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Welcome to the second lesson of the second
module on the course of Strength of
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Materials.
.
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We will be discussing the Analysis of Strain
II and in the last lesson we introduced the
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strain after discussing the different aspects
of stresses in module 1.
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..
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In this particular lesson it is expected that
once it is completed, one should be able to
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understand the concept of axial and the normal
strain in a stressed body of variable cross
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section. In the last lesson, we discussed
about the strain, the axial strain, the normal
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strain
in a bar which is of uniform cross section.
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In this particular lesson we are going to
discuss
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that if a bar is of variable cross section
then what will be the strain in that particular
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body?
To understand the concept of shearing strain
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also one should be able to understand the
concept of shear modulus and Poisson’s ratio.
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.
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.Thus the scope of this particular lesson
includes recapitulation of the previous lesson.
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In
the lesson what we have discussed, the first
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one, we look into it through the question
and
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answers. Then the evaluation of axial strain
in a body of variable cross section evaluation
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of shear strain and shear modulus and subsequently
the evaluation of Poisson’s ratio. We
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will see the meaning of Poisson’s ratio
and the evaluation of Poisson’s ratio in
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the body
which is subjected to through loading and
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undergoing stresses.
.
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Before we proceed let us look in to the questions
which I had posed last time the first
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question which was posed was what is meant
by elastic limit? Now let me explain this
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with reference to the figure.
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..
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If you can remember last time I had demonstrated
that the stress strain relation ship is of
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this particular form where this axis represents
strain and the y axis represents stress. Now,
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say that if we keep on applying load on a
bar in an axial manner if we apply a tensile
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pull
to a bar gradually, the bar will undergo deformism
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and thereby you exceed that for a bar
which is subjected to axial pull how to compute
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the strain corresponding to that port each
increment of the load.
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If we compute the strain for a particular
load and compute the stress corresponding
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to that
load, we can get a plot of stress versus strain.
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Now up to a particular limit if we keep on
applying the load, on removal of the load
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the bar is expected to come back to its original
position. But beyond a point if we are applying
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the load, the state of stress and the strain
in the bar will be such that it will not come
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back to its original position after the removal
of the load. Now the limiting point up to
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which on removal of the load the material
comes
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back to its original position, we call that
as elastic limit and this is what is indicated
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over
here and up to that limit we expect that the
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stress is proportional to the strain which
is
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popularly known as the Hooke’s Law.
Second question was what is the difference
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between nominal stress and True stress?
When you compute the stress that any point
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in the stress strain curve, the stress at
any
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point is given by the applied load divided
by the cross sectional area. When we take
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the
original cross sectional area, the stress
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thereby which we get P divided by cross sectional
area, we call that stress as nominal stress.
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Whereas we are applying the load in the
specimen and beyond elastic limit the power
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is expected to have a reduced cross sectional
area.
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If we calculate the stress based on this reduced
area will get P divided by the actual cross
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sectional area which is little higher than
the stress, this is higher than the nominal
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stress
and that stress p divided by actual cross
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sectional area is called as True stress. So
P
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.divided by the original cross sectional area
is an nominal stress, and P divided by the
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actual cross sectional area will give us the
True stress.
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Now the third question posed was, how you
will evaluate strain in a bar with gradually
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varying cross section. First let us look into
how we perform test on a tensile specimen
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in
tensile pulling equipment. This experiment
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is primarily the tensile testing of a HYSD
bar.
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HYSD stands for the High Yield Strength Deformed
bar, commonly termed as tar steel.
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We are interested to primarily evaluate three
aspects of it, the hill strength of the bar,
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the
ultimate stress and the percentage elongation
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the member undergoes. The bar will be
tested under constant axial pull and the load
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will be applied till the member fails. The
whole length of the bar has been divided into
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a number of segments and each segment
measures 40 mm. After the specimen is tested,
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after failure, we will re-measure the length
and will see the how much extension this individual
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segment has under gone which will
give us the measure of elongation that the
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material has undergone.
The objective of this particular test is to
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evaluate again the hill stress of the material,
the
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ultimate stress at which it fails and the
elongation it undergoes which gives us the
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measure of the ductility of the specimen.
This is the control panel from which the load
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will be applied to this particular specimen
at a constant rate till the member fails and
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the
type of load that is being applied on to this
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particular specimen is a tensile pull.
Now we look into how the load is being applied
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on the specimen till its failure. After the
bar has been tested if we look into the failure
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of the bar, we will see that one side has
been found shape of a cone and the other side
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is that of a cup and this kind of failure
is
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generally called as cup and cone failure.
If we look into the whole length of the bar
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you
will find that the centre part it has under
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gone extension and because of elastic deformacy
the length has been extended.
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In the beginning we had divided the whole
length of the bar and the segment of 40 mm
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now if we measure the extended part of that
particular segment now, after it has
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elongated we find that this particular length
which was originally 40 mm is now 54 mm.
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This indicates that this particular bar will
be extended by 14 mm. This gives us the
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measure of percentage elongation and thereby
the ductility of the material. This is the
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experiment which has been performed on a bar
applying a tensile pull and then snapped
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at a point and the type of failure we observed
we call as the cup and cone failure.
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If you remember in the last lesson we were
discussing the equi apply at tensile pull
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in a
bar, the body is expected to undergo deformation
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gradually and at certain points of time it
undergoes failure and that is what this is
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represented in this stress strain curve.
