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 its 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:

Stress = F/ (width * depth)

F is the axial load (force) at the fracture point

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

The equation of these two stresses (failure) yields:

Flexural Stress=3FL/(2(width * depth^2))

Usually, L (length of the support span) is much bigger than d, so the fraction 3L/(2*d) is bigger than one.

Any differences between the tensile strength and flexural (bending) strength are likely to be statistical in nature. One normally wouldn't expect to see too much difference in these strength values for a ductile, e.g., metallic, material, but this may certainly not be the case for nominally brittle materials, such as ceramics and certain composites. The strength of brittle materials is generally highly sensitive to the size of the sample; larger components will often display lower strengths because their strength is a function of their defect population and with a larger sample, there is a higher statistical chance of finding a larger defect.  Indeed, hundreds of years ago Leonardo da Vinci actually found that larger/longer wires had lower strengths than shorter ones. When one compares the strengths measured in tensile vs. bend (flexural) tests, the active statistical surface or volume that sees the highest stresses, i.e. the sampling surface/volumes, are very different. For similar sizes specimens, the entire sample surface/volume will experience the highest stresses in tension, whereas in bending the highest stresses will only be on the top loading surface, i.e.the active sampling volume id s much smaller (the precise surface/volume where these stresses are highest on this surface depends upon whether 3- or 4-point bending is employed). If one does the statistical calculations, I think that statistically, 10 tensile tests are probably equivalent to roughly 100 bend tests (for samples of similar size) in characterizing the strength properties of a nominally brittle material.  Bottomline: the tensile strength will invariably represent a more conservative measure of the strength as compared to the easier done flexural tests.

I may be missing a trick here but how can one compare tensile and bending properties since they are due to different actions being applied to the beam?

There is no trick here.  Both tension and bending testing applies tensile stresses to the sample which motivates yielding and/or fracture.  Accordingly, both types of tests can be, and are routinely, used to evaluate the strength and/or fracture strengths of materials.  The difference, of course, is that the nature of the stress and strain states in tension and bending are not the same. With a tensile test, the maximum tensile stresses are experienced throughout the entire volume (and surface area) of the test piece; in bending (where the sample sees tensile stress above the neutral axis and compressive stresses below), the maximum tensile stresses are conversely concentrated in a small region on the top surface above the neutral axis.  Accordingly for similar sized test pieces, the tensile sample sees the maximum stresses throughout its entire gauge length, i.e., over a much larger volume than the corresponding bend sample,   In brittle materials which are highly sensitive to the defect population, this change in statistical sampling volume means that strength and fracture properties measured in tensile tests are likely to be somewhat lower than the corresponding properties measured in bending, because there is a higher statistical probability of finding a larger defect.

Thank you Robert.  That all makes good sense.  I was coming at this from a view-point of ductile materials where plasticity can develop in bending so that the strength is more than that just to cause first yield.  

In bending , there is shear stress besides tensile stress . what's role shear stress play in the fracture during bending? how it influence the fracture?

Hou:

Well first it depends on the form of bending. In four-point bending, the bending moment is constant between the inner two loading rollers, and so this represents a state of pure bending where the shear stress distribution is irrelevant. However, yes, in three-point bending, shear stresses are introduced which peak at the neutral axis and are reduced to zero on the free surfaces. Clearly this is different from a state of uniaxial tension.

Does this then affect fracture in these two geometries? Well, shear stresses are obviously not unimportant but these are global stresses yetd fracture is a local event. If we look at the local field at the tip of a crack in these two geometries, i.e., three-point bending vs. uniaxial tension, the local stress and strain fields in the immediate vicinity of the crack tip, which control fracture there over appropriate (invariably nano to microscale) dimensions, will be the same, i.e., the K-field, which will be unique under small-scale yielding conditions such that fracture will be essentially the same in these two geometries. Essentially this means that fracture will be controlled by local stresses and strains in the vicinity of a crack or other defects, even though the global stress/strain fields are not the same. This holds true for linear elastic conditions, but progressively begins to break down with increasing plasticity (this is the region governed by so-called nonlinear elastic fracture mechanics), until in the limit of full plasticity, essentially there is no uniqueness between the crack tip fields in tension vs. bending.

@Robert:

The ratio of shear /normal stress is larger,the material show more ductile. Is this another possible reason that  strength  in tensile tests are lower than strength measured in bending?

Hou:

You may be correct in certain situations but personally I don't think that this factor plays much of a role in the differing strengths measured in tension and bending.  The effect is never that large in any event.  Moreover, it is most pronounced with brittle materials, due to the large difference in statistical sampling volume (or surface area) in the two geometries, and in these materials the role of shear stresses would be less significant.

ROR

@ Dear Robert:

what's about the sampling effect in ductile steel?

With ductile materials, the statistical sampling effect is less important as the strength and toughness are far less sensitive to the defect density. Under these circumstances, I would anticipate much less difference in the mechanical properties measured in flexural vs.tension.

Dear Robert:

For crack initiation of intact ductile metal sample under 3-points bending ,which surface stress must exceed, the ultimate tensile strength or fracture strength?

And when the surface material begin to neck , the adjacent material constrain it .what’s the effect of this constraint to crack initiation?

Dear Hou:

Your question is a little difficult to answer. If the material is truly ductile, one may presume that it will fail by yielding (limit load failure) rather than by fracture per se, In that case, yielding will commence on the tensile surface, when the local tensile stress exceeds the yield strength. Under pure bending, general yielding, (or limit load) when the entire cross-section plastically deforms, with occur at a bending load up to 38% higher.  However, it is possible that even in a nominally ductile material, plastic deformation in the form of a slip band impinging on a brittle particle, such as a carbide particile in a steel, could precipitate brittle fracture, at a fracture stress which will not necessarily depend upon the matrix yield strength.  However, in terms of your question for a "ductile metal sample", the former scenario is more probable.

ROR

Dear Robert:

1. could you recommend some resources about the flow stress and fracture stress in the section of pure bending from plastic deformation to fracture?

2. I have a cylindrical sample require only high pure bending fracture force. the surface defects have  been got rid of.  generally, brittle material don't use in bending or tension.Can I use a high-strength but brittle material instead of  lower strength-ductile material to pursue high bending strength in my situation?

thank you very much for your time.

Dear Hou:

1.  One of the best books on flow stresses vs. fracture stresses and their role on fracture was written by my doctoral supervisor from way back - John Knott.  The book,  "Fundamentals of Fracture Mechanics (Halstead Press, 1973), is quite old now but worth a read if you can find a copy.

2.  I'm uncertain how to respond to your second question though. I do not believe that brittle materials do not fail in bending or tension; they may be more susceptible to bending but they can certainly fail under tension and torsion too.  I strongly believe that it would be unwise to pursue materials solely with high strength at the expense of toughness.

I hope that this helps.

ROR

Probably due to the change in the stress state. As you may know, the deformation behaviors of the composites are strongly influenced by their local stress senses- tension or compression. This is also the reason why the compressive properties are always higher than the tensile properties. 

The elasticity module at a bending is always lower than at extension due to deformation of shift and deformation under support so look for a mistake, Strength at a bending is usually higher, due to large-scale effect as material volume in which tension is maximum, at a bending usually it is less than at extension