What’s the criterion of cleavage fracture in uniaxial tension test of intact sample? Is it when the stress exceed critical cleavage stress σf ,cleavage will happen?
Fracture will commence where the stress (or strain) exceeds the tensile strength (or the limiting fracture strain) of the material .
Hello! in my expereance the best creterion is that fracture occured when the local strain axseeds its critical walue. it works wery well on both intact and notched samples.
may be this article will help you
https://www.researchgate.net/publication/276269276_Determination_of_the_Local_Strains_Near_Stress_Concentrators_by_the_Digital_Image_Correlation_Technique
Article Determination of the Local Strains Near Stress Concentrators...
The established criterion for (brittle) cleavage fracture is one of a critical stress-controlled fracture, i.e., locally the maximum tensile (or tangential) stress exceeds a local critical fracture stress but over a characteristic microstructural dimension - the characteristic dimension - which is typically related to the grain size or particle size. For fracture ahead of a crack, the so-called RKR criterion best describes this (J. Mech. Phys. Sol., June 1973); there are also several statistical models based on RKR, such as by Lin et al. (J. Mech. Phys. Sol., 1986). In steels, the local critical fracture stress is generally associated with the stress to fracture the largest particle (i.e., a carbide) but for fracture from a notch or sharp crack, this is not always the case due to statistical sampling effects.
Dear Hou:
It all depends how brittle the material is. Stress-controlled cleavage fracture is more likely in a bend (particularly notched bend) test, as the constraint can elevate the tensile stresses and create a triaxial stress state, all of which promotes stress-controlled cleavage fracture. However, cleavage can still occur in uniaxial tension at the same temperature if the applied tensile stress is high enough. For a given material, this is generally the case, but as we spoke about before, if it still possible for a given material at a fixed temperature that the notched bend sample will cleave whereas the uniaxial tensile test fails by a ductile cup and cone fracture. The fracture mode for a given material/microstructure depends upon the stress (or strain for ductile fracture), the temperature and the stress-state which governs the degree of constraint.
Cheers
ROR
Dear Robert:
What’s the micro-mechanism of cleavage in uniaxial tension test of intact steel sample?
→the microcrack forms when the tensile stress exceed cleavage stress, then propagate immediately.
Or:
→the microcrack form,but can’t propagate. Then it act like the notch or crack in notch or crack sample. When the RKR criterion meet, cleavage occur.
Thank you very much.
I think the characteristic dimension that was mentioned has something to do with plastic deformation accomodation and orientation..how much crystallographic favor it is offering..A crystallographic description can be found in this paper
https://www.researchgate.net/publication/261764767_Effect_of_crystallographic_texture_on_the_cleavage_fracture_mechanism_and_effective_grain_size_of_ferritic_steel
Article Effect of crystallographic texture on the cleavage fracture ...
Dr. Hou:
In a brittle solid such as a ceramic or even an embrittled steel, the strength (which we understand to be controlled by plasticity in ductile materials) can be controlled by the stress for brittle fracture - period! Having said that, one must realize that it is not as simple as that, as brittle solids will contain a distribution of defects of varying size and this, together with the material's intrinsic strength (which presumably is controlled by the atomic bonding), will determine the brittle fracture stress and hence the strength of the material. However, there is still one more "wrinkle". The number and size of these defects, e.g., microcracks, in the brittle solid will depend on the size (strictly volume or surface area) of the sample that you are testing (and its geometry), because fracture, for example, in a uniaxial tensile specimen (which is uniformly stressed), will depend upon the existence of the largest defect - and larger specimens (and tensile as opposed to bend specimens) will have a higher statistical probability of containing such a defect. For this reason, when one quotes the strength of a brittle solid, one needs to quote the type and size of the specimen used to measure that strength. (Leonardo da Vinci, in the 15th-16th Century, even showed for brittle steel wires that short wires were stronger than long ones!). Tensile specimens are preferred to bend specimens (although they're more difficult to test) because the (active) volume of material (or if one is dealing with surface defects, the surface area) that is stressed is far larger - in statistical terms, ten or so uniaxial tensile strength tests is statistically worth something like a hundred bend tests.
So the bottom line here is that as brittle materials will invariably contain pre-existing cracks that will strongly influence the measured stress for failure, the strength of such materials will be a function of the size of specimen that you test. That's why fracture mechanics can be useful in this dilemma, because in principle, with a fracture toughness test, you put in a worst case defect with your pre-crack, and so you take the size of intrinsic defects in the material nominally out of the equation. Consequently, your pre-cracked test will given you a strength (specifically a fracture toughness value) which should be more a function of the intrinsic strength of the material.
ROR
Dr. Hou:
In many steels, the carbides or second phase inclusions are indeed the sites for the initiation of fracture, that is brittle fracture, such as transgranular cleavage. (For ductile fracture, the inclusions play a larger role generally by debonding from the steel matrix to form voids which grow under the applied stresses until the voids link up to create a larger crack; the latter coalescence is often associated with shear instabilities along the smaller carbides which crack rather than debond to form so-called void sheets).
