# Where does Second moment of area come from?

Where does the second moment of area come from and how to derive it from scratch, or what is the basic intuition behind it?

In other words, what are the origins of $$I_{xx} = \int\int yy \text{ }dxdy$$

It seems to be related to the Radius of gyration, but it's just not fully evident to me just yet.

• – Solar Mike Jul 12 '20 at 16:15
• Not helpful. They don't show the origin of the formula. – Jek Denys Jul 12 '20 at 16:24
• Plenty of useful sites to be found or even a textbook or two. – Solar Mike Jul 12 '20 at 16:25
• Your comment is not helpful. Undergrad books that I have and at least first 5 pages of google on the topic omit the origin of the equation. – Jek Denys Jul 12 '20 at 16:29

## 2 Answers

The Second Moment of Area stems from the Radius of Gyration. Radius of gyration is the root mean square distance of particles from the axis. In other words, it's an average distance to the particles of a rigid body from an axis, that is squared to eliminate the negative signs.

Looking at an Area, A, that is rotated about X-axis, the radius of gyration is used to give the mean distribution of particles in the area. Therefore, to get the resistance of this area (cross-section A) about the X-axis, we can integrate it over dx and dy.

That is why $$y$$ is squared. It is the radius of gyration or a mean squared distance perpendicular to the axis of rotation.

Giving rise to the equation: $$I_{xx}=\int_A{R_{gr}^2}dA = \int_A{y^2}dA$$

and in case of rotation about y-axis: $$I_{yy}=\int_A{R_{gr}^2}dA = \int_A{x^2}dA$$

• What is it useful for? – Solar Mike Jul 12 '20 at 17:20

Second moment of inertia is a property of a surface measuring its area distribution from its neutral axis.

In engineering it is closely related to bending moment resistant.

Let's see why:

In materials with linear stress/strain ratio, with a constant modules of elasticity like steel, wood, plastic, metals, the deformation of section along depth of the beam is linear and plain surfaces remain plain with a small angle of rotation theta.

Therefore the strain along the depth for a small differential surface is linearly increasing e= y tan(theta) hence it's proportional to y and the stress is proportional to y da.

But this differential element stress must be multiplied by its distance from the fulchrum, neutral axis, to find the contributory moment.

Hence

y.y da = y^2 da.

Sorry, but my phone doesn't have fancy fonts to display math.