Since the beam is undergoing uniform bending, it is obvious that a plane on the beam remains plane, that is:
There is a tensile (positive) strain at the top of the beam, and a compressive (negative) strain at the bottom of the beam. Therefore by the Intermediate Value Theorem, there must be some point in between the top and the bottom that has no strain, since the strain in a beam is a continuous function.
Let L be the original length of the beam. ε(y) is the strain as a function of coordinate on the face of the beam. σ(y) is the stress as a function of coordinate on the face of the beam. ρ is the radius of curvature of the beam. θ is the bend angle
Therefore the longitudinal normal strain varies linearly with the distance y from the neutral surface. Denoting as the maximum strain in the beam (at a distance c from the neutral axis), it becomes clear that:
Therefore, we can solve for ρ, and find that:
Substituting this back into the original expression, we find that:
From statics, a moment (i.e. pure bending) consists of equal and opposite forces. Therefore, the total amount of stress across the cross section must be 0.
Since y denotes the distance from the neutral axis to any point on the face, it is the only variable that changes with respect to dA. Therefore:
Therefore the first moment of the cross section about its neutral axis must be zero. Therefore the neutral axis lies on the centroid of the cross section.
Note that the neutral axis does not change in length when under bending. It may seem counterintuitive at first, but this is because there are no bending stresses in the neutral axis. However, there are shear stresses (τ) in the neutral axis, zero in the middle of the span but increasing towards the supports, as can be seen in this function;
This definition is suitable for the so-called long beams, i.e. its length is much larger than the other two dimensions.