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Lecture 33: Fracture: Part 1


Hello friends todays lecture will be on fracture
ah. So, after we do if you remember the tensile curve ok we start from elastic deformation
then we we go into the plastic deformation range and plastic deformation range you have
uniform ah deformation ok. And then you have a necking and then no uniform deformation
and at some point you will have fracture in the material ok. So, we want to understand
this phenomena of fracture that what does it mean and does it depend on that ah different
materials in different materials do we have different type of fracture ah mechanisms ok,
so all those things we would like to see. This is one of the some very famous photographs
to to bring out the importance of understanding of fracture, our and actually importance of
materials also these are some ah ships which were made in world war second these are called
liberty ships and most of this ship failed like this, you can see ah the fracture the
ship has fractured into two parts ok, I completely two part it has fractured and the big ship
like this just fractured into two parts is a kind of a very difficult thing to understand
ok. So, people found out that there were problems
in the material there were problem in the fabrication and so on ok and lot of improvements
took place after that. In fact, titanic also if you see it was a kind of a fracture problem
only that the material was not ah of good quality , lot of carbon used to be there in
the steel at that time ok and when the iceberg hit the titanic ship ok.
So, it hit fractured instead of they having a ductile fracture or ductile deformation
it actually fractured in a brittle manner ok. And then once it is the crack started
it under the ah the massive load of that ship and that propagated and the whole ah ship
got again I think divided into two parts ok. So, types of fracture if you want to see there
can be two type in which we can divide one is brittle ok, brittle is a is a very difficult
and problematic kind of fracture because you don’t get any warning. So, once it starts
and then it will go at a very fast rate so you can say it is a fast fracture also.
So, brittle fracture in this you do not have any plastic deformation or very small plastic
deformation and that is why there is no warning ok. So, you will not know when you have reached
the , the fracture stress of the material whereas, as you when you are plastic deformation
it will take ah lot of time in getting deformed and then go into the fracture mode ok. And
before that you will actually be able to see that something is deforming out of shape.
So, separation in this case is normal to tensile stress it is kind of a de cohesion between
the two atom. So, all the atom bonds kind of break ok and usually observed in b c c
and f c c material, f c c because it is very ductile ah ah f c c structure materials are
ductile you do not see there in bcc also you will see only at very low temperatures at
normal room temperature you may not see it ok
So, there is effect of temperature also here ok and hcp of course, you will see this kind
of fracture ah brittle fracture ah ductile fracture you will have appreciable plastic
deformation ah which occurs prior to or during the fracture process in a tensile test pronounced
necking before fracture cup and cone fracture. So, when we do tensile test also we will clearly
ah be able to tell you that which one is a ah ductile material which is a brittle material
depending upon how they are getting fractured. So, in if you do a tensile test of brittle
material, just taking the gate section actually your material will divide into two parts ok.
And the fracture has taken place ah normal to the tensile stress, where is in case of
ductile material the material will actually neck first ok and in this also there is a
kind of a very nice geometry to that ok. So, in one case it will be a cup and in the
other part it will be a cone ok. So, it separates like this so it is called a cup one cup and
cone fracture, fracture mode depends on ah different ah parameters. So, you can have
as a function of material. So, crystal structure or alloying ok already
we have seen that different ah crystal structure and behave differently temperature as I told
you that ah the temperature plays very important. So, in bcc material as a function of temperature
it will either show as a brittle ah fracture as a brittle material or fracture as a ductile
material ok. So, temperature plays very important role here state of stress whether it is a
plane stress condition or plane strain condition rate of loading strain rate ok.
So, if it is high strain rate then sometime material have a fracture is a brittle material,
if you are doing the same ah test on the same material, but it has lower strain rate it
will fracture in a ductile manner environment. So, basically corrosion and hydrogen is there
hydrogen embrittlement is there. So, depending upon the environment ah the fracture will
change in some cases a ductile material will ah fracture like a brittle material because
it has already corroded ok So, all this factors will determine the fracture
modes, so if you kind of want to see the whole ah.
