True stress and True Strain


We will discuss true stress and true strain;
we have already discussed stress and a strain. But those stresses and strain were engineering
stresses and engineering strain and when we define engineering stress then that was defined
as F by A naught where F was the applied force and A naught was the initial cross sectional
area , but there is a problem in this definition. You can see that suppose this was your initial
sample of length L naught and cross sectional area A naught and then you applied a force
on it such that it is elongated to a new length L , but you can see that as it elongates to
L, its cross sectional area also decreases to some new value A . So, the force the current
force F is not acting on the original area A naught
, but is on acting on the current area A . So, in this sense, this engineering stress
is not really true representation of the stress being seen at this point, we can define the
true stress simply by sigma is equal to F by A where A now is the current area on which
the forces acting not the original area , but the current area .
So, I have now put the subscript T to distinguish it from sigma defined earlier as engineering
stress . You can establish a relationship between the
two by by looking at it that sigma T was F by A, if I want to relate it to the engineering
stress, I bring in A naught . So, I simply multiply and divide by A naught to get this
expression . So, for this; for the first factor in this expression, we can write it as the
engineering stress . So, it is engineering stress times A naught by A, we can make further
progress by the assumption which is true in plastic deformation of volume constancy .
So, assuming volume to be constant assuming volume to be constant, we can write
this as here we can use the relationship . So, A the cross sectional area A times the initial
cross sectional area A naught times the initial length L naught was the initial volume and
the current cross sectional area A multiplied by the current length L is the current volume
and if the volume is constant during deformation, these two will be equal this assumption of
volume constant is quite true in plastic deformation not so true in elastic deformation .
But then significant part of deformation; if we are in the plastic deformation regime,
then this hm this assumption will be quite true . So, using using this ah constancy of
volume, we can now express A naught by A naught by A as L by L naught . So, with this substitution
we can write the true stress as engineering stress times L by L naught , but then I can
write L as the current length L as L naught plus increment in the length delta L divided
by L . So, you can see now that we will have sigma
1 plus delta L by L naught , but then delta L by L naught changing length divided by the
initial length is nothing, but engineering strain . So, we can write our final relationship
as true stress is engineering stress times one plus engineering strain this is a nice
simple relationship , but based on the assumption of volume constancy .
So, now we will look at the engineering strain which we defined as delta L by L naught . So,
this will be L minus L naught by L naught . So, delta L is the increment in length,
you can see that the engineering strain depends on what is the original length .
So, if delta L was the increment over original length L naught; I get the engineering strain
delta L by L naught , but suppose I implemented by further delta L, then now this increment
also will be divided by the original length L naught although; now this increment the
subsequent increment delta L prime; let us say is being applied on the current length
L and if I divide delta L by the delta L prime by current length, then I will get a different
value . So, we define true strain by taking the current
length into account . So, we will say true true incremental strain d E true is a small
incrementing length L divided by let me write capital; I was using capital small increment
in length L divided by the current length L . So, this will be the true increment the
true incremental strain for the same delta L the incremental engineering strain would
happen d L by L naught which will be a different quantity .
So, now to find the strain at any given length E T all we have to do is to integrate this
incremental strain from a strain of 0 to the current strain . So, if we write this as now
d L by L, then the limits change from L naught to L , but this is a simple integrand, you
know that this integrates to log of L . So, we have log of L; this is a definite integral
with limits L naught an L . So, finally, I get a value log of L by L naught
. So, the true strain is simply logarithm of current length divided by the initial length;
let us now ah try to establish a relationship between true strain and the engineering strain
. So, this is simple to do because you can see that E T is simply log of L by L naught
, but on the numerator L can be written as the initial length plus the change in length
divided by L naught . So, this then gives you log of 1 plus delta
L by L naught , but delta L by L naught was nothing, but engineering strain . So, you
can write it as log one plus epsilon . So, the relationship between true strain and engineering
strain can again be written as epsilon T is log of 1 plus the true strain is log of 1
plus engineering strain . So, let us compare the engineering strain
and true strain for an example situation here shown; it is extreme situation where I start
with an initial length and then extend it or elongate it to twice its original length
. So, we go from L to 2 L and then finally, again compress it back to L and let us see;
how the situation is described in terms of engineering strain and true strain .
So, initially of course, there is no change in length . So, delta L is ah 0 . So, the
engineering strain is 0 and again since there is no change in length . So, L and L naught
are the same . So, you have log of one which is also indicating 0 . So, we are starting
with 0 engineering strain and 0 true strain , but the intermediate stage of elongation
above up to a length of 12 is described differently by engineering and true strain .
So, the engineering strain will give me now the change in length is equal to the original
length . So, it will be L by L . So, which is one . So, engineering strain is 1 which
is 100 percent strain; whereas, true strain gives me a more conservative estimate of this
here, we have log of the final length is 12 and the initial length is L . So, you have
log 2 which is only 0.9 . So, it is saying about 69 percent strain, whereas, the engineering
strain is giving me 100 percent strain , but the interesting situation comes when we describe
the final compression . So, now in this stage from intermediate to
final the engineering strain will give me a change in length which is equal to minus
L , but the original length now is 12 because we are talking of a strain from intermediate
to final this will give me minus minus 0.5 negative sign is indicating compression; whereas,
in the case of true strain, I get log of final length which is L and the initial length which
in this case is 12 . So, we get log of half and log of half is
nothing, but minus of log 2 . So, if I add the total strain in the process of going from
initial to intermediate to final the total strain .
So, let us say E net . So, E net will come out to be if I add the two strains, then I
will get 1 minus 0.5 which will still give me 50 percent strain, whereas, ah the net
strain is 0 because there is no change in the initial and final length . So, engineering
strain is not able to capture this kind of change in length , but if you look at the
true strain epsilon 2 net, then you can see the intermediate stage was log 2 and then
subsequently we had minus log 2 . So, it truly describes that the true strain finally, is
0 . So, you can see the true strain actually is
the real strain and engineering strain can give you hm erroneous value, if the deformations
are very large and that is why for careful scientific theories of plastic deformation,
if you are trying to develop, you will always use true stress and true strain and not engineering
a stress and engineering strain , but for practical purposes, since using initial length
and initial cross sectional diameter is always easier .
So, for practical purposes engineering stress and engineering strain are quite good, we
do not get a situation of 100 percent extension followed by compression in most practical
situations . So, this ah problem will not come.

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