Flexure Formula

Stresses caused by the bending moment are known as flexural or bending stresses. Consider a beam to be loaded as shown.

Consider a fiber at a distance y from the neutral axis, because of the beam’s curvature, as the effect of bending moment, the fiber is stretched by an amount of cd. Since the curvature of the beam is very small, bcd and Oba are considered as similar triangles. The strain on this fiber is

By Hooke’s law, \varepsilon = \sigma / E, then

which means that the stress is proportional to the distance y from the neutral axis.

Considering a differential area dA at a distance y from N.A., the force acting over the area is

The resultant of all the elemental moment about N.A. must be equal to the bending moment on the section.

\displaystyle M = \int dM = \int y\, dF = \int y \, \left( \frac{E}{\rho}y \, dA \right)

\displaystyle M = \frac{E}{\rho} \int y^2 \, dA

but \int y^2 \, dA = I \,\,, then

M = \dfrac{EI}{\rho} \,\, \text{or} \,\, \rho = \dfrac{EI}{M}

substituting \rho = Ey / f_b

\dfrac{Ey}{f_b} = \dfrac{EI}{M}

then

and

The bending stress due to beams curvature is

f_b = \dfrac{Mc}{I} = \dfrac{\dfrac{EI}{\rho}c}{I}

The beam curvature is:

where \rho is the radius of curvature of the beam in mm (in), M is the bending moment in N·mm (lb·in), f_b is the flexural stress in MPa (psi), I is the centroidal moment of inertia in mm⁴ (in⁴), and c is the distance from the neutral axis to the outermost fiber in mm (in).