Consider a fracture specimen with large dimensions under load. We assume that an internal crack is nucleated and propagates over large distances; if a starter crack is initially present, the equal configuration corresponds to crack motion initiation and propagation over macroscopic distance. Then we have:

TENSION

The crack, with half length l, is perpendicular to the applied tension direction. The failure stress σT is given by

σT = √(2ϒ E/π(1-ν2)l)

where ϒ is the surface energy, E Young's modulus and ν Poisson's ratio.

COMPRESSION

The identical crack is aligned parallel to the compression direction. The failure stress σC is

σC = σT / ν ,

indicating that three times larger applied stress is required to bring to failure an isotropic specimen in compression as compared to tension.

BENDING

The identical crack will have one tip moving through the compression region (2). We assume that this is the condition G2 = 2 ϒ that controls the complete failure of the fractured specimen, where G2 is the crack extension force (per unit length of the crack front) pertaining to the tip of the crack that moves through the compression part (2) of the specimen; the other part (1) of the specimen is in tension. Consequently, we have

(a) bending with shear (case of bending by terminal transverse load)

σF / σT = e0 / 6L0

(under the condition : 6L0 / e0 >> 1) where σF is the failure stress and e0 and L0 are the thickness and length of the bending specimen measured along the applied load and specimen axis directions, respectively; in term of the bending moment MF this relation becomes

MF = 2I σT / e0

where I is the geometric moment of inertia of sample section about width axis through middle of the section;

(b) bending without shear (case of bending by terminal couple only)

MC = MF / ν

where MC is the constant bending moment; this suggests that three times larger bending moments are required to fracture isotropic samples in the latter applied loading.

The present modelling assumes fracture propagation over large distance. Why this is important?

Because:

(1) The propagating cracks have unique (single) orientations with respect to the respective applied loadings

(2) The crack lengths have similar magnitude

(3) the mechanism of failure is crack propagation and not initiation of crack motion or crack nucleation

References: please refer to our contributions in ResearchGate, for example.

Because it is anonymous, negative vote should be afforded only to researchers who provided a comment

Metals or ceramics or plastics, it goes by what is stronger, whether compression or tension. Tension is stronger  in plastics but weak in ceramics. But ceramics are stronger in compression than tension. Coming to bending, it depends on the span to depth ratio. When the ratio L/d is low , the shear component is more and when the L/d is high the pure bending mode is more. Shear as you know is only a part of the tension or compression strength. So it reduces the over all flexural strength depending on its component. coming to long span specimens where the length is very large compared to the  thickness or dia, pure bending exists where the flexural strength is not dependent on the shear component and is higher. In conventional materials, the flexural strength is between that of the tensile and compressive strengths. The shear strength of ceramics is high and the shear strength of plastics is low. So the shift towards tensile or compressive stress is again material specific.. Layered composites normally have higher flexural stengths than tensile strenghts, because of the fact that for the same probability of  failure, the fibres in the outer face of the specimen see more stresses than the ones near the neutral axis. As flexure is not a cross section property but an outer surface property unlike tension which is a cross sectional property, laminated composites show this anomaly. Thanks

Padmanabhan Krishnan

First of all, we have to specify that this Q&A is concerned only with brittle materials. No plasticity (crack-tip or yielding) is taken into account. We have to mention this condition in the Q&A.

 If crack propagation is the controlling mechanism of failure, this is the condition G2 = 2 ϒ that controls the complete failure of the fractured specimen. Why? Simply because G1 is much larger than G2 (G1 is the crack extension force at the tip of the crack in the tensile region (1) of the bent specimen).

Measurements in the literature focus only on the tensile region of the bending specimen; under such conditions, the definition of flexural strength is associated either to nucleation of crack or propagation of crack over small distance (the latter corresponds to transitory crack configurations), or both, as possible mechanisms controlling the failure of the material.

Above considerations must be taken into account in this forum.

In compression of brittle materials, there is no shear. Poisson's tension opens the faces of the propagating crack that is aligned parallel to the compression direction.

In bending of brittle materials, both tension and compression regions exist. In absence of shear (pure bending), the crack is perpendicular to the local tension direction after crack propagation over large distance, whilst in the compression region, the identical crack is parallel to the compression direction. In presence of shear, deviation of the crack from above configurations are expected in both regions and this has been taken into account in our modelling.

Our modelling assumes that crack propagation over large distance is the mechanism of failure. This is expected to be the right mechanism in large specimens.

When treating failure stresses, for clarity, it is convenient to specify whether this is under crack nucleation or motion of  transitory crack configurations (initiation of fracture motion) or crack propagation that considerations are made.

Good question… Please check references below.. hope it will be helpful for you…

Regards…

Article Strength and failure characteristics of brittle jointed rock...

http://www.dtic.mil/dtic/tr/fulltext/u2/a048188.pdf

https://nptel.ac.in/courses/105106055/Mod5/Lecture14.pdf

When the crack (be it planar or non-planar) is associated with crack-tip plasticity, the relations between the failure stresses in specimens tested in tension, compression and bending given above (see our answer 1) remain valid, except that σT now incorporates a quantity that contains the crack-tip plastic zone size, crack-front shape and orientation of average crack surface. Please see “NON-PLANAR CRACK WITH CRACK-FRONT PLASTIC YIELDING UNDER GENERAL LOADING” in our contributions in ResearchGate.