Neutral axis

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Beam with neutral axis (x).
Beam with neutral axis (x).

An axis in the cross section of a beam or shaft or the like along which there are no longitudinal stresses / strains. If the section is symmetric and is not curved before the bend occurs then the neutral axis is at the geometric centroid. All fibers on one side of the neutral axis are in a state of tension, while those on the opposite side are in compression

Since the beam is undergoing uniform bending, it is obvious that a plane on the beam remains plane, that is:

γxy = γzx = τxy = τxz = 0

Where γ is the shear strain and τ is the shear stress

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

Since the bending is uniform and pure, there is therefore at a distance y from the neutral axis with the inherent property of having no strain:

\epsilon_x(y)=\frac{L-L(y)}{L} = \frac{\theta\,(\rho\, - y) - \theta \rho \,}{\theta \rho \,} = \frac{-y\theta}{\rho \theta} = \frac{-y}{\rho}

Therefore the longitudinal normal strain εx varies linearly with the distance y from the neutral surface. Denoting εm as the maximum strain in the beam (at a distance c from the neutral axis), it becomes clear that:

\epsilon_m = \frac{c}{\rho}

Therefore, we can solve for ρ, and find that:

\rho = \frac{c}{\epsilon_m}

Substituting this back into the original expression, we find that:

\epsilon_x(y) = \frac {-\epsilon_my}{c}

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 stresses in the neutral axis.

[edit] See also

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This definition is suitable for the so-called long beams, i.e. its length is much larger than the other two dimensions.

External Link: Proof that the centroid lies on neutral axis