Mechanical Behavior of Materials (Part - II)

Prof. Sudhanshu Shekhar Singh

Department of Materials Science and Engineering
Indian Institute of Technology, Kanpur

Lecture - 09
Fracture Toughness

Welcome back class to our course on Mechanical Behavior of Materials Part 2. So, in the last
lecture we discussed about, the variation of stress intensity factor for different configurations, or
geometries of the plate ok.

(Refer Slide Time: 00:48)

So today, we will discuss about, the concept of Fracture Toughness, which is one of the most
important parameter, we use for design purpose ok. So, let us proceed. So, we will talk about,
fracture toughness ok. So, now suppose, I have a plate and the crack is ok. So, let us talk about say
we have a SENT plate, just for understanding ok. So, suppose the crack length was a 1 and it was
under an applied stress of a sigma ok. Now, due to some circumstances, suppose the crack length
increases from a 1 to a 2.

Suppose, we are doing fatigue ok; at the constant amplitude. We will discuss about fatigue later
on, ok. And now, it becomes a 2; so, the total crack length, now becomes a 1+ a2 which is greater
than a 1, right; and if you remember, the formula for 𝑘 = 𝑌𝜎√𝜋𝑎 . And for the SENT, it is y is
equal to 1.12 and if I take CCT, y will become 1, for infinite plate ok.

So, if the crack length is increasing, so this a, if it increases, sigma is constant. So, k is also going
to increase, right ok. Similarly, if we keep the a phase and increase sigma, k is also going to
increase ok. So, k can increase either by increasing sigma in keeping a fixed, or by increasing a
keeping sigma fixed ok.

So, if we consider this particular configuration and the crack is growing, say with respect to time
and number of cycles, you know at some point of time, k is going to reach to a particular value,
where; but the material is going to fail, is not it? Right. So, that particular value will be called as
fracture toughness ok.

So, again if we increase the crack length and keep sigma fixed k is going to increase, and it is
going to attain a particular value; where the material is going to fail and that particular value is
fracture toughness ok. So, as soon as, k becomes kc. So, now, I am introducing a term called kc;
the material is going to fail ok, and this kc here, is called critical stress intensity factor ok.

So, as soon as k1 becomes equal to; so as soon as k 1 becomes equal to kc right, your material is
going to fail. So, that is the critical condition, is not it? Because, material is failing at that particular
value of k. So, we call it as, critical stress intensity factor ok. So, kc we can define as stress intensity
factor, at which the crack grows in unstable manner ok. So, that is kc.

So, this means if I have now k c. So, now, let me write this particular equation, in terms of k c. Here
previously, it was only k 1 right. So, kc we will have a combination of 𝜎𝑐 √𝜋𝑎𝑐 ok. So, either sigma
c has to reach or a c has to reach; so that the stress intensity factor becomes equal to the critical
stress intensity factor.

So, if one of two are reaching the value of 𝜎𝑐 and a c, the material is going to fail; either load or in
this case a 𝜎𝑐 , because load can be converted to 𝜎𝑐 ok; or a which is a c in this case, have attained
ok, or sigma has reached to𝜎𝑐 , or a has reached to a c ok. So, either load 𝜎𝑐 or a c have attained ok;
that means, sigma has reached to 𝜎𝑐 and a has reached to a c ok and as soon as that happens, the
stress intensity factor reaches to a value of k c and material fixed ok; k c ok.

(Refer Slide Time: 07:30)

So, now I am introducing the term called plane strain fracture toughness; which is k 1 c ok. So, in
this course, I am going to make difference between k c and k 1 c ok; different books will have
different terminology; but in this course, we will have k c, which is stress intensity factor; this is I
have already mentioned, critical stress intensity factor and k 1 c in this course, we will use at plane
strain fracture toughness ok, or critical stress intensity factor in plane strain condition ok.

So, remember the difference between k c and k1c; and you will see eventually, what is the difference
between kc and k 1 c, when I will show you a plot ok. So, now, we can write k 1 c same as k c, ok;
𝜎𝑐 √𝜋𝑎𝑐 ; but here, k 1 c using, plane strain condition ok. Now this k1c is a material constant and if
this is the environment; it is not going to change ok. So, it depends upon the material and
environment.

