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Welcome to the first lesson of the second
module which is on analysis of strain. In
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the
first module we discussed about stresses and
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we have seen various aspects of stresses.
Now we are going to discuss certain aspects
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of strain.
.
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..
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After completion of this particular lesson
one will be able to understand the concept
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of
axial strain, the concept of normal strain
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and strain at a point in a stressed body.
One will
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be able to understand the relation between
stress and strain, which is very important
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when
we talk about the strength of material, is
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not only the stresses but the relation between
the
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stress and the strain is important. One will
be able to understand the concept modulus
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of
elasticity which we need for the analysis
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as we go along.
.
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We
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will evaluate normal strain and strain at
a point. We look into the relationship
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between stress and strain and thereby the
Hooke’s law which are defined for the elastic
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.body and then the different terms which we
get in the stress and strain relationship
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and
the modulus of elasticity.
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.
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Before we start looking into the aspect of
the strain let us look into the questions.
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The
first question is what are the equations of
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Cartesian co-ordinate system in a stressed
body?
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.
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..
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The equations of equilibrium in Cartesian
co-ordinate system are dell σ x dell x plus
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dell
τ xy dell y plus X is equal to 0 and dell
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τ xy dell x plus dell σ y dell y plus Y
is equal to 0
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where x and y are the body force components.
Now as you know σ x and σ y are the normal
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stresses to the x and y direction and τ xy
is
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the shearing stress component. So these two
define the equations of the equilibrium in
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two dimensional system.
.
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.The second question posed was, what are the
equations of equilibrium in a polar
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coordinate system of a stressed body?
.
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the equations of equilibrium in polar co-ordinate
system, this particular aspect was
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discussed in the sixth lesson of the module
one, wherein we defined stress components,
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the radial stress, sigma r, the circumferential
stress sigma theta and we have seen how to
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derive this equations of equilibrium.
The radial directional stress is sigma r,
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the circumferential stress sigma theta and
consequently we are sharing stress component
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which are τ rθ . And in a polar coordinate
form these are the equations of equilibrium
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dell sigma r dell r plus 1 by r dell tow theta
by dell theta plus sigma r minus sigma theta
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by r is equal to 0. Dell τ rθ dell r plus
dell
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sigma theta dell theta by r plus τ rθ by
r is equal to 0. These are the two equations
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which
defines the equations of equilibrium in polar
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co-ordinate system for two dimensional
stress analysis.
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..
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The third question was what is the value of
maximum shear stress if σ 1 is equal to 10
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MPa and σ 2 is equal to 0?
Let us look into, if you remember the Mohr’s
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circle of stress in which we had sigma axis
and the tow axis. Now in the Mohr us circle,
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the point on the maximum stress is σ 1 , and
the minimum stress is σ 2 . Now radius of
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this circle is defined as the maximum value
of
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the shearing stress, which is tow max. And
tow max is given by σ 1 minus σ 2 by 2.
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In the
given problem, σ 2 is 0, and σ 1 is 10.
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So there by maximum shearing stress is 10
by 2 is
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5 MPa. These are the three questions which
were posed and the answers for these were
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defined. Let us look forward into strain analysis.
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..
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Now we are going to look in to axial normal
strain. Now let us assume we have a body
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which is acted on by force P, length of the
body let us assume as L. When a body is
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subjected to either change in the temperature
or subjected to the forces, it undergoes
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deformations, and in strength of material
we are interested to look in to this deformation
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and we try to define a quantity in terms of
this deformation.
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Let us assume after deformation, the length
is L plus delta. Thereby delta is the
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deformation or the extension of the body.
Now we will define a quantity, which is the
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ratio of this deformation to the original
in the length and is generally designated
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as
epsilon. Epsilon as equals to delta by L and
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this is known as strain. The load, the force
which is acting on the body is in the axial
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direction and we assume that the deformation
is uniform along the length of the body. And
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thereby everywhere the strain is same.
Also, as we looked in to earlier, when the
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body is subjected to axial force, we get a
stress
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which we call as normal stress which is the
stress normal to the cross section of the
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body.
