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In the last class we had started discussing the solution of Newman and Raju, we were not
able to completely cover it, for the sake of continuity we will quickly look at what
we have discussed in the last class. And I had mention Raju and Newman in 1979 under
took a detailed, comprehensive three dimensional finite element analysis, and based on that
result, in 1981 they had come out with an empirical relation consisting of double-series
polynomials.
And the relationship looks something like this, this you have determined in the last
class. And sigma t refers to the tensile stress, and sigma b refers to the bending stress,
and we have also looked at how to calculate Q.
This is given for a by c less than equal to 1 also given for a by c greater than equal
to 1, and if you do this, with this kind of an expression the maximum error is about 0.13
percent for all values of a by c.
Then, we moved on to look at what are these functions F. This has component of M 1, M
2, M 3.
And these were also defined, all these equations were determined in the last class. We have
also looked at the functions F theta, G, as well as F w. Then, we moved on to look at
what is the function H? And you have to note, that using engineering judgment Newman and
Raju express the function H as in this form, H equal to H 1 plus H 2 minus H 1, sin power
p theta, and p is defined as 0.2 plus a by c, plus 0.6 into a by B. H 1 is defined as
1 minus 0.34 a by B, minus 0.11 a by c, multiplied by a by B, and H 2 is given as 1 plus G 1
multiplied by a by B plus G 2 a by B whole square, to this extent, I think we have seen
it in the last class and we have to know the functions G 1 and G 2, these are defined next.
The function G 1 is given as minus 1.22, minus 0.12 multiplied by a by c, then the function
G 2 is given as 0.55 minus 1.05 multiplied by a by c whole power 3 by 4 plus 0.47 a by
c whole power 3 by 2. You know this completes the definition of the components consisting
of the empirical relation, and you should note that he had done a three dimensional
final read element analysis, based on that at fit a empirical relationship and the relationship
is as follows.
K 1 equal to sigma t plus H sigma b multiplied by root of pi a divided by Q, and a function
which is dictated by a by B ratio, a by c ratio, c by W ratio and also the position
theta. And what you'll have to note is, Newman and Raju have reported that the empirical
relation is within 5 percent of the finite element calculation for the full range of
a by B, and a by c, from 0 to 1, and 2 c by W less than 0.5, we will also have to note
down the width of the specimen, that also playing a role, And you have to note, despite
their empirical origin these equations have gained wide acceptance, and are employed for
various fracture-related calculations whenever surface flaws are encountered.
So, as far as surface flaws are concerned you can always go and look at the empirical
relations of Newman and Raju, and use it for your calculations that level of confidence
in the fracture community has placed on the axis. So, we have seen surface cracks are
very important from practical point of view, and we have looked at 3 different methodologies.
One was the very simplistic approach by Irwin then, there was improvement of a front free-surface
correction factor. Then, there was also correction factor, because of plastic zone length and
you had graphs of flaw shape parameter. Then we had also looked at direct analysis of surface
cracks, where people have provided separate graphs for tension as well as bending, and
finally, we have looked at the empirical relations of Newman and Raju.
Now, another important aspect that we will have to discuss in relation to stress intensity
factor and the role, you should also look at what way we will select fracture toughness
for radius problems. Suppose, I have a through the thickness crack,
and for the purpose of discussion; the crack is shown as a very sharp corner, and you have
the crack front straight, with the simplest problem to take. And let us look at what way
the material responds, because of high localized stresses near the crack-tip region.
So, you have very high levels of stresses, this your elastic solution provides, in view
of these very high localized stresses, what happens? There is no contraction in this zone,
in this zone the material wants to contract, and we will have to see to what extent in
this, contract contraction is possible. And ahead of this zone there is little contraction,
and what happens in this zone? So, what I would do is, I would repeat the
animation and you can recapture the various sequence of ideas that we have represented.
