The neutral axis is an axis in the cross section of a beam (a member resisting bending) or shaft along which there are no longitudinal stresses or strains.
Theory If the section is symmetric, isotropic and is not curved before a bend occurs, then the neutral axis is at the geometric centroid of a beam or shaft. 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, a plane on the beam remains plane. That is:
γ x y = γ z x = τ x y = τ x z = 0 {\displaystyle \gamma _{xy}=\gamma _{zx}=\tau _{xy}=\tau _{xz}=0}
Where γ {\displaystyle \gamma } is the shear strain and τ {\displaystyle \tau } is the shear stress There is a compressive (negative) strain at the top of the beam, and a tensile (positive) 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 (span) 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 at its neutral axis. 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:
ϵ x ( y ) = L ( y ) − L L = θ ( ρ − y ) − θ ρ θ ρ = − y θ ρ θ = − y ρ {\displaystyle \epsilon _{x}(y)={\frac {L(y)-L}{L}}={\frac {\theta \,(\rho \,-y)-\theta \rho \,}{\theta \rho \,}}={\frac {-y\theta }{\rho \theta }}={\frac {-y}{\rho }}}
Therefore, the longitudinal normal strain ϵ x {\displaystyle \epsilon _{x}} varies linearly with the distance y from the neutral surface. Denoting ϵ m {\displaystyle \epsilon _{m}} as the maximum strain in the beam (at a distance c from the neutral axis), it becomes clear that:
ϵ m = c ρ {\displaystyle \epsilon _{m}={\frac {c}{\rho }}}
Therefore, we can solve for ρ, and find that:
ρ = c ϵ m {\displaystyle \rho ={\frac {c}{\epsilon _{m}}}}
Substituting this back into the original expression, we find that:
ϵ x ( y ) = − ϵ m y c {\displaystyle \epsilon _{x}(y)={\frac {-\epsilon _{m}y}{c}}}
Due to Hooke's law, the stress in the beam is proportional to the strain by E, the modulus of elasticity:
σ x = E ϵ x {\displaystyle \sigma _{x}=E\epsilon _{x}\,}
Therefore:
E ϵ x ( y ) = − E ϵ m y c {\displaystyle E\epsilon _{x}(y)={\frac {-E\epsilon _{m}y}{c}}}
σ x ( y ) = − σ m y c {\displaystyle \sigma _{x}(y)={\frac {-\sigma _{m}y}{c}}}
From statics, a moment (i.e. pure bending) consists of equal and opposite forces. Therefore, the total amount of force across the cross section must be 0.
∫ σ x d A = 0 {\displaystyle \int \sigma _{x}dA=0}
Therefore:
∫ − σ m y c d A = 0 {\displaystyle \int {\frac {-\sigma _{m}y}{c}}dA=0}
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