Kind of a overall scene or overall ah difference between different fracture mode ok. So, you
have brittle here and ductile here ok. So, this is brittle here and ductile here ok and
this is at high temperature at this is at low temperatures ok, ah at low temperatures
in brittle material you will have a fracture like this where the crack is going through
the grain ok. It is a cleavage fracture we say or you can
have intergranular brittle fracture. So, you can see is see in this case the crack has
gone through the grain boundary. So, it is called intergranular and this will be call
as intragranular, then you can have plastic growth of voids, so you can see some voids
are forming within the grain here ok. So, within the grain it will be trans granular
or intra granular and between two grains it will be intergranular ok. So, avoid can be
at the grain boundary or voids can be within the grain and this voids will then grow then
they will connect with each other and the material with fracture. And this is for ductile
material rupture ah by necking of shearing off. So, you can see that necking is taking
place and at the end you will have a fracture here as a point fracture or cup and cone fracture
or you can have shearing ah in form of shearing at high temperature again you have will have
ah formation of voids at the grain boundaries ok.
So, it is ah intra granular fracture through formation of voids or there can be formation
of wedge at the grain boundaries then it is called wedge cracks growth of voids by power
law creep another when we will see creep you will see again this type of fracture taking
place ok or it will be through ah necking as we saw and the low temperature also ok.
Now, coming to the ah brittle fracture ok it is easier to understand the brittle fracture
first ok, so we will first look at the brittle fracture ok so.
Coming to before coming to that let us see what will be the theoretical cohesive strength.
So, basically what is we are saying is when we have fracture. So, suppose this was the
material initially of course, you had atoms here ok and there was atomic bond between
them ok. So, when I am saying that it is fracturing into two parts ok, so it is fractured here
ok. So, basically I am breaking these bonds between the atoms and then it will be in the
material will be in two parts. So, I can kind of get an estimate that what
kind of ah stress is required to do this breaking of all the bonds ok. So, if you I take up
material of this size let us say there will be like millions of atoms and millions of
bonds that that I have to break to take into two parts ok. So, suppose the atoms are there
then there are bonds between them this is a fracture plane and I am simultaneously rupturing
all the atomic bonds for fracture to take place.
So, if I want to do that as the theoretical calculation the derivation I am not giving
here you can look in the book for example, by book by on mechanical metallurgy by deter.
So, sigma theoretical will be equal to e by 2 pi, it is similar to what we saw earlier
in case of shearing also that when I want to share all the atoms at one go ok I have
to break all the bonds ok, create new bond and by that I will slip it by one atomic distance
ok and what was the ah stress required to do that.
Similarly, here now in this case I am breaking all the bonds between the atoms ah simultaneously,
and this is the ah strength or stress required to do that that is equal to young’s modular
key divided by 2 pi young modulus already we know for most of the metallic material
it is in the range giga Pascal, but the engineering material when we try to find out the fracture
stresses these are around 10 to 1000 times slower than the theoretical value ok already
we have seen that it is usually in mega Pascal or something. So, maybe ah 2 or 3 order of
magnitude difference between what we get from theory and what we see in the ah experiments
ok. So, again in this case also there must be
some something in the material ok which is bringing down the ah this strength of the
material ok. So, in case of plastic deformation shearing we say said that the defect is dislocation
that defect is dislocation which is bringing down the theoretical shear strength of the
material ok, in this case also we are going to have some defects and this defects are
called cracks. So ah the reason for bringing down the this
theoretical strength is that there are fine cracks present ok. So, first time it was proposed
by Griffith ok, so and he proposed that for brittle materials because ah fracture process
is very easy to understand in brittle materials because there is no plastic deformation. So,
only the separation between the two materials through breaking of the bonds ok without any
plastic deformation. So, Griffith offered an explanation for the
discrepancy between observed and theoretical cohesive strength of course, what he offered
was only applicable for brittle material like glass ok not even brittle metals it is ah
brittle material like glass and Griffith propose that brittle material contained large number
of fine cracks ok. So, what happens when you have large number of fine cracks for example,
one crack is shown here in this figure ok you are applying a ah force f ok. So, the
stress on the system will be given by sigma with that say we call it is as nominal stress
ok that is force upon this is the area of cross sectional area of the sample ok. So,
nominal stress is only force upon per unit area ok.
Now, when you look if you have a crack in the material what will happen is that locally
the stress will be very high where the crack tip is there ok and this is called stress
concentration. So, wherever you ah have sharp ah some discontinuity like this, you are going
to have a very high for the nominal stress which you are applying the locally the stress
will be very high ok. And that is why if you remember when I was discussing the tensile
test also I said that when we prepare the sample tensile sample we do not prepare it
like this ok. The reason is this sharp corner will impose
or we will give rise to stress concentration and the locally the stress here will be very
high. So, this kind of sharp corner like this will act like a crack here ok. So, instead
of that what we do is we give ah , kind of the curvature like this ok and ah to to minimise
the stress concentration ok. So, what will happen is that where the crack tip is there
the stress is very high, so this is my sigma max ok.
