Polar moment of inertia

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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 twist 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.

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

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

In objects with significant cross-sectional variation, which cannot be analyzed in segments, and/or a non-circular cross-section a more complex approach has to be used. See 3-D elasticity.

A schematic showing how the polar moment of inertia is calculated for an arbitrary shape about an axis o. ρ is the radial distance to the element dA.

$J_x = \int r^2\, dA$

J_x = the polar moment of inertia about the axis x
dA = an elemental area
r = the radial distance to the element dA from the axis x

Or for an element of a circular section:

$J = \int_0^r 2 \pi r^3 dr$

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

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

Torsional stress:

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

where $T$ is the torque, $r$ is the radius and $J x$ is the polar moment of area.

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

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

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

The rotor of a modern steam turbine.

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 is (τ_yield) 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:

ω = 2π f

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

P = T ω

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

The maximal torque is:

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

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.

$J_{x} = \frac{\pi r^4}{2}$