Can you use then the size of the inclusion or carbide to calculate the fracture stress? Well In principle - yes - because if you know the fracture toughness Kc, the size (a) of the particle (or particles) that initiated the fracture and the applied stress (sigma), you have almost everything in the relationship Kc = Q sigma (pi a)1/2 to estimate the critical value of sigma at fracture.
However, as you can guess, it isn't always that easy. You are unlikely to know all the local details: for example, you don't know the local geometry of the fracture, which in part can be characterized by the geometry factor Q, you don't know the appropriate local stress because that will strictly depend upon the configuration of how the particle is stressed and fractures, you don't necessarily know the size of the critical particle that initiated the fracture, and in many instances, certainly for ductile fracture, where cracks initiate and then grow subcritically until instability (final fracture), you don't know necessarily how big that crack was at fracture (although presumably that could be measured macroscopically from the fracture surface. However, several of these factors could be approximated to give a "back-of-the envelope" estimate of the fracture stress.
To my mind though, the best way to effect this calculation, if that's your intent, would be to test a uniaxial tensile sample of a relatively brittle steel which fails by cleavage. As this is a uniform stress situation, you know the stress sigma, and the statistical sampling volume will be huge (the volume of the specimen) such that fracture will inevitably initiate from the largest particle (be it carbide or inclusion related). And so, if you know which type of particles (carbide or inclusions or whatever) are involved in the fracture process and you can measure the size distribution of these particles to find the largest one, you could in estimate the fracture stress. An example of such a calculation is given in the attached paper.
ROR
Dr. Professor Ritchie:
In your article"ON THE EFFECT OF SAMPLING VOLUME ON THE MICROSCOPIC CLEAVAGE FRACTURESTRESS."
Why the S(fracture strength of a particle) is constant in the temperature range ?
in a uniaxial tensile specimen, the fracture strength is given by Su . why it isn't be calculated according to KIC formula?
Thank you very much,
Dr. Hou:
Another good question. The approach that was taken in that work ["ON THE EFFECT OF SAMPLING VOLUME ON THE MICROSCOPIC CLEAVAGE FRACTURE STRESS") and the previous papers by Lin et al. (all of which are available on ResearchGate, was to be able to predict the value of KIc, which is a global parameter describing the overall toughness, on the basis of the local micromechanisms of fracture, in this case the fracture of a carbide particle in a ferritic steel.
As there is a range of sizes of carbide particles, which accordingly have a range of strengths (the strength S vs. size a was estimated using a simple Griffith criterion), the model had to be constituted statistically, i.e., what is the probability ahead of a crack tip, where the tensile stresses are high, of finding a large enough carbide, with a resulting low enough strength, to fracture locally and precipitate global fracture at KIc (using a weakest link - Weibull - assumption). So the approach was to predict the KIc of a steel on the basis of the carbide strength S using this statistical RKR-type model.
The precise purpose of the paper to which you refer is that in a uniaxial tensile specimen, the statistical sampling volume (where the active fracture processes take place) will always be large enough to statistically find the largest carbide (with the lowest - lower-bound strength Su) to cause fracture. In fact, at first sight it is somewhat amazing to note that this active sampling volume in a uniaxial tensile sample (where the stresses are nominally uniform) will be on the order of 9 or more orders of magnitude larger than the active sampling volume existing ahead of a crack tip (where the maximum stresses peaks at a distance of ~2 CTODs from the crack tip, i.e., at a distance on the order of K2/[sigma_y . E], where sigma-y is the yield or flow stress and E is Young's modulus). If you cube that to get the active volume at a crack tip, it is a very small number (as stated above, ~9 or more orders of magnitude smaller than the volume of a typical tensile specimen), and so the largest particles with the lowest strengths often do not participate in the fracture at a sharp crack because statistically they are unlikely to be found in that tiny active volume.
ROR
Dr. Hou:
There are two inherent/theoretical/ideal strengths, the ideal shear strength and the ideal cohesive strength, which are computed from the stress needed to, respectively, separate an ideal (defect-free) array of atoms in shear or to separate them normally. In simple terms, both ideal strengths are computed from the binding energies, or the bond strengths, between atoms and are typically calculated to be on the order of G/10 for the ideal shear strength (where G is the shear modulus) and ~E/10 for the ideal cohesive strength (where E is Young's modulus). They respectively represent the upper-bound values for the shear yield strength and fracture strength of any material (in the absence of defects). In reality though, because of the inevitable presence of defects, the ideal shear strength is 1 to 2 orders larger than macroscopically/experimentally-measured shear yield strength values; if we are referring to metallic materials, the most relevant defect here is, of course, the presence of dislocations. Similarly, for the ideal cohesive strength, most experimentally measured fracture strengths are 1 to 2 orders of magnitude smaller because of the inevitable presence of cracks.
And so if you can make a defect-free material, you can realize much enhanced properties, both for deformation (yielding) and brittle fracture, but unless you test smaller and smaller samples approaching the nanoscale, it is practically impossible to realize these upper-bound strengths. For the former case, this has led to the catch phrases like "small is better" or "size matters", but in realistic and practical engineering terms these phrases are only of academic relevance.
ROR