The stress strain curve says that if we apply
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load gradually up to elastic limit, if we
release the load the load the member comes
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back to its original position. But we keep
on
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applying the load it undergoes through the
maximum stress and then fails at a lower
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value. This part we call it as a failure stress
and this part the maximum stress we call as
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ultimate stress. Also, in the last lesson
we said that in a particular area in the bar
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we
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.generally reduce the cross sectional area
from the entire original area. The purpose
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of that
is to concentrate the failure in that particular
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zone so that we can observe how the failure
is occurring in that particular member.
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In this particular experiment there we have
not reduced the cross sectional area as such
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but we are applied a pull to the bar but the
whole bar was marked with segments of 40
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mm. After the application of the tensile pull
the segmental lengths which was originally
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40 mm was measured again and the segment in
which the failure had occurred we
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measured the extension which we found to be
around 14 mm. Now this gives us a
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measure of how much elongation the member
can undergo before it really fails.
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.
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Here is
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the third question:
What happens if we apply axial pull in a bar
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for which the cross section is no longer
uniform but varies gradually as that of this
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particular one as it is indicated in this
particular figure.
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In the previous lesson we said that if we
apply a load axial pull p in a bar it undergoes
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deformation and from its original length L
it becomes L plus delta and this extension
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delta we had seen how to compute. Now if instead
of having such uniform cross section
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if the cross section of the bar is varying,
say we have an area here which is A 1 and
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the
cross sectional area here is A 2 and this
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bar is having subjected to an axial pole p
we
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assume that this area is varying linearly
between A 1 and A 2 over the length L. Now
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this
cross sectional area of this bar can be of
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cylindrical type thereby we can compute
diameter, diameter could be varying, or we
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can have a rectangular cross section in which
the thickness is uniform and thereby because
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of change in the width, the cross sectional
area is varying.
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.Now at any cross section which is at the
distance of x from the left end, the cross
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sectional area of this bar is going to be
equal to, this is A 1 plus if we take a line
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down this
is also A 1 so this particular length is a
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A 2 minus A 1 by 2 so this is (A 2 minus A
1 by 2) x
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by L and since we have this unit plus this
unit which will give me the total area over
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here
so star 2 so this is equal to A 1 plus (A
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2 minus A 1 )x by L.
Hence the equation is:
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A is equal to A 1 plus (A 2 minus A 1 by 2)x
by L into 2 is equal to A 1 plus (A 2 minus
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A 1 )x by L. So this is cross sectional area
over this particular segment. Now, if we like
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to
find out the elongation of this particular
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bar which is deltaAs we have seen in the last
lesson delta is equal to P l by AE since here
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the cross section is varying at every segment
we may take a small segment which is dx is
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equal to ∫ P by AE dx. AE is function of
x
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now d x which is varying from 0 to l now we
substitute the value of a here which is the
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function of x and integrate it over the length
which will give me the deformation delta.
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Also, sometimes we can compute the extension
by considering the whole of the bar into
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number of segments. If we divide the whole
bar into a number of segments and if we
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consider one of the segments on an average,
this is area A 1 this is area A 1 ‘. On
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an
average here we can say the area is average
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of this area and this area by two which is
constant over this particular segment.
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Here, we are adding something and here we
are subtracting something. Therefore on an
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average this satisfies for the whole segment.
For these if we compute delta for this
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particular segment likewise we can compute
delta for each of these segments and if we
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sum them up we get the total delta for the
whole of the bar. But since we are dividing
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the
whole bar in to a number of segments, it is
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expected that and we are approximating that
the area is average of the two ends, so it
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is expected that we may get a little error
in to it .
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However, we can compute instead of going for
a rigorous calculation, we can sometimes
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compute the elongation in the bar in this
manner. However, for e precise analysis the
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deformation delta should be computed in this
expression.
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..
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If a bar is fixed at one end and it is been
pulled by an axial pull P this is of length
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L.
When the bar is been pulled it is undergoing
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extension this is the extension delta and
we
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have seen how to compute this extension as
a function of this axial pull P in terms of
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the
cross sectional parameter A and the length
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L. Now when this is being pulled, the bar
undergoes deformation in other directions
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as well. In fact when it is being pulled or
elongated in axial direction it undergoes
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contraction in other directions.
.
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If it is stretched by an axial force then
there is a contraction in the transverse dimensions
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as well. Poisson demonstrated that the ratio
of this transverse strain to the axial strain
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is
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.constant for stresses within proportional
limit. So within elastic limit, within proportional
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limit where stress is proportional to the
strain the transverse strain which we get
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because
of these axial pull, the ratio of this transverse
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00:20:25,390 --> 00:20:33,500
strain to the axial strain or longitudinal
strain is constant and these particular constant
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is called as Poisson’s ratio.
.
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Therefore when the bar
is being pulled by axial force p it undergoes
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deformation in the
axial direction. If we call this as x direction
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then it undergoes deformation in the x
direction and thereby the strain which it
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will designate by epsilon we call the x direction
strain as epsilon x which we can compute in
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00:21:10,780 --> 00:21:19,220
terms of Pl cross sectional area and modulus
of elasticity of that member. Also, it undergoes
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deformation from its original, if these dot
line is that the deformed configuration so
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from its original depth the depth also deduces
thereby it undergoes some deformation over
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here delta prime. So the deformation as
Poisson had observed is that these deformations
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in the transverse strain the transverse
strain to the axial strain remains constant
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00:21:53,020 --> 00:21:57,030
for stresses up to the proportional limit
and this
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is what we designate as Poisson’s ratio
commonly designated as mu sometimes it is
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designated as the Greek symbol mu as well.
So mu is called as Poisson’s ratio
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..
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For uniaxial loading when the bar is subjected
to uniaxial pull in the x direction then as
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defined by the Poisson’s ratio it is equal
to the lateral strain to the longitudinal
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strain.