In line with that, we define that the strain
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or the deformation which is along the axis
has
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the axial strain, since to compare with the
normal stress we call this strain as normal
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strain. So the strain associated with the
normal stress is called normal strain and
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we
define this as elongation per unit length.
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Now as this is the ratio of two lengths, basically
this is a dimensionless number quantity and
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thereby there are no units as such.
However, it is customary to define strain
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in terms of the ratio of the lengths, say
meter by
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meter or millimeter by millimeter for a body.
Accordingly some times we say the units of
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strain as millimeter by millimeter. And this
strain is defined in terms of a number and
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sometimes we define this in terms of percentage
as well.
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..
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Now the strain when we are calculating for
the deformation delta, we are assuming that
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this uniform deformation in the entire length
of the bar. If the deformation is not uniform,
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if we compute the strain from delta by L,
then we are assuming the strain on an average
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sense over the entire body.
Now if the deformation is not same everywhere
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and if the deformation varies along the
length of a bar then the way we have computed
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stress at a point in a stress body, we
compute strain also in a stress body. Now
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let us look into this particular figure where
a
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bar which is fixed at one end is subjected
to a pull P. Now let us assume that we are
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interested find out the strain at a point,
within this body where this is A which is
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at a
distance of x from the fixed end.
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To evaluate strain at a point in a stress
body, what we do is consider an imaginary
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fiber,
let us say A B, which is of length dx or delta
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x. This bar when it is subjected to pull P
the
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fiber also is stretched or deformed and let
us assume this is the stretched fiber which
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was
originally delta x, let us say this as A prime
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B prime and the length of this stretched fiber
is delta x plus dell delta. Thereby the extension
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of this fiber is dell delta.
And as we have defined, the strain epsilon
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for this particular fiber is dell delta by
dell x.
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And this strain at this particular point A
you can define as; strain is equal to delta
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x tends
to 0, this is dell delta by dell x. This we
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can write as d delta by dx. Hence d delta
is equal
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to epsilon dx. So what is the length L if
this is the length of the member which is
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defined
as L, then over the entire length the deformation
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delta which is integral 0 to L d delta is
equal to integral 0 to L epsilon dx and this
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is the deformation if the strain varies along
the
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length of the bar.
If we have uniform strain everywhere then
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this is equal to epsilon integral 0 to L dx
which is L and epsilon we get as delta by
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L as we have seen earlier. If the strain is
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.constant everywhere then it is uniform and
we get delta by L. If it is not then delta
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is
integral 0 to L epsilon dx. That is how we
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compute the strain at a point.
.
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In Strength of Materials, we have seen how
to calculate the stress at any point in a
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stressed body and we have seen various components
of the stresses.
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Now we have defined a quantity which we call
as strain, strain at a point or strain on
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an
average sense, if it is uniform over the length
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of the bar. Now what we need to do which
is of relevance in Strength of Materials is
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that the relationship between stress and the
strain. Now, for evaluating the relation between
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the stress and strain, take a body, apply
a
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tensile pull in the body, and we apply this
load gradually over the bar, for each increment
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of the load, we try to find out how much deformation
the body undergoes.
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Here, if we look, if this is a bar we have
formed a section in different cross sectional
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forms. This part is called the grip which
is inserted in the tensile stress equipment
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and the
whole bar is pulled. Now on this bar we fix
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up a length which is in between this bar,
which is little away from this grip, so that
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the length on which we focus our attention
is
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not affected by the variation of the load
in the grip zone. And this length on which
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we
focus our attention we call this as the gauge
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length. So this is the initial distance between
the two predefined bars and this we call as
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gauge length.
And as I said this length we consider between
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the bar and little away from the grip zone
so that this particular zone is not affected
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by the force distribution or the stress
distribution in the grip zone. Now if we apply
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a pull on this bar gradually, as we have
seen the bar will under go deformation and
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there by there will be change in the gauge
length and if we measure that increment or
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the deformation, then we can compute the
strength.
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.And as we have seen earlier the stress for
a body which is subjected to an axial pull,
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the
normal stress is the axial pull divided by
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the cross sectional area. So for each increment
of the load, we can compute the stress, we
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00:17:58,580 --> 00:18:05,520
can compute the correspondingly the strain
and then we can plot a graph to establish
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the relationship between the stress and the
strain.