So, k 1 c varies with material and environment. So, if you fix the material and environment, k 1 c
will remain fix ok. Now, when we design, we can design based on either the stress or we can
design based on the slot size in the material ok. So, we can have two condition, one is maximum
flaw size dictates design stress. So, what does it mean? That we are keeping flaw size fixed; that
means, we have done say NDT; and we know what is the maximum flaw size there, ok.

And based on that, the design stress should be calculated. So, k 1 c is 𝑌𝜎𝑐 √𝜋𝑎𝑐 ok. So, sigma
design we can write, as 𝑘1𝑐 = 𝑌𝜎√𝜋𝑎𝑚𝑎𝑥

and ideally it should be less than this particular value ok. So, if we know what is the maximum
flaw size in a given plate; we know what is k1c, that value is given. So, based on that, we can
calculate what is the stress requirement and remember a max; now we are denoting that as say a c
and k 1 c is known. So, we are actually calculating 𝜎𝑐 .

So, we have to keep the stress value less than 𝜎𝑐 . So, that your material does not fracture, right. If
you are higher than 𝜎𝑐 or sigma design here, then your material is going to fail. So, if I plot, sigma
design versus a max. So, for a particular a max, we are going to obtain a particular sigma design
value, right; of this design value, say for this a max we got sigma design value of this, for a max
of this particular value; we got something like this and so on.

So, we can obtain a plot like this ok and if we are above this line, so if we are in this particular
region, the material is going to fail, ok and if we are below this, if we are here, safe region ok. So,
given a plate, we know what is k 1 c and we have done NDT, we know what is the maximum flaw
size; we can calculate what is sigma ok. And if you are above that particular value, material will
safe and if you are below this, material is not going to safe.

So, this is what this plot shows. Now, we can do the opposite; where we know what is the sigma.
So, we can calculate what is the maximum allowable flaw size in the material ok.

(Refer Slide Time: 13:22)

So, design stress dictates, maximum flaw size; in case of sigma is known, we are calculating what
would be the maximum flaw size so that it does not fail. So, again𝑘1𝑐 = 𝑌𝜎 √𝜋𝑎𝑚𝑎𝑥 and let us
say this is not design ok. So, now we can write, this equation as
1 𝑘 1𝑐 2
𝑎𝑚𝑎𝑥 = 𝜋 ( 𝑌𝜎 )

And if the material does not fail, then

1 𝑘 1𝑐 2
𝑎𝑚𝑎𝑥 < 𝜋 ( 𝑌𝜎 )

So, if you plot this again, like we did last time; so here we have now a max and then the bottom
will be sigma design. So, for a particular sigma design, say this particular design; we are now
getting a max value here, similarly for this particular design we are getting this and so on.

So, we can actually, get a plot like this ok. Now again, which region is safe and this region is
unsafe. So, same as the last time, the top part here, if we are in this particular regime, the here
material will fail and the bottom region, material is safe, or material will not fail ok. So, we can
use k 1 c or plane strain fracture toughness data; to calculate either sigma design or the allowable
maximum flaw size in the material so that, it does not fail.

So, you know plane strain fracture toughness is being used for designing purpose a lot ok; that is
why I already mentioned that this is one of the most important parameters, we use ok. So, why did
I distinguish between kc and k 1 c here? So, let us understand that, you already know about the
plane stress and plane strain, we have discussed; so typically, if we have a plate and if it is very
very thin, for all practical purpose, we assume that it is in plane stress condition and if the material
plate is thick; it is in plane strain condition ok.

Now, if I take a particular plate, we can also say that the surface; so, suppose this is the plate here
the surface, both the surfaces this side and this side, that will be plane stress condition ok and as
we move inside, it is going to be under plane strain condition ok. And if that is the case your plastic
zone size is also going to change and that the crack growth behavior is also going to change ok. I
will show you now, the variation in plastic zone size and the crack growth behavior we will discuss,
when we talk about the fracture mechanisms ok.