Here we have introduced a negative sign in
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this particular expression and this indicates
that when the bar is being pulled in the axial
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00:22:43,960 --> 00:22:49,590
x direction in the other direction it is
undergoing contraction and that is why this
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00:22:49,590 --> 00:22:55,170
negative symbol has come over here. The
negative sign indicates that an elongation
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in the axial direction causes contraction
in the
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transverse directions and the transverse strain
is uniform throughout the cross section and
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00:23:05,980 --> 00:23:09,240
is the same in any direction in the plane
of cross section.
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An important point to be noted is that the
transverse strain which could be the y and
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z
direction is uniform throughout the cross
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section and is the same in any direction in
the
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plane of cross section. So this leads to the
fact that for uniaxial
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loading that means a bar
when it is loaded in the x direction with
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an axial pull p the contraction or the deformation
in the y and z direction if we call this as
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x direction this as y direction and this as
z
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direction so the strain in the y direction
which is deformation to its original length
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in the
y direction epsilon y is the same as that
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of epsilon z as we said that in all directions
it is the
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same is equal to minus mu star epsilon x where
mu is the Poisson’s ratio. So lateral strain
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to the longitudinal strain is equal to the
Poisson’s ratio so there by the lateral
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00:24:28,600 --> 00:24:33,020
strain the y
or in the z direction are equal to the Poisson’s
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ratio times the axial strain.
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00:24:36,370 --> 00:24:37,370
..
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In the previous lesson we have seen that when
we relate the stress to the strain after the
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proportional limit or after the elastic limit
the stress sigma is proportional to the strain
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or
sigma x is equal to E epsilon x thereby the
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00:25:01,020 --> 00:25:08,250
strain epsilon is equal to sigma x by E and
the
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00:25:08,250 --> 00:25:14,890
strain in the transverse direction epsilon
y or epsilon z in trans of the axial strain
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00:25:14,890 --> 00:25:20,090
is equal
to minus mu sigma x by y in place of epsilon
214
00:25:20,090 --> 00:25:29,420
x we substitute sigma x by E which gives as
mu sigma x by E. Now this particular concept
215
00:25:29,420 --> 00:25:47,330
can be extended to a planar body.
If we take a planer body in which we have
216
00:25:47,330 --> 00:25:57,720
the stresses normal stresses sigma x and sigma
y
217
00:25:57,720 --> 00:26:05,000
and if we say that it is original dimension
is deltax and deltay, now when there have
218
00:26:05,000 --> 00:26:08,370
been
pulled they will undergo extension. When it
219
00:26:08,370 --> 00:26:10,180
is being pulled in the axial direction that
is
220
00:26:10,180 --> 00:26:14,710
the x direction it will undergo elongation
in the x direction and it will undergo
221
00:26:14,710 --> 00:26:20,030
contraction in the y direction. When it is
being pulled in y direction it will undergo
222
00:26:20,030 --> 00:26:24,320
elongation in the y direction and there will
be contraction in the x direction.
223
00:26:24,320 --> 00:26:30,750
Now, if we take independently that when it
is undergoing elongation in the x direction
224
00:26:30,750 --> 00:26:40,270
which is epsilon x is equal to sigma x by
E and correspondingly the strain in the y
225
00:26:40,270 --> 00:26:46,090
direction
epsilon y is equal to minus mu sigma x by
226
00:26:46,090 --> 00:26:52,120
E as for the definition of the Poisson’s
ratio.
227
00:26:52,120 --> 00:27:00,080
This is the case when sigma y is not there
the planar element is being pulled by the
228
00:27:00,080 --> 00:27:05,230
axial
pull in the x direction. When the plate element
229
00:27:05,230 --> 00:27:10,300
is been pulled in the y direction and
having no load in the x direction it will
230
00:27:10,300 --> 00:27:13,330
undergo elongation in the y direction and
thereby
231
00:27:13,330 --> 00:27:21,840
there will be contraction in the x direction
as per the Poisson’s definition. So when
232
00:27:21,840 --> 00:27:25,710
it is
being pulled by sigma y we have the strain
233
00:27:25,710 --> 00:27:36,030
epsilon y is equal to sigma y by E and
correspondingly epsilon x is equal to minus
234
00:27:36,030 --> 00:27:43,620
mu star sigma y by E.
Now if we consider the case when these plate
235
00:27:43,620 --> 00:27:52,170
element is been subjected to sigma x and
sigma y simultaneously when it is under the
236
00:27:52,170 --> 00:28:00,000
action of whole in the x as well as y direction
then we can superpose the effect of this individual
237
00:28:00,000 --> 00:28:05,510
results because we are within the
proportional limit or within the elastic limit
238
00:28:05,510 --> 00:28:08,480
and as a result we an superpose the results
of
239
00:28:08,480 --> 00:28:16,110
.the two and thereby the epsilon x finally
is for simultaneous action of sigma x and
240
00:28:16,110 --> 00:28:22,830
sigma y
is equal to epsilon x from sigma x by E which
241
00:28:22,830 --> 00:28:32,700
is the direct effect and because of y is equal
to minus mu star sigma y by E and likewise
242
00:28:32,700 --> 00:28:46,600
epsilon y is equal to sigma y by E minus mu
star sigma x by E. These are the strain in
243
00:28:46,600 --> 00:28:53,090
the simultaneous action of axial pull both
in the
244
00:28:53,090 --> 00:29:00,940
x and y.
.
