The polar moment of inertia can be calculated by adding the moment of inertia about two mutually perpendicular axes that lies into the plane of the cross-section and concurrent at the centroid.

The above figure shows the plane lamina with two perpendicular axes (x & y) lying into the plane and the z-axis is perpendicular to the plane and all three axes (x, y, z) are concurrent at ‘O’.

Thus by using the perpendicular axis theorem, the polar moment of inertia about the z-axis (**Jₒ**) is given by,

**Jₒ = 𝙸𝘇 = 𝙸𝘅 + 𝙸𝘆**

Where**𝙸𝘅** = Moment of inertia about the x-axis**𝙸𝘆** = Moment of inertia about the y-axis.

Therefore by finding the moment of inertia about the centroidal x and y-axis and adding them together we can find the polar moment of inertia.

## Polar moment of inertia for different shapes:

The polar moment of inertia for some different shapes are discussed below,

**1] For solid circular shaft:-**

The moment of inertia of the circular profile about the centroidal x-axis is given by,

𝙸𝘅 = `\frac{\pi }{64}` d⁴

The moment of inertia of circular profile about the centroidal y-axis is given by,

𝙸𝘆 = `\frac{\pi }{64}` d⁴

Thus the polar moment of inertia about ‘O’ is given by,

Jₒ = 𝙸𝘅 + 𝙸𝘆

Jₒ = `\frac{\pi }{64}` d⁴ + `\frac{\pi }{64}` d⁴

**Jₒ =** `\mathbf{\frac{\pi }{32}}` **d⁴**

**This is the equation for the polar moment of inertia for the circular shaft.**

**2] For hollow circular shaft:-**

The above figure shows the cross-section profile of a hollow circular shaft with an outer diameter (do) and inner diameter (di).

The moment of inertia of the hollow circular shaft about the centroidal x-axis is given by,

𝙸𝘅 = `\frac{\pi }{64}`.`d_{o}^{4}` – `\frac{\pi }{64}`.`d_{i}^{4}`

𝙸𝘅 = `\frac{\pi }{64}`. (`d_{o}^{4}` – `d_{i}^{4}`)

The moment of inertia of the hollow circular shaft about the centroidal y-axis is given by,

𝙸𝘆 = `\frac{\pi }{64}`.`d_{o}^{4}` – `\frac{\pi }{64}`.`d_{i}^{4}`

𝙸𝘆 = `\frac{\pi }{64}`.`(d_{o}^{4} – d_{i}^{4})`

Now the polar moment of inertia for the hollow circular shaft about point ‘O’ is given by,

Jₒ = 𝙸𝘅 + 𝙸𝘆

Putting the values of 𝙸𝘅 and 𝙸𝘆,

Jₒ = [`\frac{\pi }{64}`. (`d_{o}^{4}` – `d_{i}^{4}`)] + [`\frac{\pi }{64}`. (`d_{o}^{4}` – `d_{i}^{4}`)]

**Jₒ =** `\mathbf{\frac{\pi }{32}}`.`\mathbf{(d_{o}^{4} – d_{i}^{4})}`

**This is the equation of polar moment of inertia for a hollow circular shaft.**

**3] For thin rectangular plate:-**

The above figure shows the profile of the thin rectangular plate with width of ‘b’ and height of ‘d’.

The moment of inertia of rectangular shape about centroidal x-axis is given by,

𝙸𝘅 = `\frac{bh^{3}}{12}`

The moment of inertia of rectangular shape about the centroidal y-axis is given by,

𝙸𝘆 = `\frac{hb^{3}}{12}`

Now the polar moment of inertia for the rectangular shape is given by,

Jₒ = 𝙸𝘅 + 𝙸𝘆

Jₒ = `\frac{bh^{3}}{12}`+ `\frac{hb^{3}}{12}`

**Jₒ =** `\mathbf{\frac{1}{12}}` **[bh³ + hb³]**

**This is the equation of polar moment of inertia for a rectangular shape.**

## Solved example:

**The hollow circular pipe has an outer diameter of 40 mm and an inner diameter of 35 mm. Find the polar moment of inertia for the pipe.**

Given :-

dₒ = 40 mm

d𝐢= 35 mm

**Solution:**

By using the formula of the polar moment of inertia for a hollow circular cross-section.

Jₒ = `\frac{\pi }{32}` x `[d_{o}^{4} – d_{i}^{4}]`

Jₒ = `\frac{\pi }{32}` [40⁴ – 35⁴]

⁴Jₒ= 104003.89 mm