(Refer Slide Time: 17:47)

So, we are not talking about variation of fracture toughness, with thickness ok. So, depending upon
the plate thickness, your material fracture toughness is going to change and at a particular thickness
value, you know below which, your material is going to be under plane strain condition and that
is when we call it k 1 c ok. So, if we have a crack; so, suppose this was my plate.

So, it is like a crack ok; I am showing it crack in three dimensions ok. So, this is your crack here.
And this is the cracks, front; crack front means, the front part here ok. So, this part here is my crack
front ok. So, now, if I see the variation of stress through the thickness, right this particular; now if
I just let me draw surfaces here. So, these are my surfaces of the plate right.

So, this is under plane stress, the other surface is under plane stress and inside, we are going to
have plane strain, correct, ok; because, on the surface it is free. So, 𝜎𝑧 , right; or inside the plane
stress is going to be 0. Similarly, on the other surface it is going to be 0. So, both the surfaces are
under plane stress and inside you are going to have all the 3 stresses. So, you are going to have
plane strain condition, right.

And then remember, we had discussed, in the plane stress condition; it is biaxial and the plane
strain condition, it is triaxial bond ok. So now, if I plot 𝜎𝑧 , which is through the thickness or sigma
3, with respect to distance from the center of the plate ok. So, this is 0. So, this is the mid-point.
So, we are say here. So, this is your midpoint. And then we are moving from midpoint towards the
surface. So, here we have, distance from surface ok.

If your plate thickness is B, so what will be the distance from the midpoint towards the surface?
That will be B/2. So, here right. So, this becomes the surface. So, that is under plane stress
condition. So, sigma z is 0, ok and as we move inside, you are going to see plane strain condition
right. So, we can draw something like this. The variation of 𝜎𝑧 ; something like this ok, where here
you have, plane strain condition ok.

So, you have very high triaxiality at B/2 position which is the surface, you have plane stress
condition, which is in biaxial mode; here distance right; plane stress is also biaxial and it makes
sense because sigma is 0. So, you have only 2 stresses. And in between, you are going to have
again triaxial stresses, but it will be slightly less. So, lower triaxiality we can see ok.

So, your 𝜎𝑧 is varying, like this through the thickness; where, it is going to be 0 at the surface and
then it will reach to a maximum which is corresponding to the plane strain condition and the same
thing will happen on the other side of the midpoint ok. So, it will go like this, and then, it will go
down to 0 value on the other side of the midpoint of the plate ok.

Now, what will be the plastic zone? So, again plastic zone is higher than which condition? It is
higher in the plane stress condition and then it is the lower, in the plane strain condition right. So,
if I again see the variation of plastics on sides, from the surface to the other surface; it is going to
be the highest at the surface, then it will decrease right and then again it will increase and then it
will be again highest on the other surface; because, both the surfaces are in plane stress condition
ok. So, let me plot that.

(Refer Slide Time: 24:09)

So, if you have a crack ok. So, you are going to see a larger plastic zone, at one side of the surface
ok and then it is going to decrease ok and then it is again going to increase, to other surface ok.
So, this is your one surface here, this one is your another surface of the plate and both the surfaces
are under plane stress condition. So, your plastic zone size is larger; this is also under plane stress
condition ok.

And as you move down, your triaxiality is going to increase ok, and that means, your plastic zone
size is going to decrease. So, here you are going to have plane strain condition, very high triaxiality
ok. So, you have a smaller plastic zone side. So, this is how your plastic zone size varies from
plane stress to plane strain condition ok throughout the thickness; and if I just see the projection
of it is going to look like this, where you have the smaller version, and then the larger one.

So, I am seeing the projection ok. So, this will correspond to plane stress on both the surfaces and
this will be inside which is plane strain ok. So, this is how the plastic zone size changes through
the thickness, right. Now, this will lead to the variation of fracture toughness values also through
the thickness ok.