245
00:29:00,940 --> 00:29:09,270
Now having looked into the strength in the
axial direction when it is acting only in
246
00:29:09,270 --> 00:29:12,610
one
direction or acting in the two directions,
247
00:29:12,610 --> 00:29:19,350
where in we have the strains epsilon x epsilon
y .
248
00:29:19,350 --> 00:29:27,370
We discussed the stresses we encounter, the
stress which is in the x direction in the
249
00:29:27,370 --> 00:29:32,240
y
direction and also the body undergoes at a
250
00:29:32,240 --> 00:29:40,160
point a stress which we call as a shearing
stress which acts between the two planes.
251
00:29:40,160 --> 00:29:47,400
If we look into such a case where a particular
body is subjected to, for example this body
252
00:29:47,400 --> 00:29:54,410
is fixed at this particular end and it is
subjected to a horizontal force p which is
253
00:29:54,410 --> 00:29:59,670
acting on
this plane let us assume that this plane has
254
00:29:59,670 --> 00:30:09,170
deformed as something like this may be this
length is L, this width is B hence the force
255
00:30:09,170 --> 00:30:18,330
which is acting on this plane will cause a
shearing stress which is P by deltaLB. This
256
00:30:18,330 --> 00:30:23,480
is the shearing stress.
Now in this, this particular body is fixed
257
00:30:23,480 --> 00:30:28,780
at this particular end and this horizontal
planar
258
00:30:28,780 --> 00:30:36,500
force will try to cause a deformation in this
particular body and this is expected that
259
00:30:36,500 --> 00:30:39,880
it is
deform state it will undergo deformation as
260
00:30:39,880 --> 00:30:49,260
shown in this dotted line. Now why this
deformation the angle which we get here this
261
00:30:49,260 --> 00:30:58,610
we call as the shearing strength and denoted
by the symbol gamma. So gamma is the shearing
262
00:30:58,610 --> 00:31:06,841
strength
and this is what is defined here,
263
00:31:06,841 --> 00:31:13,490
the shear strength is defined as the change
in right angle of a body. You see that when
264
00:31:13,490 --> 00:31:19,790
this particular body which has right angle
over here after the action of this planner
265
00:31:19,790 --> 00:31:22,860
force
which is causing deformation in this body
266
00:31:22,860 --> 00:31:29,280
undergoes a deformed state and thereby this
right angle undergoes the deformation and
267
00:31:29,280 --> 00:31:35,350
this angular deformation is turned as the
shearing strength and generally it is measured
268
00:31:35,350 --> 00:31:41,470
in radians due to the applied shear stress.
269
00:31:41,470 --> 00:31:42,470
..
270
00:31:42,470 --> 00:31:48,710
As we had observed in case of the stress and
the strain, in an actual pull up to the
271
00:31:48,710 --> 00:31:53,700
proportional limit or up to the elastic limit
the stress is proportional to the strain.
272
00:31:53,700 --> 00:32:03,570
Likewise here the shear strain also is proportional
to the applied shear stress. So in an
273
00:32:03,570 --> 00:32:11,830
elastic body up to the elastic limit the shearing
strain is proportional to the applied shear
274
00:32:11,830 --> 00:32:22,380
stress or the shear stress we designate it
by tau and shear strain we designate that
275
00:32:22,380 --> 00:32:30,000
by
gamma. So tau is proportional to gamma within
276
00:32:30,000 --> 00:32:38,090
elastic limit and thereby if you remove
this proportionality constant we can say tau
277
00:32:38,090 --> 00:32:48,920
is equal to G gamma and this proportionality
constant G is commonly known as shear modulus
278
00:32:48,920 --> 00:33:00,500
or many a times we called that as
Rigidity modulus. So the G is the shear modulus
279
00:33:00,500 --> 00:33:07,770
is equal to tau by gamma so the shearing
stress by the shearing strength gives as the
280
00:33:07,770 --> 00:33:22,980
shear modulus G. This is an important result
where we look into how as we go along to apply
281
00:33:22,980 --> 00:33:29,450
this or to evaluate the stress or the strain
in a body when they are subjected to different
282
00:33:29,450 --> 00:33:34,110
kinds of loading.
283
00:33:34,110 --> 00:33:35,110
..
284
00:33:35,110 --> 00:33:51,250
Here are some example problems based on the
aspects discussed earlier. An aluminium
285
00:33:51,250 --> 00:33:59,010
bar which is having a cross sectional area
of 160 mm square carries the axial load as
286
00:33:59,010 --> 00:34:03,980
shown over here. We will have to compute the
total change in length of the bar and the
287
00:34:03,980 --> 00:34:10,740
modulus of elasticity E is given as 70 GPa
the Gigapascal.
288
00:34:10,740 --> 00:34:18,220
If you remember, we had solved one problem
which is similar to this but we had different
289
00:34:18,220 --> 00:34:26,260
cross sectional areas for the different segments
in which we had computed the extensions
290
00:34:26,260 --> 00:34:33,399
of the different segments and finally we added
the extensions to get the final elongation
291
00:34:33,399 --> 00:34:39,330
in the bar. In this particular case also since
there are variations of the loads at different
292
00:34:39,330 --> 00:34:47,260
points the first thing that should be checked
is whether the applied loads keep the body
293
00:34:47,260 --> 00:34:49,220
in
equilibrium or not.
294
00:34:49,220 --> 00:34:58,660
If we look into that 35 Kilo Newton is acting
in the negative x direction 10 kilo Newton
295
00:34:58,660 --> 00:35:04,779
is acting in the negative x direction so total
force that is acting in the negative x direction
296
00:35:04,779 --> 00:35:12,390
is 45 kilo Newton and the positive x direction
15 kilo Newton and 30 kilo Newton to
297
00:35:12,390 --> 00:35:19,019
forces erecting so thereby this body is in
equilibrium under the applications of the
298
00:35:19,019 --> 00:35:22,710
forces
in the bar.