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.
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Now if look in to the plot between the stress
and the strain, the plot is something like
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this.
In this particular figure, this axis represents
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the strain and the y-axis represents the stress.
Now plot each increment of the load, we compute
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the deformation, there by we get the
strain. And for several such points we plot
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it, on a steel bar this is the profile we
get a
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different stage of loadings.
Now for this particular figure there are various
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terms indicated with it. One we have
called as proportional limit, this we have
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called it elastic limit, this particular point
we
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00:19:12,970 --> 00:19:20,320
have called it a yield point, this is the
ultimate strain or ultimate stress, this particular
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point, actually this is wrong. This is not
normal, this should be nominal and this is
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the
nominal failure strength. Now here if you
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00:19:34,760 --> 00:19:39,450
note it that we compute the stresses here
with
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reference to the original cross section of
the bar which we have computed.
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If you can visualize, when we are pulling
this bar or when we are applying a tensile
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pull,
as it deforms the cross sectional area of
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the bar reduce, now the stress a we know is
stress
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00:19:57,940 --> 00:20:03,840
divided by the cross sectional area, the cross
sectional area as we take the original cross
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sectional area, and accordingly we get the
profile of this curve. If we take the actual
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area,
load divided by the actual area then the stress
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00:20:15,381 --> 00:20:20,590
value will be different from this. In fact
from this particular point configuration is
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00:20:20,590 --> 00:20:24,040
some thing like this. Now the stress when
you
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00:20:24,040 --> 00:20:31,280
compute reference to the original cross sectional
area of the bar, we call those stresses as
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the nominal stresses and correspondingly the
nominal strain. Otherwise if we compute
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the stresses with reference to the change
cross sectional area of the bar, we call that
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00:20:45,070 --> 00:20:46,070
stress
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00:20:46,070 --> 00:20:52,400
.as true stress and the corresponding strain
as true strain. So here what we have plotted
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00:20:52,400 --> 00:20:55,680
is
the nominal stress and the strain.
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00:20:55,680 --> 00:21:03,490
Now, the meaning of this proportional limit
is, up to the level of proportional limit,
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00:21:03,490 --> 00:21:09,650
the
stress is proportional to the strain. So we
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00:21:09,650 --> 00:21:14,130
say the stress, sigma is proportional to the
strain
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00:21:14,130 --> 00:21:25,450
epsilon. Now this particular point elastic
limit is the point, up to which if the load
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00:21:25,450 --> 00:21:29,640
is
applied on the body and if it is released
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00:21:29,640 --> 00:21:34,340
the body comes back to its original state
and that
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we call as the elastic limit of the body.
But beyond elastic limit, if we apply load
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00:21:42,460 --> 00:21:45,290
and go
beyond the elastic limit, and if we release
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00:21:45,290 --> 00:21:49,460
the load the specimen will not come back to
its
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00:21:49,460 --> 00:21:56,360
original place and some amount of deformation
is permanently set in the body and that
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00:21:56,360 --> 00:22:06,980
we call as permanent state. And yield point
is the point, when the body starts yielding.
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00:22:06,980 --> 00:22:11,810
It
goes beyond the elastic limit, the plasticity
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starts forming in the section.
And if we keep on applying the load at that
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00:22:18,640 --> 00:22:23,820
particular point in fact increasing the load,
the
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00:22:23,820 --> 00:22:31,660
bar deforms and the extension, the deformation
becomes excessive and reaches o the
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stress which we call as loading down on the
load when the bar fails and that how we get
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00:22:40,230 --> 00:22:43,360
failure stress which is less than the ultimate
stress.
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00:22:43,360 --> 00:22:49,320
What is the maximum stress a bar can attain?
We call that as ultimate stress and this particular
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00:22:49,320 --> 00:22:59,060
point is the failure stress. Now in the bar
these two points proportional limit and the
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00:22:59,060 --> 00:23:02,820
elastic limit is vary difficult to distinguish
and
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00:23:02,820 --> 00:23:14,530
we consider that up to this level, this up
to the elastic limit, the stress is proportional
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00:23:14,530 --> 00:23:18,860
to
the strain. If we remove this proportionality
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00:23:18,860 --> 00:23:26,500
constant, sigma is written as constant times
E, which is constant for that material and
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00:23:26,500 --> 00:23:32,500
this is what we know as Hooke’s law.
