SI unit for polar moment of inertia, like the area moment of inertia, is metre to the fourth power (m⁴).

Conversion from Area Moment of Inertia

By the perpendicular axis theorem, the following equation relates J_z to the area moments of inertia about the other two mutually perpendicular axes:^[2]

$J_{z}=I_{xx}%2BI_{yy}$

Application

NOTE: The following is only true for a circular or hollow circular section.

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

Torsional stress:

$\tau = \frac{T r}{J_{z}}$

where $T$ is the torque, $r$ is the distance from the center and ${J_{z}}$ is the polar moment of inertia.

In a circular shaft, the shear stress is maximal at the surface of the shaft (as that is where the torque is maximal):

$T_\max = \frac{{\tau}_\max J_{z}}{r}$

Most frequently the inverse problem is solved, in which one solves for the radius.

Sample calculation

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 x 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.

The angular frequency can be calculated with the following formula:

$\omega=2 \pi f$

The torque carried by the shaft is related to the power by the following equation:

$P=T \omega$

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

The maximal torque is:

$T_\max = \frac{ { \tau }_\max J_{z} }{r}$

After substitution of the polar moment of inertia the following expression is obtained:

$r=\sqrt[3]{\frac{2 T_\max}{\pi {\tau}_\max}}$

The radius is 0.200 m. If one adds a factor of safety of 5 and re-calculates the radius with the maximal stress equal to the yield stress/5 the result is a radius of 0.343 m, or a diameter of 69 cm, the approximate size of a turboset shaft in a nuclear power plant.

Comparing various moments of inertias for a solid cylinder