299
00:35:22,710 --> 00:35:28,690
Last time when we had solved a similar example
if you remember what we need to do is
300
00:35:28,690 --> 00:35:36,230
we have also from this point to this point
let us say this point is A, this is B, this
301
00:35:36,230 --> 00:35:44,850
is C, this
is D. Now from A to B just a little below
302
00:35:44,850 --> 00:35:49,769
B the effect of force which is existent is
this 35
303
00:35:49,769 --> 00:35:58,760
kilo Newton so if I take a cut here and make
a free body of this of the part A B just tried
304
00:35:58,760 --> 00:36:06,680
to the application of this 15 kilo Newton
load, then this is 35 kilo Newton. So, to
305
00:36:06,680 --> 00:36:10,980
keep
this bar in equilibrium we will have to have
306
00:36:10,980 --> 00:36:14,400
the resisting or the reactive load which is
35
307
00:36:14,400 --> 00:36:25,310
kilo Newton of this application of the load,
it extends or elongates the bar so the
308
00:36:25,310 --> 00:36:34,180
extension of this bar under the application
of 35 kilo Newton will be delta is equal to
309
00:36:34,180 --> 00:36:45,359
PL
by AE is equal to P here is 35 kilo Newton
310
00:36:45,359 --> 00:36:55,690
so 35 into 10 cube so much of Newton L of
this segment is 0.8 meter which is 800 mm
311
00:36:55,690 --> 00:36:59,589
divided by the cross sectional area. The cross
312
00:36:59,589 --> 00:37:08,789
.sectional area of the segment is 0.60 mm
square so 160 into E. E for this particular
313
00:37:08,789 --> 00:37:27,349
material is 70 GPa is 70 into 10 to the power
9. If you compute it comes to as 2.5 into
314
00:37:27,349 --> 00:37:34,270
10
to the power minus 6 mm. Now this particular
315
00:37:34,270 --> 00:37:44,490
bar since it is undergoing axial pull, the
tensile pull so this is a positive elongation.
316
00:37:44,490 --> 00:37:50,900
Now let us look into the free body of this
particular segment. Just after the application
317
00:37:50,900 --> 00:37:55,060
of
this force if we cut the bar then we have
318
00:37:55,060 --> 00:37:58,750
a forcing system something like this, here
we
319
00:37:58,750 --> 00:38:09,859
have 35 kilo Newton at this point we have
15 kilo Newton. Now 15 kilo Newton is acting
320
00:38:09,859 --> 00:38:14,220
in the positive x direction 35 kilo Newton
is acting in the negative x direction so the
321
00:38:14,220 --> 00:38:20,920
balance is 20 kilo Newton on this side. So
to make this part in equilibrium look to 20
322
00:38:20,920 --> 00:38:24,029
kilo
Newton often on this side so this will make
323
00:38:24,029 --> 00:38:28,150
the body of this particular free body in
equilibrium.
324
00:38:28,150 --> 00:38:34,670
So this particular segment, the part of the
body immediately after application of this
325
00:38:34,670 --> 00:38:38,509
load
to the point prior to the application of this
326
00:38:38,509 --> 00:38:47,130
load will have a force system which is
something like this. Since 20 is acting in
327
00:38:47,130 --> 00:38:50,210
this direction so the redistrict 20 will be
acting
328
00:38:50,210 --> 00:39:04,059
in this and so for balancing this it has to
act in this form thereby this bar will undergo
329
00:39:04,059 --> 00:39:14,539
stretching because of this 20 kilo Newton
load.
330
00:39:14,539 --> 00:39:22,700
For the third segment for this particular
segment if I take a free body here at this
331
00:39:22,700 --> 00:39:27,950
point
then we have the segment in which the force
332
00:39:27,950 --> 00:39:34,339
here is 35 Kilo Newton at this point, it is
15
333
00:39:34,339 --> 00:39:42,980
kilo Newton at this point, it is 30 kilo Newton
so we have 30 plus 15 is equal to 45 is
334
00:39:42,980 --> 00:39:48,000
acting in this direction, 35 which is acting
in this direction so to balance this we will
335
00:39:48,000 --> 00:39:51,940
have
to have 35 and the 10 acting in this particular
336
00:39:51,940 --> 00:39:54,990
direction so in this particular body turn
the
337
00:39:54,990 --> 00:40:03,170
Resistive force would be turned in this direction.
This particular segment is segment 2.
338
00:40:03,170 --> 00:40:08,109
In the first segment we have seen that has
undergone an extension of 2.5 into 10 to the
339
00:40:08,109 --> 00:40:14,569
power minus 6 mm, the second segment which
is under the application of 20 kilo Newton
340
00:40:14,569 --> 00:40:25,549
extension is delta for the second part equal
to P which is 20 kilo Newton into 10 cube
341
00:40:25,549 --> 00:40:30,440
into
L which is 1 meter. This is 100 mm divided
342
00:40:30,440 --> 00:40:42,819
by the cross sectional area which is 160 into
E which is 70 into 10 to the power 9 and this
343
00:40:42,819 --> 00:41:01,359
if we compute it comes as 1.785 into 10 to
the power minus 6 mm. So this is again an
344
00:41:01,359 --> 00:41:04,280
extension because it is a tensile pull which
is
345
00:41:04,280 --> 00:41:11,339
acting in this particular segment so this
is going to give the extension of the bar.