And as you go away from the crack tip the stress is equal to whatever you will get from
the force per unit area and the dimension of the crack is also defined here. So, the
total dimension is 2 c ok, we will if it is inside crack we will take only the half of
that that is c if it is a surface crack suppose it can be here. Then we will just take the
whatever is the total ah length of the crack which is going to be c ok. So, if it is inside
crack then whatever crack length is there half of that, so we will always say that it
is 2 c. So, half of that will be c if it is surface
crack then we will just take the c here ok. So, sigma max will be equal to ah you can
have an expression also here it will be equal to 2 sigma nominal and 2 under root of c by
rho ok I think that. So, when you have fine crack it produces a
stress concentration ok and its magnitude is closer to cohesive strength in localized
reason. So, what will happen that the over all the stress is very small ok, and we are
seeing that ok the material has fractured at this is stress the locally the stress will
be very high that can be close to the cohesive strength of the material. So, locally here
it will reach the what we have calculated for given material. So, sigma max can reach
e by 2 pi locally ok where is what you are measuring it is on the sigma nominal that
is why you feel that ok it is fracturing at very lowest stress ok.
So, as you can see this sigma max depend upon the crack length. So, if you have a larger
crack length then the stress will be also be higher the stress concentration at that
crack length will be higher ok. So, it will be easier for you to fracture the material
if the crack is sharp the radius of curvature is small ok. So, for a blunt crack the radius
of curvature will be more ok for a sharp crack the radius of curvature will be small ok.
So, because it is coming in the denominator ok, so let us call this rho one this rho 2,
so row 1 is greater than rho 2. So, when my rho is or crack is getting sharper
and sharper this value will be smaller and smaller. So, my because it is in the denominator
the sigma max value will be even higher ok. So, bigger crack and sharper crack ok you
make the sigma max even higher ok you will have a higher local stress, and it will be
easier to fracture the material ok. So, over all the stress is much lower than the local
stress ok, so this is what ah you have ah ways we discussed in the.
So, Griffith criteria propose that or states that that are crack will propagate when the
decrease in elastic strain energy is at least equal to the energy required to create the
new crack surfaces ok. So, what will happen when you have suppose material there are atoms
here and you are applying a stress on the system ok.
So, all the bonds are under stress conditions. So, there is a elastic strain energy, if you
are not applying any stress the atoms at the equilibrium condition at minimum potential
energy. So, when I am applying any stress on the system ok all these bonds are now stretched.
So, they will have some elastic strain energy, so they will have some potential energy associated
with them ok. So, when you create a crack, so basically
you are breaking the bonds here. So, you are releasing this elastic energy my atoms are
again going into the relaxed state ok and this bonds are now broken. So, there is not
going to plane any elastic energy, but the problem is that when I fractured the material
I am creating two new surfaces. So, there will be some surface energy associated with
this is with this which is called gamma ok. So, my strain energy is release, but I am
creating two new surfaces. So, that is what he proposed that a crack will propagate when
the decrease in elastic energy, whatever is the decrease in elastic energy due to the
breaking of bond is at least equal to the energy required to create the new crack surfaces
when it is equal to the energy surface energy of the two new surfaces. So, suppose ah crack
of length 2 c is present inside the plate or c at the edge ok and.
We are taking the thickness of plate is negligible, so it is a plane stress condition ok. So,
for that the decrease in elastic energy per unit of plate thickness due to formation of
crack is given by a relationship like this, u e pi c square sigma square by e and surface
energy increase due to presence of crack will give you new because of two new surfaces the
this is surface energy now, it will be equal to 4 c into gamma s ok and the total change
in potential energy due to formation of crack will be.
Delta u the change in potential energy, so u s plus u e, so according to the Griffith
criteria the crack will propagate under constant applied stress if an incremental increase
in crack length produces no change in total energy of the system. So, when I am increasing
the crack length ok if there is no net change in the energy of the system the crack will
will be a stable crack and it will propagate ok. If ah surface energy is high then it will
not propagate, because creation of surface energy makes the total energy of the system
will increase ok if the ah elastic energy is higher than the surface energy then; obviously,
the crack will be a stable crack. So, the minimum criterion for stable crack
is that at least they are both these two energies are equal or there is no net addition of energy
during the this process of crack propagation ok if the energy is negative it is it is it
is even better ok, but at least it should be there should not be any net ah addition
to the ah energy to the system ok. So, basically the change in energy delta u as a function
of increase in the crack length that should be 0 ok.