There is one more reason to it, the variation of fracture toughness with thickness and I will discuss
that when we will talk about the fracture mechanisms ok. But for now, this is one of the reasons,
because of the variation of plastic zone side; your fracture toughness is also going to vary through
the thickness ok not through the thickness, vary with respect to the thickness ok. So, if you have
very thin material, then your material is under plane stress condition. So, your fracture toughness
is going to be higher right.

Because your plastic zone is also higher right. And if you decrease the material thickness, you are
going to have more amount of or more magnitude of plane strain condition ok and your fracture
toughness is going to decrease and eventually, you are going to reach to a particular thickness
value, after which it is predominantly under plane strain condition and not in plane stress condition
ok.

So, at that particular point your fracture toughness value remains constant. So, let me plot it. So,
that you can understand it better. So, it is going to vary something like this ok. So, this is your k c;
Now, this is critical stress intensity factor ok, or we can call it also fracture toughness right; but
not plane strain fracture toughness, ok. And, this particular point here, this is corresponding to k 1
c, which is plane strain fracture toughness. And now this becomes, material property you can see
with even at higher thickness of B is here ok.

And k c varies along y axis. So, if we are having B higher than a particular value, it does not change
with thickness of the plate and it becomes the material property. So, we call it plane strain fracture
toughness and it is a material property ok. And if the thickness is lower than this, lower than this
value. So, this is I would say let us, let us say critical thickness ok. So, if the thickness of the plate,
is lower than the critical thickness of the plate, then it is the fracture toughness or critical stress
intensity factor is changing right; it is increasing, as you reduce the thickness.

So, it is not constant; because it is depending upon the thickness. So, here you have plane stress
condition. So, your material is very thin. So, that your material is predominantly under plane stress
condition here it is plane stress condition sorry plane strain condition, the material is sufficiently
thick; so that, it is predominantly plane strain condition, and in between you are going to have
contribution from both plane stress and plane strain condition right.

So, you are going to have mixed mode ok. So, your k c value changes with thickness and after a
particular value of thickness, it remains constant; that value is called k1c; plane strain fracture
toughness and that is a material property ok. So, let me show you from one more schematic right.
So, suppose I have a thickness of the plate like this; then I am increasing the thickness of the plate
like this and then I am making it you know very large.
So, here the surface in both the cases. So, let us say this is plate A, plate B, and plate C ok. So, in
plate A, if you see both the surfaces are under plane stress and as you go inside, you can have plane
strain also, but if it is very thin it is predominantly plane stress. In plate B, both the surfaces are
again under plane stress, and as you move inside you are going to see plane stream condition ok.

Now come to see, you also have plane stress condition at the surface, but it is so thick, that it is
predominantly under plane strain condition and not a combination right. So, this will fall, I would
say, you know looking at the schematic and it should be greater than b c; if I say you know b c is
critical thickness of the plate ok.

So, depending upon the thickness of the plate, your fracture toughness is going to vary, and you
know we are going to eventually talk about the determination of fracture toughness from a stream
standard; and there is a criteria for thickness of the material to be used to be to which can satisfy
the condition of plane strain ok, we will discuss that. So, now, let us solve some problems, so that
you can understand how to determine the stress intensity factor etcetera ok. So, let us solve one or
two problems related to that.

(Refer Slide Time: 32:58)

So, the question one is find critical crack length for different configuration or geometries, when k
1 c is given. So, is the fixed material, ok; and environment is also fixed. What we are doing? We
are just changing the geometry, of the plate. So, k 1 c is fixed, because your material is fixed ok;
applied stress, in 100 MPa and k 1 c it is fixed and it is given as 50 MPa √𝑚.

So, we have been given 2 values; k 1 c is given an applied stress is given. So, we need to calculate
what is the critical crack length. So, we have 3 configuration ok. So, the 1st one is, what we have
been using for long time now. So, this is say A, plate A where you have central through crack ok.
So, this is CCT and it is an infinite plate ok. So, here your crack length is 2a. Now, second is semi-
infinite SENT ok. So, crack length here is a. So, this is condition 2 and the last one is penny shape
crack. This is new to you. So, I will just draw, again infinite plate.
So, if you remember, I am drawing zig-zag line; and I have already told that this, in this force this
belongs to the infinite. So, this is a 2 D crack, like this ok. And the crack length is denoted like a
here. So, this is penny shaped, ok; again infinite. This is semi-infinite, and this is also infinite ok.
So, we have these 3 conditions, these three configurations.

