I have some intuition about the (second) moment of inertia, and there is some motivation to define this concept if we think about the $KE$ of a rotating body or the torque $\tau$ applied, for example, on a circular object with radius $r$:

$\tau = r \times F = rm\Delta p/\Delta t = mr^2\Delta w/\Delta t = I\alpha $

Then moment of inertia could be expressed as $I = \int y^2dm$, where $y$ is the distance between a point on the object and it's rotational axis.

I want to know the motivation for defining the (second) area moment $\int_{A}y^2dA$. What is the relationship between this moment and the moment of inertia?

Here it's clear that if we consider the density of the object $\rho_0$ to be uniform and equal to one ($\rho_0 = 1$), then we could say that both moments are the same. But why do we use area moment on problems regarding bending moments on beams instead of inertia moment?

Through this reasoning when I'm calculating the tensions that result from the bending moment $M$ in the formula $σ=(M×y)/I$, then I'm assuming the material is made out of water? Wouldn't $\rho$ be an important factor to take into consideration?

I still cannot relate the inertia moment (related with an object's rotation) with area moment that is used when we take into account a bending moment on an object. If possible use some graphics and intuitive examples and avoid using tensor calculus.

  • $\begingroup$ Mass moment of inertia is used for rigid body motion, and 2nd area of moment is used for beam problems. Only when dealing with stresses due to a dynamic loading on a heavy beam you may end up with both quantities in the same expression. $\endgroup$ Dec 31, 2014 at 1:00
  • 1
    $\begingroup$ The second moment of area is often referred to as a 'moment of inertia', but it has no real connection with the mass moment of inertia used to analyze the dynamics of a system. $\endgroup$
    – user16622
    May 17, 2016 at 15:05

2 Answers 2


In beam bending, the bending stress in the cross section is proportional to the location where the stress is calculated from neutral axis of the beam, where by definition the bending stress = 0.

The neutral axis of the cross section is found by calculating the first moment of area of the cross section, and then finding the centroid using that moment.

In other words, the neutral axis is the line about which the tensile force and the compressive force produced in bending are balanced.

If we continue with the analysis of the internal forces created by the bending of a beam, the moment of these forces about the neutral axis of the beam can also be calculated, and this calculated moment must equal the bending moment at that location of the beam. One of the integrals which results from this moment calculation is the second moment of area of the cross section, which is usually called I. The value of I depends on the geometry of the cross section and not the value of the bending moment, which is why I can be calculated and tabulated for sections of various types.

The mathematical development of I can be found in the following article:

Beam Bending

See equations 7.4.18 thru 7.4.20 for details.


Take an arbitrary cross section with normal stress $\sigma(x,y)$ as a function of location and sum up all the forces and moments

$$\begin{align} \sum F & = \int \sigma \,{\rm d}A \\ \sum M & = \int y \sigma\,{\rm d}A \end{align} $$

You can find the neutral axis when only bending stress exists (no axial forces) by taking the first equation $\sum F =0$. The solution is found if $$\sigma(x,y) = k\, y$$ with $y=0$ on the neutral axis (stress symmetry line). Now use the linear stress distribution along the section in the moments

$$ \sum M = \int k\,y^2 {\rm d}A = k\, \int y^2 {\rm d} A $$

$$ \boxed{ k = \frac{\sum M}{\int y^2 {\rm d}A} }$$

and $$ \sigma_{max} = k y_{max} = \frac{ (\sum M) y_{max}}{ \int y^2 {\rm d}A} $$


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