So, I have the plate with a crack pull, you see a very high level of stresses then, because
of a consequence of the stresses the material wants to contract, and you consider a cylinder
of material in this zone. And we will also have to qualify this result for plates of
various thicknesses. What happens in a very thin plate? And what
happens in a thick plate? These are two extremes that we will look at.
What you can always notice is, when you have a stress concentration zone normally you come
across a uniactive field changing into biaxial field but, in the case of crack problems for
thick plates you have to also consider one more aspect, we have already seen, when you
consider the crack-tip is very sharp sigma x equal to sigma y, if the crack is blunt,
at the crack-tip you will have sigma x as 0, for the purpose of discussion we, let us
consider crack-tip is sharp. So, you have sigma x equal to sigma y, and
if the plate is sufficiently take, you will also have stresses developed in the thickness
direction of the plate, and that is given as nu times sigma x plus sigma y. And if you
look at the strain, strain will be 0, if you look at the stress between nu times sigma
x plus sigma y and, because of very high localized stresses you will have essentially plastic
deformation at the crack-tip. So, when we have plastic deformation at the
crack-tip, it is prudent to take the Poisson ratio as 0.5. So, in a sense, in thick plates
you have triaxial state of stress near the crack and that zone is indicated here.
So, in the case of crack problems near the vicinity of the stress concentration, you
could have under suitable circumstances a triaxial state of stress. And that is what
is depicted for a thick plate and for a thin plate. And you can make a neat sketch of this
so, in a thick plate leaving the surfaces, the zone interior to that will be in a triaxial
state of stress. And you find there is negligible contraction in the case of a thick plate.
So, in such problems from fracture instability point of view, what you will have to do? You
will have to calculate stress intensity factor when you have a crack in a thick plate, from
fracture instability point of view, you will have to know what value of fracture toughness
that you will have to take, though, we will have a separate chapter on fracture toughness
testing, certain concepts related to that we may have to discuss even in earlier chapters.
So, in such a problem where you have a thick plate, you have to use the plane strain fracture
toughness. And I have already mentioned, the plane strain fracture toughness is lower than
the plane stress fracture toughness. So, the combination of stress intensity factor value,
and the selection of fracture toughness ultimately dictates the fracture instability phenomenon.
On the other hand in the case of a thin plate you have a free contraction, and what is recommended
is, you have to use plane stress fracture toughness for your fracture instability calculations.
After having looked at through the thicknesses cracks, we will have to go and look at how
you are going to handle these surface cracks?
What happens in the case of a surface and corner cracks? We all know that, it has a
curved crack front and a stress intensity factor varies along the curved crack front,
and if you consider a cylinder of material close to the curved crack front the inner
zone is experiencing a triaxial state of stress. You know this is depicted by a shaded region
so, whenever you have a triaxiality constrain the recommendation is, you have to use plane
strain fracture toughness for your instability calculations, and when you have to do the
use of plane strain fracture toughness, you have to recognize that surface flaws are always
dangerous, because you are always comparing the stress intensity factor of a surface flaw
for a through the thicknesses edge crack but, for an edge crack in a reasonably thin plate
you may use the plane stress fracture toughness but, if you have a surface crack, then you
have to use a plane strain fracture toughness. So, because of that you have to keep in mind,
the surface cracks are always dangerous. See, now we have look that stress intensity factor
for a variety of problems, and it is better that, we also find out a sort of a thumb rule
in react into different crack situations.
So, we will go and review those results. And we have started the discussion with equally
spaced cracks in an infinite strip, and we had discussed that from this, you get the
center crack specimen, from this solution you are also able to get the single edge notch
specimen, and also a double edger notched specimen or crack specimen.
And when you have a center crack, we are looked at in an infinite plate it is K 1 equal to
sigma root pi a, for a finite plate you will have a function of a by W, it may be a by
W or 2 a by W, depending on how the result is reported.
