Flexural Strength - Flexural Versus Tensile Strength

Flexural Versus Tensile Strength

The flexural strength would be the same as the tensile strength if the material were homogeneous. In fact, most materials have small or large defects in them which act to concentrate the stresses locally, effectively causing a localized weakness. When a material is bent only the extreme fibers are at the largest stress so, if those fibers are free from defects, the flexural strength will be controlled by the strength of those intact 'fibers'. However, if the same material was subjected to only tensile forces then all the fibers in the material are at the same stress and failure will initiate when the weakest fiber reaches its limiting tensile stress. Therefore it is common for flexural strengths to be higher than tensile strengths for the same material. Conversely, a homogeneous material with defects only on it surfaces (e.g. due to scratches) might have a higher tensile strength than flexural strength.

If we don't take into account defects of any kind, it is clear that the material will fail under a bending force which is smaller than the corresponding tensile force. Both of these forces will induce the same failure stress, whose value depends on the strength of the material.

For a rectangular sample, the resulting stress under an axial force is given by the following formula : . This stress is not the true stress, since the cross section of the sample is considered to be invariable (engineering stress).

• is the axial load (force) at the fracture point
• b is width
• d is thickness

The resulting stress for a rectangular sample under a load in a three-point bending setup (Fig. 3) is given by the formula below (see "Measuring flexural strength").

The equation of these two stresses (failure) yields :

Usually, L (length of the support span) is much bigger than d, so the fraction is bigger than one.