Up to the proportionality limit or up to the
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00:23:32,500 --> 00:23:35,450
elastic limit, the stress is proportional
to the
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00:23:35,450 --> 00:23:43,510
strain. Or stress equal to E times strain,
where E is called as the constant of
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00:23:43,510 --> 00:23:52,980
proportionality or the modulus of the elasticity,
which is an important parameter in the
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00:23:52,980 --> 00:24:02,610
Strength of Materials and this is constant
for the material which we are considering
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00:24:02,610 --> 00:24:08,799
during the evaluation of the stress and the
strain known as Hooke’s law which is at
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00:24:08,799 --> 00:24:11,230
the
elastic limit, the stress is proportional
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00:24:11,230 --> 00:24:22,570
to the strain or sigma is equal to E times
epsilon.
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00:24:22,570 --> 00:24:23,570
..
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So we have seen the Hooke’s law which is
up to the proportionality limit, the stress
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00:24:31,690 --> 00:24:34,420
is
proportional to the strain and there by we
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00:24:34,420 --> 00:24:39,160
get sigma is equal to E times epsilon. We
have
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seen the elastic limit, if the bar is loaded
and is allowed to extend and if the bar is
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00:24:48,770 --> 00:24:51,870
within
the elastic limit, if we release the load,
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00:24:51,870 --> 00:24:56,660
the bar is expected to come back to its original
state.
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The permanent set is, as the member yields
as it reaches to yield stress then plasticity
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00:25:05,340 --> 00:25:11,650
forms in the section, then if we release the
load, as we have seen in the case of elastic
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00:25:11,650 --> 00:25:17,530
material when the bar is still within its
elastic limit the bar comes back to its original
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00:25:17,530 --> 00:25:24,841
position but once it starts yielding, then
if we release the load, the bar does not comes
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00:25:24,841 --> 00:25:30,890
back to its original position and some amount
of deformation sets permanently in the
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00:25:30,890 --> 00:25:39,790
body and that is what we call as permanent
set. Yield point again is the point where
215
00:25:39,790 --> 00:25:43,350
the
material starts yielding or it goes beyond
216
00:25:43,350 --> 00:25:47,970
the elastic limit and plasticity starts setting
in
217
00:25:47,970 --> 00:25:51,040
the member.
218
00:25:51,040 --> 00:25:52,040
..
219
00:25:52,040 --> 00:26:01,520
Well, these are some of the stress strain
relationship for different materials. If we
220
00:26:01,520 --> 00:26:06,400
take
concrete specimen, apply tensile pull in tensile
221
00:26:06,400 --> 00:26:14,071
stress equipment, and then we get the
profile similar to this, which is the stress
222
00:26:14,071 --> 00:26:18,010
strain relationship. If we take materials
like
223
00:26:18,010 --> 00:26:24,110
aluminum, cast iron or high carbon steel we
get these kind of profile and these are
224
00:26:24,110 --> 00:26:28,870
necessary to know the relationship between
stress and strain by the modulus state of
225
00:26:28,870 --> 00:26:34,820
elasticity. So that we can compute the stress
and can compute the strain and we can
226
00:26:34,820 --> 00:26:40,390
establish the relationship between stress
and strain in a body when the material is
227
00:26:40,390 --> 00:26:45,580
used
either for some equipment or the machine part
228
00:26:45,580 --> 00:26:52,030
or in structural body where we interested
to find stress and strain, we need to know
229
00:26:52,030 --> 00:26:58,910
the relationship between stress and strain
material that with which the machine part
230
00:26:58,910 --> 00:27:07,860
or the structure is composed of.
231
00:27:07,860 --> 00:27:08,860
..
232
00:27:08,860 --> 00:27:26,590
Now as we have seen in the previous figure,
in the first figure, in this particular figure
233
00:27:26,590 --> 00:27:32,710
as
you have seen that we have defined yield point.