The moment you go to an edge crack, we have noted from our discussion, edge cracks have
a higher stress intensity factor than a center crack. We had a factor of 1.12, for an infinite
plate it will be like this, for a finite plate you will have a function related to a by W
or 2 a by W, depending on how the result is reported. On the other hand, when I have an
embedded circular flaw, what we noted that stress intensity factor remain constant on
the crack front, and in comparison to a center crack in an infinite plate the stress intensity
factor is lower, it is 0.64 times sigma root pi a.
After the circular flaw, we had moved on to an elliptical flaw and only in an elliptical
flaw we noted that, the stress intensity factor can change from point to point on the crack
front. And that definition is given here, you have a sigma root by a divided by I 2,
and you have a function related to theta, and we have also noted when you specify theta,
how to locate the point on the ellipse.
If I have theta like this, you drop a normal and this hits this, ellipse at this point
so, for this point the stress intensity factor is given. From an embedded elliptical flaw
we moved on to a surface crack, and here we had extrapolated what all the understanding
of that elliptical flaw, and also for an edge crack and the simplest solution was simply
change the solution by 1.12 times sigma. Later, on we improved upon a definition of crack
length modified for plastic correction, and then we had also looked at empirical relation
of Newman and Raju. But basically you have to recognize, compare
to an embedded elliptical flaw a surface flaw is little more dangerous. And then we moved
on to a corner crack, and for the corner crack we recognize you have a free surface on this
side, as well as another free surface and we realize that, you will have K 1 is much
higher for this and you have this as simplified to 1.2 times sigma root pi a.
So, this discussion in a sense brings out a relative appreciation of how you should
react to cracks of various types; whether they are through the thickness crack, or interior
to the object, or on the surface, or it is an embedded crack, you have a rough approach
on how to look at the influence of the stress intensity factor on the overall structural
behavior.
Now, what we will do is, we will move on to modeling of plastic zone near the vicinity
of the crack-tip. And we will have to look at what is the motivation for all this? You
know, essentially we have discuss fracture mechanics in the context of brittle materials,
that is how Griffith started, later on it was extended by Irwin and Orowan for handling
ductile materials, and when you go and look at that what you will have to learn is, we
also want to look at materials that fracture with limited plastic deformation, at applied
stress levels less than those producing net section yielding.
This is how we have started the extension of fracture mechanism, brittle materials to
ductile materials, and you have to recognize initially fracture mechanics was focused on
linear elastic material behavior. And we had sufficient success in this, the theory is
matched with experimental observation so, with the success of linear elastic fracture
mechanics people also thought, for materials for which such an approximation would be invalid
also became of interest so, that is the motivation. See, you cannot keep away from your understanding
of what happens at the crack-tip, because of un-elastic deformation, some kind of modeling
is always needed, because you have to graduate from brittle materials to high strength alloys,
from high strength alloys to even intermediate alloys which exhibit reasonable levels of
plasticity. So, here what you will look at is, what are
the tricks that, they develop within the domain of linear elastic fracture mechanics? To extent
this analysis to certain kind of a ductile materials, later on if there is extensive
plastic deformation, you have to bring in the concepts of elastoplastic fracture mechanics.
What we will now recognize is, what is the motivation for finding out all these aspects?
So, the motivation is extension of linear elastic fracture mechanics. And as I mention
with the success of LEFM, materials for which such as approximation would be invalid also
became of interest. And there is on the another observation, this also has prompted people
to go in that direction. See the structural component itself would likely to obey LEFM
but, the small scale laboratory experiments needed to provide the fracture properties,
would not behave in such a fashion. So, this is another angle to it, people want
to find out the relevant fracture parameters to characterize the particular material in
service. So, this is another dimension why people were
looking for newer ways of handling the plastic zone near the crack-tip. What we will have
to note is materials that are sufficiently ductile and tough, that the extent of plastic
yielding accompanying the crack growth would be comparable to the specimen dimensions.
So, this is the goal, if you have this kind of a situation, how will you go and do it?
And what was initially done was, that was the work of Irwin Kies and Smith in 1958,
to broaden the applicability of the linear elastic approach, and proposed a plasticity-enhanced
stress intensity factor in which the crack lengths were slightly enlarged suitably.