So, we have to calculate what is the critical crack length ok and you will see that depending upon
the configuration, the crack length is going to change, the critical crack length; although we have
fix the stress and k 1 c ok. So, it depends upon the configuration of the plate also.

So, here 𝑘1𝑐 = 𝑌𝜎𝑐 √𝜋𝑎𝑐 . So, Y for CCT, it is 1. And y for semi-infinite SENT is 1.12 ok and for
penny shape; let me write down, it will be,
2
𝑘1𝑐 = 𝜋 𝜎𝑐 √𝜋𝑎𝑐

So, we know what is the value of k 1 c. So, if I just see the first equation, we know what is k 1 c, we
have been given sigma applied stress here ok. So, we have to calculate what is A, in all the cases
ok. So, let us do one by one.

(Refer Slide Time: 37:48)

So, plate A, which is CCT. So,

𝑘1𝑐 = 𝑌𝜎𝑐 √𝜋𝑎

k 1 c is given as 50 ; y is 1, sigma is 100 MPa. So, now, you know all the values; we have to
calculate, what the value of a c is, and a c will be coming out to be, 79.6 millimeter ok. But
remember, this was in this is CCT right. So, the crack length is not a, it is 2a. So, we have to
calculate, what is 2 a c? And, it will come out to be 159.2 millimeter you can calculate ok. Please
be very careful with the unit; see k 1 c is MPa root meter.
And a here, whatever you will calculate, in this particular equation will come out to be in meter;
you have to convert it to millimeter. Similarly, if I give you the value of a in millimeter and you
are using k 1 c in MPa root meter, first you have to convert a in meter ok. So, that do not make that
mistake ok. So, just check the unit of a sigma and k 1 c should be of same unit ok. If k 1 c is MPa
root meter, a should be in meter.

So, you have to convert, if it is given in millimeter. Now plate B, which is SENT,

𝑘1𝑐 = 𝑌𝜎𝑐 √𝜋𝑎

Y is 1.12, sigma is 100 ok. And, if you calculate ac will come out to be 63.4 millimeter. So, here
we will just use ac because, we do not need to convert it to 2; because we in SENT we use a ok.
Now plate C, which is penny shaped,

2
𝑘1𝑐 = 𝜎 √𝜋𝑎𝑐
𝜋 𝑐
we can calculate what is a c and a c will come out to be 196 millimeter ok. So, you have three
configuration, applied distress is same, material is same. So, k 1 c is fixed. So, you know applied
stretch is same, but you see the critical crack length, right or the maximum allowable crack length
of flaw size in the given plate is varying; it is the lowest in SENT and the highest in penny shaped
crack; that means, SENT is more dangerous, right because, the a c value is smaller.

So, it can allow very small crack length as compared to CCT and penny shaped right. So,
depending upon the configuration, your maximum allowable crack size changes and vice versa
also; you know if we fix the crack length, the maximum allowable design stress will also change
right; the plot if you remember ok. So, overall the above problem, provides which type of crack or
configuration I would say, configuration would be more dangerous, when other parameters and
what are the other parameters here? Sigma and k 1 c are fixed ok.

So, from this problem, we understand the effect of configuration ok or geometry of the plate. So,
now, let us solve another problem, where we will understand the effect of crack length or a/w.

(Refer Slide Time: 43:18)

So, question 2; for a given material, the material is fixed; that means, k 1 c is going to be fixed, k 1
c is 50 MPa root meter right, width of the plate is 30 centimeter ok. Central crack or through
thickness crack, this is 3 millimeter ok. So, we have a plate of thickness not thickness width of 30
centimeter. So, this is given, k 1 c is given 50 MPa root meter ok.