234
00:27:32,710 --> 00:27:41,150
In this particular zone, this particular
curve shows that we have defined yield point.
235
00:27:41,150 --> 00:27:46,760
So corresponding to this stress we know
that this is the yield stress we defined as
236
00:27:46,760 --> 00:27:52,990
σ y but as we have seen in the subsequent
figure
237
00:27:52,990 --> 00:28:02,910
that we do not have any defined yield point
and in strength of material when we deal
238
00:28:02,910 --> 00:28:08,560
within the elastic limit we need to know the
what is the value of the yield stress beyond
239
00:28:08,560 --> 00:28:14,450
which the member will start yielding. So we
try to limit our stress when we deal with
240
00:28:14,450 --> 00:28:19,900
elastic level of analysis.
We like to limit over self to the elastic
241
00:28:19,900 --> 00:28:22,900
level, so we need to know what is the value
of the
242
00:28:22,900 --> 00:28:27,220
yield stress?
But from this kind of stress strain distribution,
243
00:28:27,220 --> 00:28:34,500
it is difficult to know the value of the yield
stress. Now, to compute the value of yield
244
00:28:34,500 --> 00:28:43,010
stress we do is to observe that the strain
corresponding to the yield stress value is
245
00:28:43,010 --> 00:28:49,380
the order of point two percent, this is point
0, 0
246
00:28:49,380 --> 00:28:57,250
two is the strain. Now, if we draw a tangent
to the curve at this particular point, which
247
00:28:57,250 --> 00:29:01,030
we
call it as initial tangent and if we draw
248
00:29:01,030 --> 00:29:07,010
a line at point two percent strain and if
draw a line
249
00:29:07,010 --> 00:29:13,600
parallel to initial of the tangent, the point
were it cuts the curve, the stress strain
250
00:29:13,600 --> 00:29:17,660
curve,
the corresponding stress we call as yield
251
00:29:17,660 --> 00:29:27,690
stress and this yield stress is normally we
designate as proof stress. So for the material
252
00:29:27,690 --> 00:29:36,580
where we apply a tensile pull and plot a
stress strain curve and corresponding to that
253
00:29:36,580 --> 00:29:43,390
stress strain curve if we do not get a defined
yield point, corresponding to which we are
254
00:29:43,390 --> 00:29:50,090
interested to find yield stress then we compute
the yield stress in the direct way corresponding
255
00:29:50,090 --> 00:29:57,980
to point two percent of strain and this
stress we call as proof stress or the yield
256
00:29:57,980 --> 00:30:05,340
stress of that particular material.
257
00:30:05,340 --> 00:30:06,340
..
258
00:30:06,340 --> 00:30:14,360
Well as we have seen earlier, in case of the
stresses when a bar is subjected to axial
259
00:30:14,360 --> 00:30:18,400
pull
we have computed the stress which we have
260
00:30:18,400 --> 00:30:23,960
defined as the normal stress, the stress which
is normal to the cross section of the bar
261
00:30:23,960 --> 00:30:27,940
and the load is acting perpendicular to the
cross
262
00:30:27,940 --> 00:30:36,240
section through the axis of the bar.
Now for an axially loaded bar, if we are interested
263
00:30:36,240 --> 00:30:49,430
to compute the strain, let us assume
that this particular bar say is fixed at this
264
00:30:49,430 --> 00:31:05,720
end, the length of this bar is L and is subjected
to pull P, thereby if the delta is the extension
265
00:31:05,720 --> 00:31:16,930
then the strain is equal todelta by Land we
consider that for this particular bar a this
266
00:31:16,930 --> 00:31:20,770
analysis you carry out is within the elastic
limit
267
00:31:20,770 --> 00:31:28,360
of the bar. And we have seen right now in
the stress strain relationship that within
268
00:31:28,360 --> 00:31:31,410
the
elastic limit of the bar is the stress is
269
00:31:31,410 --> 00:31:44,750
proportional to the strain. And we write sigma
equals to E(epsilon). So, in place of epsilon,
270
00:31:44,750 --> 00:31:56,161
you can write E(delta by L). Hence from this
equation we can write the deformation delta
271
00:31:56,161 --> 00:32:10,030
as sigma times L by E. As we know that this
is a bar which is axially loaded and the load
272
00:32:10,030 --> 00:32:14,960
is acting through the axis of the bar, any
cross section.