See, you have to look at this was the kind of work in 1958, when fracture mechanics was
in the initial stages, where significant contributions and new ideas have been proposed by the group
lead by Irwin. And they had come out with a very simple modification, what they have
said is; you have to get up plasticity-enhanced stress intensity factor, in this what they
had done is; we have already look at in the contest of a surface crack instead of taking
the crack length as a, they have taken a small extension of crack length which is dictated
by the plastic deformation near the crack-tip. This was one kind of an approach, another
kind of approach was promoted by Wells and Cottrell, they have done it independently
in 1961. They advanced an alternative concept in the hope that, it would apply even beyond
general yielding conditions. We would take this, when we discuss j integral and another
measure for elastoplastic fracture mechanics is the COD approach. We will just see that,
these two approaches were there to understand the plastic deformation near the crack-tip.
And from the result of Irwin wells evaluated the crack opening displacement and here it
is used in a different context. We would have a pictorial representation now,
we will also spend some time a little while later, from the point of view of Irwin you
imagine the crack is to be longer than the actual crack. The actual crack is only this
much so, at the tip the would be opening, which is labeled as COD.
So, people had coin new parameters, particularly from the point of view of particularly from
the point of view of taking the laboratory results, useful for analyzing actual structural
components. So, in this chapter you will confine our attention
to, what was the discussion done by Irwin? What is the approximate shape of plastic zone?
And also what is the model given by ductile in finding out extent of plastic deformation
ahead of the crack-tip? And before we proceed into that, it is worthwhile to recall, how
plane stress and plane strain terminologies are used in fracture mechanics?
If you look at in applied mechanics these have very precise meanings, we know what is
the plane stress, we have looked at the stress tensor as well as strain tensor. The moment
you come to fracture mechanics these are applied in somewhat looser ways, you have to recognize
that and keep that in your mind. In the contest of applied mechanics, plane
stress rigorously means that the principal stress acting in the direction normal to the
plane of interest is negligibly small. In fracture what do you call as plane stress,
it is referred for thin components with in-plane loading, and we also do one more thing, we
say the surface layer of thicker components behave in a plane stress fashion.
You know this is something unusual, this is so convenient for us to develop certain concepts
in fracture mechanics.
So, we have adapted a slight variation in the definition of, what is plane stress and
plane strain in the context of fracture mechanics. So, you have to recognize that, there is a
difference. And you will also have to note, for a state of plane stress to occur the stress
gradients in the direction of normal to the plane must also be negligibly small.
This condition is not completely satisfied even in thin plates, it is only approximately
satisfied for a thin plate. See, because of plastic deformation you have the neighboring
elastic region will refuse to deform to that extent. So, you will have sort of a tension
compression type of situation in the plastic zone as well as the elastic zone.
So, you will have gradient of stresses developed in the thickness direction so, in the contest
of fracture mechanics though we say thin plates subjected to in-plane loading can be equated
to a plane stress situation, from the definition of applied mechanics the definition is strictly
not corrected, it is only approximately satisfied in a thin plate.
Certainly it does not satisfy the conditions in the surface of a thicker body. You have
gradients existing into thickness direction that is quite alright, because we are handling
a very complex problem situation, we have to carry forward.
So, we need to make certain approximations that make our life simpler. And before we
get into the discussion of plastic zone, we have already seen the range of linear elastic
fracture mechanics and elastoplastic fracture mechanics, based on the plastic zone.
And you have to recognize the plastic zone is very highly localized only when you magnify
you are able to see, and this is the situation that exist in high strength materials in plane
strain. And in the case of high strength material in plane stress, you have little more plastic
zone but, compare to the plate dimensions and also the crack length the size of the
plastic zone is very small. So, only for these class of problems we will make suitable approximations
and find out certain modifications to our way of calculating stress intensity factor.