And you have a very small crack here, 3 millimeter; again not up to the scale, but you can
understand ok. So, the plate is under applied stress sigma and we have to figure out what is the
maximum allowable sigma, or design stress ok; what is the maximum allowable stress ok. So, if I
have this plate, what is the maximum allowable stress ok. So, we know k 1 c. So, let us solve.

So, this is CCT specimen right. Now, this is not infinite. So, you cannot take y directly equal to 1
is not it; but for a particular case, if you can calculate what is a/w and if it is very small for all
practical purpose, you can take y is equal to 1 and you will see this now ok. So, let us solve
assuming that y is not equal to 1, we do not know anything and then we are just; we know the
expression we are solving the problem.

So, 𝑘1𝑐 = 𝑌𝜎√𝜋𝑎

a is given, k 1 c is given, we have to calculate y and sigma; y is not given, it is not equal to 1. So,
if you remember, 2 w here, right; you remember what we discussed for CCT, we take width as 2
w. So, 2 w is given as 30 centimeter and 2 a is given as 3 millimeter ok. And we also know for
𝑎 𝑤
CCT, 𝑌 = 2𝑓(𝑤)√𝜋𝑎 see the last lecture ok. So, from here this particular equation.

So, let me write down equation 1 and equation 2. So, from equation 2, we can find out what is the
value of y ok. So, to do that, we need to know what is f (a/w) and for CCT, f(a/w) is given as
1
𝑎 𝜋𝑎 𝑎 𝑎
𝑓 (𝑤) = [sec (2𝑤)]2 [1 − 0.025(𝑤)2 + 0.06(𝑤)4 ].

We know a, we know w. So, we know what is a by w right.

So, you can calculate and this will come out to be 0.088 ok. Now use equation 2 and figure out
what is y, so from equation 2, we can calculate what is y and it will come out to be 0.99297 ok.
Now, see if 0.99297, it is very nearly equal to 1 for all practical purpose right, and if you see a by
w here also a is what; 1.5 millimeter.

And w is 15 centimeter ok. So, a by w will be 0.01 which is very small as compared to 1. So, for
all practical purpose, although we can calculate y, but you know it is nearly equal to 1. So, we will
just take it as 1 and then we can solve equation number 1. So, from equation 1, k1c required
to 𝑌𝜎√𝜋𝑎 , k 1 c is 50, y is 1, sigma and then √𝜋𝑎. So, if we calculate, sigma; it will come out to
be, 728 MPa ok.

So, this is condition, where for all practical purpose, we are assuming that a by w is, it very less
which is 0.01. So, y is approximately equal to 1. Now, let us increase the crack length space, the
width of 30 centimeter and increase the crack length and see what happens to value of y ok; and
then sigma also. So, previously it was 3 millimeter crack length.

(Refer Slide Time: 49:52)

Now, let us assume, the crack length has increased. And now, this becomes 3 centimeter instead
of 3 millimeter. So, 3 centimeter this is 2 a equal to 3 centimeter. So, now you have a plate, of 30
centimeter ok, and then crack length again not up to the scale of 3 centimeter. And then you have
stress a. So, you have to calculate sigma right. So, we are just increasing the crack length from 3
millimeter to 3 centimeter. And let us see now what happens ok.

So, here 2 a/2 w is a/w or this will come out to be 0.1; previously it was 0.01 right. We can calculate
again f (a/w), and if you do that, this will come out to be 0.2819 and then we can calculate y from
𝑎 𝑤
2𝑓 (𝑤) √𝜋𝑎. So, y will come out to be, 1.0059 ok. Again, very near to the value of 1 and if you
use this value of y, sigma will come out to be; now let us see, condition C. Now suppose crack
length has increased to 10 centimeter.
So, 2 a equal to 10 centimeter ok. So, a by w is 1 by 3 which is 0.33. Now, it is not very less than
1. So, let us calculate what is f (a/w) so you have to use, this formula for f (a/w) ok. So, if you use
that formula, expression f (a/w) will come out to be 0.545 ok. And now we know the expression
for y. So, let us use that. So,

𝑎 𝑤
𝑌 = 2𝑓 (𝑤) √𝜋𝑎 = 1.0707

and if we calculate y will come out to be 1.0707 and sigma will be approximately ok.

