# 9+ Polar Moment Of Inertia Formulas [2023]

The polar moment of inertia, quantifies an object/section’s resistance to rotational motion about a specific axis.

This property plays an important role in structural and mechanical engineering, ranging from bolted steel connections to complex machinery analysis.

Every shape has a different polar moment of inertia. In this blog post, we’ll show the formulas for different sections, so you don’t have to calculate them yourself.

Let’s get started. đźš€đźš€

## What Is The Polar Moment Of Inertia?

The second polar moment of inertia, often denoted as I_{p} or J, is a measure of an object’s resistance to torsional, or twisting, deformation around an axis perpendicular to its longitudinal axis.

It is commonly used in engineering and mechanics to analyse and design structures, components and connections subjected to torsional loads, such as shafts, steel tubes and bolted connections.

## Formula Of Polar Moment Of Inertia

The formula for the 2nd polar moment of inertia is the following:

$$I_p = \int \int_{A}^{} r^2 \,dA $$

With,

dA | Infinitesimal element |

r | Radial distance to element dA |

I know, I know, we all don’t like integrals. They are just too complicated, right?

So, the formula for the 2nd polar moment of inertia that we all understand better is:

$$I_p = I_y + I_z$$

Where

I_{y} | Moment of inertia strong axis |

I_{z} | Moment of inertia weak axis |

Now, the formula has become a lot easier. You can check out our blog post about moment of inertial formulas to calculate I_{y} and I_{z} quicker.

## 1. I- and H Section

**Polar moment of inertia**

$I_p = \frac{w \cdot h^3}{12} – \frac{(w-t_w) \cdot (h-2\cdot t_f)^3}{12}$

$ + \frac{(h-2\cdot t_f) \cdot t_w^3}{12} + \frac{2\cdot t_f \cdot w^3}{12}$

### Example calculation

$h$ = 300mm, $b$ = 150mm, $t_f$ = 10mm, $t_s$ = 7m

$I_p = \frac{w \cdot h^3}{12} – \frac{(w-t_w) \cdot (h-2\cdot t_f)^3}{12} + \frac{(h-2\cdot t_f) \cdot t_w^3}{12} + \frac{2\cdot t_f \cdot w^3}{12} $

$I_p = \frac{150mm \cdot (300mm)^3}{12} – \frac{(150mm-7mm) \cdot (300mm-2\cdot 10mm)^3}{12} \frac{(300mm-2\cdot 10mm) \cdot (7mm)^3}{12} + \frac{2\cdot 10mm \cdot (7mm)^3}{12} $

$I_p = 7.59 \cdot 10^7 mm^4$

## 2. Square Section

**Polar moment of inertia**

$I_p = \frac{w^4}{6} = 0.1667 w^4$

### Example calculation

w = 120 mm

$J = \frac{w^4}{6} = \frac{(120mm)^4}{6} = 3.46 \cdot 10^7 mm^4$

## 3. Rectangular Section

**Polar moment of inertia**

$I_p = \frac{wh (w^2 + h^2)}{12}$

### Example calculation

w = 120 mm, h = 240mm

$J = \frac{wh (w^2 + h^2)}{12} = \frac{120mm \cdot 240mm \cdot ((120mm)^2 + (240mm)^2)}{12} = 1.73 \cdot 10^8 mm^4$

## 4. Circle/ Circular Section

**Polar moment of inertia**

$I_p = \frac{\pi \dot D^4}{32}$

### Example calculation

D = 200 mm

$J = \frac{\pi \dot D^4}{32} = \frac{\pi \dot (200mm)^4}{32} = 1.57 \cdot 10^8 mm^4$

## 5. Circular Tube Section

**Polar moment of inertia**

$I_p = \frac{\pi}{32} \cdot (D^4 – d^4)$

### Example calculation

D = 200 mm, d = 180mm

$J = \frac{\pi}{32} \cdot (D^4 – d^4) = \frac{\pi}{32} \cdot ((200mm)^4 – (180mm)^4) = 5.4 \cdot 10^7 mm^4$

## 6. Circular Section With Square Cutout

**Polar moment of inertia**

$I_p = \frac{\pi \cdot D^4}{32} – \frac{w^4}{6}$

### Example calculation

D = 200 mm, w = 100 mm

$J = \frac{\pi \cdot D^4}{32} – \frac{w^4}{6} = \frac{\pi \cdot (200mm)^4}{32} – \frac{(100mm)^4}{6} = 1.4 \cdot 10^8 mm^4$

## 7. Hexagon Section

**Polar moment of inertia**

$I_p = \frac{5 \cdot \sqrt{3}}{8} \cdot s^4$

### Example calculation

D = 200 mm, w = 100 mm

$J = \frac{5 \cdot \sqrt{3}}{8} \cdot s^4 = \frac{5 \cdot \sqrt{3}}{8} \cdot (100mm)^4 = 1.08 \cdot 10^8 mm^4$

## 8. Hollow Rectangular Section

**Polar moment of inertia**

$I_p = \frac{WH (W^2 + H^2)}{12} – \frac{wh (w^2 + h^2)}{12}$

### Example calculation

H = 240 mm, W = 120 mm, h = 200 mm, w = 90 mm

$J = \frac{WH (W^2 + H^2)}{12} – \frac{wh (w^2 + h^2)}{12} = \frac{120mm \cdot 240mm ((120mm)^2 + (240mm)^2)}{12} – \frac{90mm \cdot 200mm ((90mm)^2 + (200mm)^2)}{12} = 1.0 \cdot 10^8 mm^4$

## 9. Triangular Section

**Polar moment of inertia**

$I_p = \frac{\sqrt{3}}{48} \cdot s^4$

### Example calculation

s = 100 mm

$I_p = \frac{\sqrt{3}}{48} \cdot s^4 = \frac{\sqrt{3}}{48} \cdot (100mm)^4 = 3.6 \cdot 10^6 mm^4$

## 10. Circular Section With Socket Head Cutout

**Polar moment of inertia**

$I_p = \frac{\pi D^4}{32} – \frac{5 \sqrt{3}}{8} \cdot s^4$

### Example calculation

D = 300 mm, s = 100 mm

$I_p = \frac{\pi D^4}{32} – \frac{5 \sqrt{3}}{8} \cdot s^4$

$I_p = \frac{\pi (300mm)^4}{32} – \frac{5 \sqrt{3}}{8} \cdot (100mm)^4$

$I_p = 6.9 \cdot 10^8 mm^4$

## Conclusion

The polar moment of inertia is an important parameter in structural design. Therefore, it’s recommendable to know how to calculate it for different cross-sections. You do that by adding I_{y} and I_{z} together.

In the following blog posts, we show step-by-step, how to calculate these moments of inertias.

If you are new to structural design, then check out our design tutorials about other structural elements. đź‘‡đź‘‡

Do you miss any 2nd polar moment of inertia formula for any shape or cross-section that we forgot in this article? Let us know in the comments. âśŤď¸ŹâśŤď¸Ź