273
00:32:14,960 --> 00:32:22,700
Normal stress sigma P is divided by the cross
sectional area. So sigma is equal toP
274
00:32:22,700 --> 00:32:33,690
divided by A. So if we substitute the value
of sigma in this we get, delta is equal toPL
275
00:32:33,690 --> 00:32:41,700
by
AE. So for an axially loaded bar if we know
276
00:32:41,700 --> 00:32:47,440
the axial pull, if we know the cross sectional
area, if we know the length and if we know
277
00:32:47,440 --> 00:32:53,580
the material with which this bar is made of,
for which we know the modulus of elasticity
278
00:32:53,580 --> 00:33:00,180
then we can compute what will be the
deformation in the bar.
279
00:33:00,180 --> 00:33:11,690
Considering that every where the strain is
same, however the strain is not everywhere,
280
00:33:11,690 --> 00:33:16,490
if
there is the variation of the strain everywhere,
281
00:33:16,490 --> 00:33:25,720
in the previous calculation that the strain
or
282
00:33:25,720 --> 00:33:36,130
the deformation, delta as we have seen at
deformation delta is equal to the integral
283
00:33:36,130 --> 00:33:47,840
0 to L
epsilon dx and as we have seen in the stress
284
00:33:47,840 --> 00:33:56,810
strain relationship, sigma is equal to E times
epsilon so this is equal to integral 0 to
285
00:33:56,810 --> 00:34:02,200
L, in case of epsilon, if you write this is
sigma by
286
00:34:02,200 --> 00:34:18,940
.E dx and sigma if we write as P by A
287
00:34:18,940 --> 00:34:27,190
then we have deltas equals to integral 0 to
L P by A
288
00:34:27,190 --> 00:34:39,750
E dx. Now if P and cross sectional area are
different then we get deformation different
289
00:34:39,750 --> 00:34:44,639
at
different points. But if the axial load P
290
00:34:44,639 --> 00:34:48,530
and the cross sectional area A they remains
same,
291
00:34:48,530 --> 00:34:56,329
if there is no change then we get the same
expression which is P by AE, if the P by AE
292
00:34:56,329 --> 00:35:01,490
is
constant then integral dx will be L. So delta
293
00:35:01,490 --> 00:35:05,960
PL by AE is constant for the cross sectional
area A.
294
00:35:05,960 --> 00:35:12,180
However, if there is variation in the P, area
A and the length of the bar then we get
295
00:35:12,180 --> 00:35:16,019
different deformations at different points
and correspondingly the strain value will
296
00:35:16,019 --> 00:35:19,549
be
different. That is how we calculate deformation
297
00:35:19,549 --> 00:35:27,240
in the axially loaded bar.
.
298
00:35:27,240 --> 00:35:39,490
Having defined this strain we have looked
in to the concepts of stress in the module
299
00:35:39,490 --> 00:35:44,430
one,
now we defined he quantity called strain,
300
00:35:44,430 --> 00:35:50,779
we have established the relationship between
stress and strain and within elastic limit
301
00:35:50,779 --> 00:35:55,109
we have seen in the axially loaded bar, how
we
302
00:35:55,109 --> 00:36:01,109
can evaluate the deformation if we know the
load and if the strain is uniform and what
303
00:36:01,109 --> 00:36:03,671
is
the relation between deformation and the load
304
00:36:03,671 --> 00:36:06,240
corresponding to the cross sectional area
of
305
00:36:06,240 --> 00:36:13,279
the member and in terms of the modulus of
elasticity of the material and if there is
306
00:36:13,279 --> 00:36:15,650
the
variation of the load, if there is the variation
307
00:36:15,650 --> 00:36:21,650
in the cross sectional area, then it is expected
that the deformation to be different and accordingly
308
00:36:21,650 --> 00:36:31,259
the strain will be different at different
point. Let us once again look at an example
309
00:36:31,259 --> 00:36:36,940
related to evaluation of stresses at different
points.