346
00:41:11,339 --> 00:41:13,539
Lastly,
this particular segment which is under the
347
00:41:13,539 --> 00:41:23,820
action of 10 kilo Newton compression will
cause the deformation which is in the negative
348
00:41:23,820 --> 00:41:31,280
direction is the contraction which we will
have in the bar.
349
00:41:31,280 --> 00:41:41,511
Now if we compute the deformation corresponding
to this delta 3 is equal to 10 kilo
350
00:41:41,511 --> 00:41:50,950
Newton into 10 cube into segmental length
here is 0.6 m which is 600 by 160 into 70
351
00:41:50,950 --> 00:41:56,450
into
10 to the power 9 and this if you compute
352
00:41:56,450 --> 00:42:03,460
it comes as 0.535 into 10 to the power minus
6
353
00:42:03,460 --> 00:42:09,730
mm and this is negative because it is a compressive
force acting in this particular segment
354
00:42:09,730 --> 00:42:16,250
and thereby it is undergoing a contraction.
Finally the total elongation of the bar in
355
00:42:16,250 --> 00:42:18,151
the first segment, it is undergoing elongation
and
356
00:42:18,151 --> 00:42:22,960
in the second segment it is undergoing elongation
and in the third segment it is
357
00:42:22,960 --> 00:42:29,240
undergoing contraction. So the final delta
will be some of this three segmental deltas
358
00:42:29,240 --> 00:42:30,240
is
359
00:42:30,240 --> 00:42:36,050
.equal to delta1 plus delta2 plus delta3 and
if you compute this will give you the final
360
00:42:36,050 --> 00:42:46,289
extension in the bar.
.
361
00:42:46,289 --> 00:42:55,190
Now let us look into another example problem
which is quite interesting. Here we have a
362
00:42:55,190 --> 00:43:06,240
rigid bar AB which is being supported by a
steel rod AC which is fixed at or used at
363
00:43:06,240 --> 00:43:10,730
this
particular point and cross sectional area
364
00:43:10,730 --> 00:43:18,619
of this particular steel rod is 100 mm square.
Now what we are interested in is to find out
365
00:43:18,619 --> 00:43:24,980
that how much vertical displacement this
particular point undergoes because of the
366
00:43:24,980 --> 00:43:35,749
application of this load 10 kilo Newton, this
angle is 45 degrees and this length of the
367
00:43:35,749 --> 00:43:46,970
rigid bar is given as 2.5m. Now this particular
bar being a rigid one because of the application
368
00:43:46,970 --> 00:43:55,509
of this load, this bar, the steel rod will
undergo extension or contraction which will
369
00:43:55,509 --> 00:44:01,200
cause the movement of this particular point
from which you need to find out the vertical
370
00:44:01,200 --> 00:44:08,990
displacement.
To know how much deformation the steel rod
371
00:44:08,990 --> 00:44:17,690
undergoes we need to take a free body of
this particular joint and compute the forces.
372
00:44:17,690 --> 00:44:20,859
Now if we take the free body of this joint
A
373
00:44:20,859 --> 00:44:35,930
we have we have the 10 kilo Newton force which
is acting. We have the bar AB and we
374
00:44:35,930 --> 00:44:46,029
have the steel there. Let us call this force
as F AC and this as F AB and this angle is
375
00:44:46,029 --> 00:44:54,059
45
degrees, now if we resolve this F AC in the
376
00:44:54,059 --> 00:44:59,740
vertical direction taking the summation of
vertical forces has equals to zero that will
377
00:44:59,740 --> 00:45:08,240
give us the F AC sin 45 degrees is equal to
10
378
00:45:08,240 --> 00:45:22,580
Kilo Newton thereby F AC is equal to 10 square
root of 2 Kilo Newton.
379
00:45:22,580 --> 00:45:26,940
If you take the component of this or if you
take the summation of horizontal force as
380
00:45:26,940 --> 00:45:32,880
0 if
you take the summation of horizontal force
381
00:45:32,880 --> 00:45:45,400
as zero then F AC cos 45 degrees is
equal to
382
00:45:45,400 --> 00:45:55,800
F AB and F AC is 10 root 2 cos 45 degrees
is 1 by square root of 2 so 10 square root
383
00:45:55,800 --> 00:46:00,500
of 2
into 1 by square root of 2 is F AB and thereby
384
00:46:00,500 --> 00:46:08,050
F AB is equal to 10 kilo Newton and the each
of this forces which are acting in the bar.
385
00:46:08,050 --> 00:46:16,190
.In this bar it will be F AB is a compressive
force and F AC in the joint the forces acting
386
00:46:16,190 --> 00:46:18,789
in
this direction so in the bar it is acting
387
00:46:18,789 --> 00:46:22,589
in this form so it is a tensile force so this
is tensile
388
00:46:22,589 --> 00:46:30,720
pull in the bar F AC , because of this strain
which has been applied on this bar the body
389
00:46:30,720 --> 00:46:34,059
is
expected to undergo elongation now our job
390
00:46:34,059 --> 00:46:37,349
is to evaluate this elongation because of
the
391
00:46:37,349 --> 00:46:43,420
application of this tensile pull which is
tend to group to kilo Newton. So let us compute
392
00:46:43,420 --> 00:46:58,869
that how much elongation this bar AC undergoes
393
00:46:58,869 --> 00:47:11,339
now the deltaA C is equal to P L by AE
the extension P just now we have computed
394
00:47:11,339 --> 00:47:20,920
which is equal to 10 root 2 Kilo Newton into
10 cube N.