So that, LEFM could be applied to a broader class of high strength ductile alloys, the
moment you have a very high plastic zone you have to going for elastoplastic analysis,
and if the plastic zone is much larger compare to the crack and other specimen dimensions
you will have to going for analysis based on plastic collapse.
This we have already seen, this is just reviewed for the sake of in which domain we are going
to leave, we are not even going to look at such large plastic zone, you are going to
look at very small plastic zone, that is why we have characterize this as small scale yielding,
whatever the discussion that we do it is applicable only when this aspect is satisfied.
Usually the plasticity effects are assumed to be negligible in the highly stressed crack-tip
vicinity. This is the reasonable assumption for thick sections with small scale yielding,
and we will also take up in detail. What is small scale yielding? After looking at Irwin's
correction, for the time being you can consider, in small scale yielding the singular stress
field determine by the stress intensity factor is assumed to prevail outside the zone of
plasticity. This is the reasonably a good assumption,
we are making an assumption to make our life simpler, as long as this assumption is valid
certain improvements in our stress intensity factor calculation can accommodate this class
of materials, that is the way you have to look at it. And what is reminded here is the
fracture mechanics based engineering design looks at brittle failure of structures, structure
has a whole fails in a brittle fashion but, what happens at the crack-tip? Certain amount
of plasticity effects also dictate what happens, and how this is handle? That is the way we
have to look at it.
And how to evaluate the plastic zone, it is a very difficult aspect, it is difficult to
give a proper description of plastic zone shape and size. We will look at simplistic
models and the determination of shape has a connotation, because you would be able to
quickly compare, what happens in aplane strain situation and a plane stress situation. And
in all the models to simplify the analysis usually the material is assumed to be elastic-perfectly
plastic. In reality many materials exhibit strain hardly, we are not considering that
for a simplistic analysis, we simply say the material is elastic-perfectly plastic. And
what happens, because of the plastic zone at the crack-tip, the stiffness of the component
decreases that is the compliance increases. So, only to accommodate this kind of an observation
Irwin said, that the crack is to be mathematically modeled to be longer than the actual length.
So, the question is how to find out the extension of appropriate crack length? This could be
done in many days, we would look at a very simple model and then improve our calculations.
So, one of the quickest methods but, definitely a very crude approach is to find the extent
of plastic zone along the crack axis by simply finding the point at which one of the yield
criteria is satisfied. We will not do any elaborate elastoplastic
calculation, at the beginning of the course I had given you a review of solid mechanics,
there you had look that, when the stresses are equal, when the two principal stresses
are equal at the crack-tip for a plane stress situation as well as the plane strain situation
for which you have Poisson ratio as 1 by 3, I have asked to calculate, what are the levels
of stress given by the yield theories? So, if you perform a simple tension test you
are doing a uniaxial test and you have the stress develop is sigma y s, But if I have
multi-axial situation the individual stress magnitudes will be much higher than the yield
strength, this you had actually look that as part of a review of solid mechanics, and
I am going to use that result right away. And if I use this kind of an approach of simply
finding out the point at which on set of yield criteria satisfied, which is very crude. Nevertheless,
this helps us to compare the relative plastic zone size in plane stress and plane strain.
And I had mentioned earlier for rest of the course we will worry only about the singular
stress field, and you know this singular stress field for mode one crack, and if I have to
go and find out the plastic zone the first step is you have to determine the principal
stresses.
Now, you have the stress field, you know how to find out the principal stress from this,
I would like you to work it out and then calculate the expression for principal stresses near
the crack-tip for the mode one situation. You have the expressions for the stress field
and it easy to calculate the principal stresses. So, we will now look at the expression for
the principal of stresses, sigma 1 is given as K 1 by root of 2 pi r, cos theta by 2 multiplied
by 1 plus sin theta by 2. And sigma 2 is K 1 by root of 2 pi r, cos theta by 2 multiplied
by 1 minus sin theta by 2, and in the case of a plane stress situation sigma 3 is 0,
in the case of plane strain situation, which is seen in thick plates, I have sigma 3 given
as 2 nu K 1 divided by root of 2 pi r cos theta by 2.