So, see how it is decreasing, the maximum allowable stress is decreasing as you increasing, you
are increasing the crack length ok. So, we are fixing the k 1 c. So, you can calculate what is the
allowable stress, with respect to different crack length ok. So, that is problem 1, 2nd problem ok.
So, we have understood the concept of fracture toughness; how it varies with thickness ok.

And I also mentioned that at particular point on the fracture toughness or stress critical stress
intensity factor becomes equal to plane strain fracture technique which is the material property ok.
So, I will quickly tell you about the ASTM standard of plane stream structure toughness
measurement and in the next lecture, we will talk about how to determine plane strain fracture
toughness ok.

(Refer Slide Time: 54:11)

So, plane strain fracture toughness. So, k 1 c ok. So, you know depending upon the material I told
k 1 c is going to vary. So, the critical thickness at which k c becomes k 1 c is not constant for all the
material and changes. So, let me show you one or two examples. So, we have say 2 steel. So, this
is steel 1 and then another one is steel 2 ok. So, k c on the y axis on both sides.

So, in the first steel, we are having values of say something like this and so on and the second steel
is not very tough. So, it is smaller value. So, 20, 30, 40, 50, 60 and so on and the thickness here is,
5, 10, 15, 20, 5, 30, 20, then 50, 25 and then you know a dot, dot, dot, 50, then 75, and then 125.
So, not up to the scale, initially it is small and then you know there is a variation. So, it is not up
to the state. So, that is why I have given dot, dot, dot here ok.

And here, it is 5, 10, 15, 20, 25 and then again dot, dot, dot and then we see 50 here right. So, this
thickness, on the x axis, this is in millimeter. And again thickness here in millimeter ok. So, we
have 2 steels. So, in the 1st steel, k 1 c is 120 MPa root meter, and in the 2nd steel k 1 c is 30 MPa
root meter ok. So, this is the cut off now. So, in 1 steel, we have k 1 c here and in another steel it is
ok. So, in 1 steel, the steel 1, your k c is decreasing and then finally, it becomes constant; let us say
75 millimeter. So, here B c will correspond to nearly about 75 millimeter.

And here B c will correspond to the smaller question, it means something like this ok. So, this is
the variation of k c; and here B c will correspond to nearly 15 millimeter ok. So, we have 2 steels
different material, different composition. So, k 1 c in the steel 1 is 120 and it is of the B c is higher
thickness ok. So, if you want to obtain k 1 c value, you have to have thickness of the specimen in
this regime here; in steel 1. In the 2nd steel, k 1 c is smaller and thickness is corresponding to k 1 c
is also smaller the critical thickness.

So, you have to have thickness of the specimen to conduct plane strain fracture toughness
measurement, in this particular regime ok. So, in steel 2 the requirement of the thickness, for
fracture toughness measurement; plane strain fracture toughness measurement is smaller than steel
1 ok. So, depending upon the material, the thickness requirement for to conduct, the plane strain
fracture toughness measurement is different.

(Refer Slide Time: 59:15)

And to measure the plane strain fracture toughness we use ASTM standard; which is E 399 ok;
and this is for plane strain fracture toughness measurement ok. And what does it say? It says, the
crack extension, resistance under conditions of crack tip plane stream, in mode 1. So, it is in
opening mode ok. So, mode 1 for slow rate of loading, under predominantly linear elastic
conditions and negligible plastic zone ok. So, this also means that, it provides, the measurement
of crack extension, extension resistance at the onset 2 percent or less of cracked extension ok.
So, we will use ASTM standard which is E 399 ok and we will use compact tension specimen for
in E 399 ok and when we talk about this ASTM standard, we will assume the linear LEFM
condition to be valid; that means, the plastic zone size is negligible or very very small, it is under
mode 1 condition ok and it will provide the measurement of crack extension resistance
corresponding to the crack extension of 2 percent or less ok.

So, in the next lecture, we will start discussing about, how to measure plane strain fracture
toughness using E 39 399 ASTM standard, ok.

Thank you, we will meet next week.