395
00:47:20,920 --> 00:47:27,799
We will have to compute the length AC, now
we know length AB we know this angle so
396
00:47:27,799 --> 00:47:41,880
AC length AC into cos 45 degrees is equal
to 2.5 m so L AC is equal to 2.5 square root
397
00:47:41,880 --> 00:47:49,160
of
2m so L into 2.5 square root of 2 into 10
398
00:47:49,160 --> 00:47:59,150
cube mm divided by the cross sectional area
which is 100 mm square into E which is 2 into
399
00:47:59,150 --> 00:48:17,480
10 to the power 5 MPa Mega Pascal. This
gives us a value of square root of 2, this
400
00:48:17,480 --> 00:48:22,640
2 gets canceled with square root of 2 10 to
the
401
00:48:22,640 --> 00:48:28,930
power 5, 10 to the power 7 this is 10 to the
power 6 so this is 25 so this 10 goes up 106
402
00:48:28,930 --> 00:48:39,200
10 to the power 7 so we have 2.5 so deltaA
C is equal to 2.5 mm so the extension of this
403
00:48:39,200 --> 00:48:52,460
particular steel rod is equal to 2.5 mm.
Now the interesting part is that these particular
404
00:48:52,460 --> 00:49:03,119
bar A B being rigid, now when this rod
extends it will try to pull this particular
405
00:49:03,119 --> 00:49:06,259
joint or displace this particular joint from
this
406
00:49:06,259 --> 00:49:14,790
position to some other position and this bar
being rigid this will rotate considering these
407
00:49:14,790 --> 00:49:20,460
particular point as the center of rotation.
So if this particular point rotates, it is
408
00:49:20,460 --> 00:49:35,230
expected that it will take circular r. Now
what we
409
00:49:35,230 --> 00:49:42,069
are interested is to find out the vertical
displacement of this particular point. Now
410
00:49:42,069 --> 00:49:48,509
assuming that this rigid bar undergoes a movement
in a circular path and this
411
00:49:48,509 --> 00:49:55,980
displacement the deformation being small this
circular r we assume as a straight one.
412
00:49:55,980 --> 00:50:05,549
If we take this deformation if you call as
deltaA if we take the component of this deltaA
413
00:50:05,549 --> 00:50:15,710
along AC and perpendicular to AC then this
perpendicular segment will indicate the
414
00:50:15,710 --> 00:50:24,920
rotation of this deformed AC and this particular
part will reduce stretching of the bar or
415
00:50:24,920 --> 00:50:31,650
the steel rod AC. So this is the extension
of the steel rod AC this is the final position
416
00:50:31,650 --> 00:50:35,150
and
this particular movement indicates the rotation
417
00:50:35,150 --> 00:50:43,619
of the bar AC so this particular joint we
will finally come down to this point. Now
418
00:50:43,619 --> 00:50:49,240
we have already computed that how much
extension this particular steel bar AC has
419
00:50:49,240 --> 00:50:59,499
undergone which is equal to 2.5 mm.
This is our deltaA C and what we are interested
420
00:50:59,499 --> 00:51:08,640
is the value deltaA now this angle is 45
degrees so this is 45 degrees this angle also
421
00:51:08,640 --> 00:51:12,540
is 45 degrees and thereby this angle is also
45
422
00:51:12,540 --> 00:51:21,849
degrees so from this particular triangle where
in this arm is deltaA this is deltaA C so
423
00:51:21,849 --> 00:51:37,859
deltaA sine45 degrees will give deltaA C so
deltaA sine 45 degrees is equal to deltaA
424
00:51:37,859 --> 00:51:46,930
C
which is equal to 2.5 mm hence the deltaA
425
00:51:46,930 --> 00:51:53,910
is equal to 2.5 star root 2 which is equal
to
426
00:51:53,910 --> 00:52:09,510
3.54 mm. So the vertical displacement of point
A or this joint A is equal to 3.54 mm.
427
00:52:09,510 --> 00:52:17,869
Here you can see that how we can apply the
expression for extension. Because of the
428
00:52:17,869 --> 00:52:24,049
application of the load in our system where
two bars are connected a rigid bar is held
429
00:52:24,049 --> 00:52:28,619
by
an extensible bar which is producing the reflexion
430
00:52:28,619 --> 00:52:34,239
of this particular joint.
431
00:52:34,239 --> 00:52:35,239
..
432
00:52:35,239 --> 00:52:49,240
Let us look into another problem where the
bar is no longer uniform but it has a variation
433
00:52:49,240 --> 00:52:59,739
along the length of that bar on this end the
diameter of the bar is given as 20 mm the
434
00:52:59,739 --> 00:53:09,109
other end it is given as 30 mm and it is being
pulled by a load 30 kilo Newton, what we
435
00:53:09,109 --> 00:53:14,589
need to do is determine the elongation of
the tapered cylindrical aluminum bar caused
436
00:53:14,589 --> 00:53:18,700
by
this 30 kilo Newton axial load where E is
437
00:53:18,700 --> 00:53:26,470
given as 72 GPa.
We have to discuss on how to compute the elongation
438
00:53:26,470 --> 00:53:31,420
in a bar where the cross section is
no longer uniform, but it is varying along
439
00:53:31,420 --> 00:53:37,569
the length of the bar so applying that concept.
Now what we do is that at any cross section
440
00:53:37,569 --> 00:53:40,180
which is at a distance of x along the length
of
441
00:53:40,180 --> 00:53:47,170
the bar we try to find out what is the diameter
so the diameter d at this particular point
442
00:53:47,170 --> 00:54:05,510
which at a distance of x is equal to (20 plus
30 minus 20) x by L. Here L is equal to 400
443
00:54:05,510 --> 00:54:18,730
mm is equal to 20 plus 10 by 400 so x by 40
mm. This is the diameter which we have at
444
00:54:18,730 --> 00:54:25,099
this particular location and since it is varying
at every point this diameter is changing as
445
00:54:25,099 --> 00:54:26,099
a
function of x.