When theta equal to 0 you can find out when nu equal to 0.5, it will be same as your sigma
1 sigma 2, what you have here. And this explains the triaxial situation near the crack-tip,
we had seen it pictorially earlier now we are looking at a mathematically so, the moment
you have a crack you have to recognize that triaxial situation is possible.
This is something very significant, we have a plate with a hole you essentially say you
have a biaxial stress field, a uniaxial stress field changes to biaxial stress field. The
moment you have a crack it can become a triaxial situation.
Suppose, the crack-tip is blunt instead of happening at the crack-tip the triaxial situation
will take place slightly ahead of the crack-tip - that is the way you have to look at it.
And what is the material model that we have used? Elastic-perfectly plastic, and the yield
criteria are two you know this, just for review we are just looking at a the von Mises criterion
utilizes all the three principal stresses. The condition is sigma 1 minus sigma 2 whole
square, plus sigma 2 minus sigma 3 whole square, plus sigma 3 minus sigma 1 whole square, greater
than or equal to 2 sigma y s square. This is the yield strength, and you want this
to be within sigma y s, if it is greater than that yielding would occur. And when you go
to Tresca criterion you have to recognize it is not sigma 1 minus sigma 2, it is written
as sigma max minus sigma minimum divided by 2, and this becomes important in fracture
mechanics problem, because at the crack-tip if you consider the crack-tip is sharp, sigma
x equal to sigma y you are getting it, not only this, both of them are the same sign,
when both of them are the same sign the other minimum stress is zero so, the zero plays
a very important role. So, when you are applying Tresca yield criteria
to plane stress situation fracture problem you have to calculate the maximum shear stress
carefully.
So, once you know the values of sigma 1, sigma 2, and sigma 3, either by using this criteria
or this criteria you can always find out, what is the plastic zone size. And this is
just to give a pictorial representation, what do we mean by elastic-perfectly plastic? You
do not consider any strain hardening of the material, because this is simple for us to
do the calculation.
Now, what is plotted is you have a crack and I have the values of stresses put here sigma
y, and if you go by your plane strain situation, when you use the yield criteria ,what you
find is, the sigma y can take a very high value of 3 times sigma y s, when you have
Poisson ratio is 1 by 3. See, in a simple tension test when the stresses reach sigma
y s, yielding takes place. We have seen in the case of a crack problem you have a triaxial
loading situation. In a triaxial loading situation particularly,
when you consider plane strain situation the individual stress magnitudes can be very high.
It can be as high as three times the values of the yield strength, and you also define
there is a length, you are looking at what happens along the crack axis, simply mark
this point as a length. This is erroneous, we would improve upon it
later first we have said, we would simply find out the yield strength value, based on
that find out what is the length, based on this find out what happens as a function of
theta, that would give the plastic zone shape which is very approximate, because we are
not considering redistribution of load. We are simply plugging in mathematically whichever
the points which reaches the yield strength value - that is not going to happen physically.
But definitely it gives an approximate shape of the plastic zone near the crack-tip, and
the value of r p, turns out to be 1 by 18 pi multiplied by K 1 by sigma y s whole square,
and this is particularly for the value of Poisson ratio equal to 1 by 3. And the moment
you come to a plane stress situation, I had mention that sigma y equal to sigma x, and
the other stress is what? Other stress sigma 3 is 0.
So, that tells you the maximum stress that you will have to look at is only sigma y s.
And if you plot in this graph and locate the point where it reaches the sigma y s, it is
here. So, in the case of a plane stress situation the plastic zone length ahead of the crack
is very long, and this is given as 1 by 2 pi multiplied by K 1 by sigma y s whole square
so, a very simplistic analysis has shown the relative importance of plastic zone in the
case of a plane stress as well as plane strain.
