# what is the physical meaning of area moment of inertia?

In our applied mechanics class, we studied about area moment of inertia. Our teacher only explained the mathematical relation of this term that is product of area and square of the perpendicular distance from the axis. But I cannot understand the physical meaning. Help me out.

The notion arises fairly naturally when one is thinking about how a constant cross section beam reacts to forces and bending moments.

You imagine a beam to be made up of elastic fibers. When it is bent in a plane, nearest the center of curvature, the fibers must "squash", so they are in compressive strain. On the side outermost from the curvature center, they are in tensile strain. Somewhere in between there is a neutral axis in the cross section that is the intersection between the cross section and the fibers that are neither in compressive nor tensile strain

Indeed, if you sketch the geometry, the amount of strain in the fibers at a normal distance $x$ from the neutral axis is

$$\epsilon(x) = \frac{x}{R} \tag{1}$$

and so the stress normal to the plane is:

$$\sigma(x) = E\,\frac{x}{R} \tag{2}$$

where $R$ is the bending curvature at that cross section and $E$ the Young's modulus. This is because the length of a fiber a normal distance $x$ subtending an angle $\delta\theta$ at the center of curvature is $(R+x)\,\delta \theta$. The same angle is subtended by a length $R\,\delta \theta$ at the neutral axis. Since all the fibers in the arc are the same length when the beam is relaxed, this means that the proportional change in length is $(R+x)/R$, whence (1) (see my sketch below for intuition; the bending is in the plane of the page):

What this means is that if you imagine the beam to be "cut in two" by any cross sectional plane and draw a free body diagram of the two halves, each side exerts an equal and opposite torque on the other.

So, in the light of (2), the contribution to the torque from a fiber with cross-sectional area $\mathrm{d} A$ at distance $x$ from the neutral axis is:

$$\mathrm{d}\tau = \text{lever arm} \times \text{normal force} = x\times E\,\frac{x}{R}\,\mathrm{d} A = \frac{E}{R}\,x^2\,\mathrm{d}A\tag{3}$$

whence one sums up all the contributions to get the total torque of one beam section on the other:

$$\tau = \frac{E\,I}{R}\tag{4}$$

The Punchline

So here is a evocative meaning from (3) and (4): the torque exerted across the cross section by a fiber is proportional to the second moment of the fiber's cross sectional area: there's one $x$ for the lever arm, multiplied by another $x$ because, through the strain geometry, the strain is proportional to the fiber's normal distance from the neutral axis.