## Section10.5Polar Moment of Inertia

The polar moment of inertia is defined by the integral quantity

\begin{equation} J_O = \int_A r^2 dA\text{,}\tag{10.5.1} \end{equation}

where $$r$$ is the distance from the reference point to a differential element of area $$dA\text{.}$$ The polar moment of inertia describes the distribution of the area of a body with respect to a point in the plane of the body. Alternately, the point can be considered to be where a perpendicular axis crosses the plane of the body. The subscript on the symbol $$j$$ indicates the point or axis.

There is a particularly simple relationship between the polar moment of inertia and the rectangular moments of inertia. Referring to the figure, apply the Pythagorean theorem $$r^2 = x^2 +y^2$$ to the definition of polar moment of inertia to get

\begin{align} J_O \amp = \int_A r^2\ dA\notag\\ \amp = \int_A (x^2 + y^2) \ dA\notag\\ \amp = \int_A x^2 dA + \int_A y^2 dA\notag\\ J_O \amp = I_x + I_y\text{.}\label{Jo}\tag{10.5.2} \end{align}

The polar moment of inertia is an important factor in the design of drive shafts. When a shaft is subjected to torsion, it experiences internal distributed shearing forces throughout its cross-section which counteract the external torsional load.

This distributed shearing force is called shear stress, and is usually given the symbol tau, $$\tau\text{.}$$ Shear stress is zero at the neutral axis and increases linearly with $$r$$ to a maximum value, $$\tau_\text{max}$$ at the outside surface where $$r=c\text{,}$$ so

\begin{equation*} \tau = \tau_\text{max}\frac{r}{c} \end{equation*} Figure 10.5.2. Section-cut view of a shaft, showing shearing forces developed to withstand external torsion $$T\text{.}$$

The force at any point is $$dF = \tau\ dA\text{,}$$ and the moment $$dM$$ exerted at any point is $$dF$$ times the moment arm, which is $$r\text{.}$$ The total moment is the integral of this quantity over the area of the cross section, and is proportional to the polar moment of inertia.

\begin{align*} T \amp= \int dM\\ \amp = \int r\ dF\\ \amp= \int_A r\ \tau_\text{max} \frac{r}{c} \ dA\\ \amp = \frac{\tau_\text{max}}{c} \int_A r^2\ dA\\ T \amp = \frac{\tau_\text{max}}{c} J\\ \tau_\text{max} \amp = \frac{Tc}{J} \end{align*}

This is the relationship between the maximum stress in a circular shaft and the applied torque $$T$$ and the geometric properties of the shaft, $$J_O$$ and $$c\text{.}$$