The plastic zone is very large in the case of plane stress; it is very small in the case
of plane strain. See, these discussions are needed even to decide, what is the appropriate
thickness to conduct fracture toughness testing. Definitely, this is the very simplistic model,
but the advantage of a simplistic model is we can also go and estimate, what is the shape
of the plastic zone. Let us look at that and even to start with the title is given as approximate.
So, whatever the methodology that we have adopted to get the length of the plastic zone
along the crack axis, extended to get the shape of the zone. And when you want to get
the shape of the zone, make it as a polar plot. We have the expressions for sigma 1
and sigma 2 in terms of theta. So, we will express the value of r p, in the range minus
pi to pi as a function of theta. And draw the polar plot that will help us to relatively
compare the plastic zones for plane stress and plane strain.
This is the purpose, see if you have to really obtain the plastic zone, either you should
go to an experimental approach, metal edges have done it, we will also see in the context
of ductile model, they have arrived at shape of the plastic zone experimentally. Otherwise,
you will have to go to a finite element calculation, exhaustive elastoplastic analysis you perform,
and you do this for every kind of specimen, because all this things are depends on the
kind of specimen as well as the loading, what we are now trying to look at is, a sort of
an approximate understanding on relative shapes of these in plane stress and plane strain,
you have to keep the focus in mind. So, it gives only a first order approximation of
the shape, because the methodology does not attempt to redistribute the load.
So, this is the key aspect. And for the case of plane stress, I had already drew your attention
and that is drawn in the case of Mohr circle, you have the zero stress, this is to be taken
into account. And you have the value of r p, as the function of theta is given like
this and these expressions would change for the different yield criterion.
If I use von Mises, I get r p as 1 by 4 pi multiplied by K 1 by sigma y s whole square,
multiplied by 1 plus 3 by 2 sin square theta plus cos theta. See, even for our well-developed
mechanics of solids, different yield criteria provide different results.
So, in fracture mechanics we are dealing very complex situations so, you have to accept
multiple solutions, you have to pick out which one is suitable for it, because in mechanism
solids what we understand certain materials obey Tresca yield criteria. So, for those
materials apply Tresca yield criteria, certain materials are obey von Mises yield criteria,
this is an accepted practice and what we are now trying to look at is, how do the plastic
zone shape differs with respect to plane stress as well as plane strain, and also with respect
to the invocation of yield criteria by von Mises or yield criteria by Tresca? And for
Tresca this turns out to be 1 by 2 pi K 1 by sigma y s whole square multiplied by cos
theta by 2, 1 plus sin theta by 2 whole square, I am not reading the brackets you put the
brackets as for the equation. Since you have these expressions, it is possible for you
to do the polar plot. So, what I would appreciate is you make an attempt, you go to your rooms,
you have the expression, you calculate for few values of theta, how does the shape look
like? And the next class we will see, and what I would also appreciate is we have develop
this for mode one situation, get similar expressions for mode two as well as mode 3 and try to
come with polar plots of approximate shape of plastic zone.
So, in this class what we have discuss was, we have look at the final expressions given
by Newman and Raju on the empirical relation of surface cracks, then we look at how triaxiality
happens near the crack-tip. And what way we use that terminologies plane stress and plane
strain in the context of fracture mechanics, what is the difference between the applied
mechanics definition and a looser definition in the case of fracture mechanics.
Then we also looked at an inside into stress intensity factors for various situations,
though the stress intensity factor for a surface crack is compare to through the thickness
crack, I pointed out, because of triaxiality from fracture instability point of view you
will have to use plane strain fracture toughness for surface cracks so, you have to keep it
at the back of your mind, that surface cracks are always dangerous.
Then we moved on to the motivation for determining plastic zone a kind of the crack-tip, the
essential idea is to extent fracture mechanics to larger class of material and people have
looked at simpler modifications, and to that extent we have gone and looked at one of the
very simple model to find out the extension of plastic zone along the crack axis, and
we have also obtain the relationship as the function of theta and I have asked to plot
this and come for the next class. Thank you.