# Why do we always need to calculate the 'Moment of Inertia' about neutral axis for bending?

I was having a look at the bending stress derivation in few of the books. The derivation indicates how each of the element in the bending stress equation is derived into that equation. However, for the moment of inertia (i.e. I), it is not indicated that about which axis are we taking the moment of inertia about. Moreover, the 'c' (i.e. the distance from the neutral axis) is also taken from the neutral axis. Can anyone explain to me that why is it necessary to take the moment of inertia about the neutral axis, and why 'c' should be taken about the neutral axis?

The derivation for bending stress is depended on the assumption that the strain distribution across the thickness is linear. i.e.

$$\epsilon(z) = a_1 z + a_0$$

where: $$a_0$$ $$a_1$$ are coefficients of the slope.

Because in the linear region the constitutive equation is $$\sigma(z) = E\cdot \epsilon(z)$$, the development of stress at distance z is proportional to the strain. i.e.

$$\sigma(z) = E (a_1 z + a_0) = E a_1 z + E a_0$$

Because stress in a small area $$\Delta A$$ is defined as $$\sigma(z) =\frac{\Delta F}{\Delta A}$$, in that small area $$\Delta A$$ which is at distance z you are getting a Force :

$$\Delta F(z) = E\cdot \Delta A\cdot \epsilon(z)$$

Those forces produce bending moments the further away they are from the neutral axis (about which the cross-section rotates), the contribution of the bending moment is greater.

$$\Delta M(z) = \Delta F(z) \cdot z= E\cdot \Delta A\cdot \epsilon(z)\cdot z$$

So at a distance (z) any small area $$\Delta A$$ will produce a bending moment equal to

$$\Delta M(z) = E\Delta A\cdot z\cdot (a_1 z + a_0)$$ $$\Delta M(z) = E\Delta A\cdot z^2\cdot a_1 z + E\Delta A\cdot z\cdot a_0$$

At this point it probably becomes more obvious why the neutral axis is selected. The neutral axis is defined as setting z=0 at the location where the integral of $$\int z dA$$ becomes to zero.

However that has the benefit that when you integrate over the entire area the previous equation: $$\int \Delta M(z) dz= \int_A \left(E \cdot z^2\cdot a_1 + E\cdot z\cdot a_0 \right)dA$$

$$M(z) = E\cdot a_1\int_A \left(z^2 \right)dA + E\cdot a_0 \color{red}{\int_A z dA}$$

$$M(z) = E\cdot a_1 \cdot \int_A z^2 dA + E\cdot a_0 \color{red}{0}$$

So the problem simplifies to: $$M(z) = E\cdot a_1\cdot \int_A z^2 dA$$

where:

• $$E$$ is the modulus of elasticity
• $$\int_A z^2 dA = I$$ is the second moment of area

(and from there you can calculate $$a_1$$ which you can find that needs to be equal to $$a_1 = \frac{1}{\rho}$$ for pure bending, and that is equal to curvature of the bending beam which simplifies to $$w''$$, so you eventually get the $$w'' ='\frac{M}{EI}$$)

Because we're not interested in the moment of inertia or the "c" ordinate of a particular area element of the beam for their own sake - we're interested in them as intermediate steps on the way to computing the bending moment and/or the axial load. That involves multiplying by axial longitudinal stress and integrating. If you take the origin anywhere other than on the neutral axis, then the formula describing the axial longitudinal stress is more complicated, and you've got a harder calculation.