446
00:54:26,099 --> 00:54:27,099
..
447
00:54:27,099 --> 00:54:46,180
So we compute delta is equal
to P L by A E now, here this bar is undergoing
448
00:54:46,180 --> 00:54:51,789
an axial pull
with a load of 30 kilo Newton so we have restricted
449
00:54:51,789 --> 00:55:05,539
this over the length L which is 0 to
400 P is 30 kilo Newton. Since we are taking
450
00:55:05,539 --> 00:55:13,250
the diameter of small segment which is dx
so A is a function of x and we have a value
451
00:55:13,250 --> 00:55:30,689
of E. This if we integrate then we get the
extension of the bar.
452
00:55:30,689 --> 00:55:31,689
.
453
00:55:31,689 --> 00:55:41,109
I have another problem which is the applications
of the concept of shearing strain. There
454
00:55:41,109 --> 00:55:54,921
are 2 rubber pairs, there are 3 steel plates
of 10 mm thickness and 2 rubber pairs are
455
00:55:54,921 --> 00:56:00,490
attached to this steel plates in between the
steel plates which is of the rubber pair
456
00:56:00,490 --> 00:56:07,569
.thickness is 50 mm and the length of these
plates and the rubber pairs are 200 mm the
457
00:56:07,569 --> 00:56:18,609
width of this rubber pairs which we look into
in the third dimension this distance is 100
458
00:56:18,609 --> 00:56:27,720
mm.
Here when this central bar is being pulled
459
00:56:27,720 --> 00:56:33,060
the rubber is undergoing deformation. Here
it
460
00:56:33,060 --> 00:56:39,369
has been told that, determine the displacement
of the metal plate considering the
461
00:56:39,369 --> 00:56:50,460
deformation of rubber pad only. Since this
rubber pad is connected with this when this
462
00:56:50,460 --> 00:56:55,739
is
being pulled the pad is undergoing deformation
463
00:56:55,739 --> 00:57:05,920
and thereby this will try to be pulled in
this form.
464
00:57:05,920 --> 00:57:11,970
Now add these interfaces between the steel
plate and the rubber pad, there will be a
465
00:57:11,970 --> 00:57:17,329
shearing force which would be acting on this
plane which will try to cause. If we take
466
00:57:17,329 --> 00:57:21,070
a
line here this particular line we will try
467
00:57:21,070 --> 00:57:29,840
to undergo a movement which will cause a
shearing strength in the body. Now 10 kilo
468
00:57:29,840 --> 00:57:35,769
Newton force is acting on this steel plate
thereby we have two surfaces on which this
469
00:57:35,769 --> 00:57:39,569
can the shearing force can get distributed
so
470
00:57:39,569 --> 00:57:42,890
half the force will be acting on this surface
and half of the force will be acting on this
471
00:57:42,890 --> 00:57:50,020
surface and thereby there will be a strain
angle and if we can compute this strain angle
472
00:57:50,020 --> 00:58:00,450
then we can compute this deformation.
The stress which will be acting is equal to
473
00:58:00,450 --> 00:58:07,069
the force which is acting in this place divided
by the area and the horizontal pull which
474
00:58:07,069 --> 00:58:10,000
is acting at the interface is equal to the
half of
475
00:58:10,000 --> 00:58:16,780
this load which is 5 kilo Newton so the shearing
stress tau is equal to 5 star 10 cube
476
00:58:16,780 --> 00:58:27,369
Newton divided by the area which is 200 star
100 which is equal to 0.25 MPa.
477
00:58:27,369 --> 00:58:34,970
So this is the shearing stress which is acting
and as we know that shearing stress divided
478
00:58:34,970 --> 00:58:40,980
by shearing strength is equals to a shear
modulus G, a shear modulus G is given over
479
00:58:40,980 --> 00:58:47,030
here. So we can compute gamma the shearing
stress which is equal to tau by G the tau
480
00:58:47,030 --> 00:58:53,930
is
0.25 and G is 150 Mpa. So this gives as a
481
00:58:53,930 --> 00:59:02,009
value of shearing strain gamma and once we
know gamma, gamma times this thickness will
482
00:59:02,009 --> 00:59:09,710
give the movement that it undergoes. So
thickness 50 mm into gamma will give us the
483
00:59:09,710 --> 00:59:17,799
deformation this bar will undergo because
of the deformation in the rubber band.
484
00:59:17,799 --> 00:59:18,799
..
485
00:59:18,799 --> 00:59:24,380
Let us summarize: This particular lesson included
the concept of axial strain in a body
486
00:59:24,380 --> 00:59:29,680
with variable cross section, the concept of
Poisson’s ratio and then the concept of
487
00:59:29,680 --> 00:59:32,789
shear
strain and shear modulus and some examples
488
00:59:32,789 --> 00:59:36,400
to demonstrate the evaluation of
deformation in a stressed body.
489
00:59:36,400 --> 00:59:37,400
.
490
00:59:37,400 --> 00:59:42,309
Here are some questions:
What is Poisson’s ratio and what is the
491
00:59:42,309 --> 00:59:47,470
range of its values?
What is shear modulus and what is its unit?
492
00:59:47,470 --> 00:59:58,029
How will you evaluate strain in a three dimensional
stressed body?
493
00:59:58,029 --> 00:59:58,029
.