So, d by d c of this ah particular equation, so u s you are adding the energy ue is you
are you are taking out the energy ok by breaking the bonds. So, that is why the negative sign
is here and what will be the change in that as a function of change in the crack length
ok. So, simple ah differentiation you have to do, so it will d by dc of four and gamma
s will be constant. So, it will be dc by dc, so it will be four
gamma s minus ah all this things will remain constant. So, this is c square, so it will
become 2 c ok the 2 pi c gamma sigma square by e ok equal to 0 and that will give you
the required stress for fracture it will be 2 e gamma s upon pi c to the power half ok.
So, this is the stress required to propagate a crack in brittle material as a function
of the size of the micro crack ok. So, if the crack is bigger you will have small
fracture stress the fracture will take place at a small stress for a thick plate plane
strain condition ok the equation will be something like.
This slight change from the previous condition, so a 2 e gamma s by c. So, 2 e gamma s by
1 minus new pi c, so there is a additional contribution of poisons ratio here because
we are considering the thick plate ok. So, thickness is more in this direction ok.
So, it is a plane strain condition and for that we will consider the ah poisons ratio
also here. So, Griffith equation shows strong dependence of fracture strength on crack length
ok, it is said it is able to predict satisfactorily fracture strength of perfect brittle material;
however, fails to predict in case of metal even brittle metals the reason metal that
fail in a completely brittle fashion undergo some plastic deformation prior to fracture
ok. So, this equations are valid only for perfectly
brittle material in the sense if you see a stress strain perfectly brittle material.
So, it is this is stress this is strain. So, ah for a perfectly brittle material it will
fracture without any plastic deformation. So, it is it will takes place in the elastic
range only ok that is why you see we are using poissons ratio here ok. So, this whole analysis
of Griffith ah criterion or Griffith theory is for the elastic ah range only or elastic
deformation only, so there is not going to be any plastic deformation.
Where as in metals ok for any metal though it may be a brittle metal if you see the stress
strain curve ok it will be something like this ah plastic part small elastic part , some
plastic deformation will always take place locally and then it will fracture. So, this
deformation plastically kind of modify this whole ah analysis ok and that is why it is
not able to predict very well in case of ah metals brittle metals ok the reason is this
as you can see that we said that where the crack tip is you will have.
This is very high ok, so your sigma max is going like this. So, this is the elastic stress
distribution just we saw this is my crack tip ok and because of that is my stress increase
in stress ah next to the crack tip, but in ah my metals where there is some plastic deformation
what will happen as soon if suppose this is my yield stress of the material ok, then what
will happen as soon as this is locally the stress go beyond that instead of ah restraining
the the bonds what will happen is there will be shearing of bonds ok basically shearing
of the planes, so material will shear. So, instead of this stress going beyond that
what will happen is locally the material will deform ok and instead of this stress ah distribution
you will have a stress distribution like this it will not go beyond the yield stress elastic
distribution after local yielding will be like this ok and you can understand that by
modifying the crack tip. So, instead of crack tip like this the the overall notional crack
tip can be something like this. So, actually what we have done is we have increase the
this is ah this was the ah radius of curvature now the radius of curvature is increased it
is blunted the crack tip. So, instead of very sharp crack tip now you
have a more blunt crack tip ok. So, and if you remember the expression which we use the
rho comes in the denominator. So, if your radius of curvature is increasing then the
maximum stress in front of stress also will come down ok.
So, plastic deformation at the root of the crack result in blunting or increase in the
radius of the crack tip and that increases the fracture strength of the material. So,
now, it will fracture at a higher stress. So, thank you ah this we have covered the
ah basically the fracture introduction and we have you seen the Griffith criterion ok.
So, in the next class we will conclude the fracture lecture and we will see that how
you can do the analysis in case of slightly brittle metal ah for for the crack and we
will also see that how the different fracture will look like the ductile in brittle fracture
when we look it into the microscope ok so. Thank you .

Cesar Sullivan

One thought on “Lecture 33: Fracture: Part 1

  1. What is the name of the author mentioned at 10:25 ? Writting it in the slides would have been better.
    Thanks !

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