310
00:36:36,940 --> 00:36:48,220
Now this is the truss member in which, which
is supporting the bill board. Billboard is
311
00:36:48,220 --> 00:36:55,559
supported by two which are stress and cross
sectional area of all the members. Call this
312
00:36:55,559 --> 00:37:01,519
as
A, this point as B this is then C and this
313
00:37:01,519 --> 00:37:08,089
is D and this as E. Now we are interested
to
314
00:37:08,089 --> 00:37:20,880
know the stress in each member when the particular
board is subjected to a load and the
315
00:37:20,880 --> 00:37:26,730
area, they are the forces which are acting
at 3 Kilo Newton here and 6 Kilo Newton here,
316
00:37:26,730 --> 00:37:35,029
.so what we need to do is to evaluate first
the forces in each member and once we
317
00:37:35,029 --> 00:37:43,170
compute the forces in each member, so when
the supports are pin jointed.
318
00:37:43,170 --> 00:37:49,079
We have the axis, axial force in each member
and this actional force at each member area
319
00:37:49,079 --> 00:37:54,279
will give out the stress, that us what we
are interested in calculate the stress in
320
00:37:54,279 --> 00:38:00,249
each
members. Now from this triangular configuration
321
00:38:00,249 --> 00:38:09,420
this length is six meter, this is given as
4m and this is given as 4m. So the distance
322
00:38:09,420 --> 00:38:14,060
BC is also going equals to this divided by
this
323
00:38:14,060 --> 00:38:18,769
is equal to the force divided by A which is
½, so this is 3m.
324
00:38:18,769 --> 00:38:31,359
So, if we call this as theta. Now if we take
the cross section area here, then we have
325
00:38:31,359 --> 00:38:39,560
the
member which is like this. This is theta,
326
00:38:39,560 --> 00:38:42,550
and we have the horizontal force is acting
here,
327
00:38:42,550 --> 00:38:53,279
which is 3 Kilo Newton, now the direction
of the force is this and the direction of
328
00:38:53,279 --> 00:38:57,180
the
force is in this direction. Now if we take
329
00:38:57,180 --> 00:39:00,519
the horizontal component of this force, which
is
330
00:39:00,519 --> 00:39:15,480
FAC, so FAC sin theta is equal to 3 Kilo Newton.
FAC sin theta is equal to 3 Kilo
331
00:39:15,480 --> 00:39:24,339
Newton. Now sin theta over here is equal to
3 by 5, this is 4 and this is 3 hence this
332
00:39:24,339 --> 00:39:28,700
is
going to be 5. Sin theta is equal to 3 by
333
00:39:28,700 --> 00:39:39,380
5. So FAC is equal to 5 Kilo Newton. If we
take
334
00:39:39,380 --> 00:39:46,309
the vertical equilibrium of the forces so
FAC cos theta and if we call this force as
335
00:39:46,309 --> 00:39:52,920
FAB
then FAC cos theta is equal to FAB or FAB
336
00:39:52,920 --> 00:39:57,990
cos theta is equal to minus FAB is equal to
0,
337
00:39:57,990 --> 00:40:11,329
that is the vertical forces in the equilibrium.
So FAB is equal to FAC cos theta. And value
338
00:40:11,329 --> 00:40:21,420
of cos theta is 4 by 5 and FAC being 5, so
the value of FAB is 4 Kilo Newton.
339
00:40:21,420 --> 00:40:28,829
Now, if we look in to the direction of the
forces assumed over here, this is the force
340
00:40:28,829 --> 00:40:33,319
in the
joint. Member is subjected to like this, which
341
00:40:33,319 --> 00:40:38,369
is compressive in nature. So the member
AC the joint force direction in this will
342
00:40:38,369 --> 00:40:49,010
be subjected to a compressive
force so in the
343
00:40:49,010 --> 00:40:56,270
member the force is in this direction and
thereby the member AB is on the tension and
344
00:40:56,270 --> 00:41:03,589
so
this is the tensile force. The member AC and
345
00:41:03,589 --> 00:41:09,660
member AB is subjected to a force of 5 Kilo
Newton and 4 Kilo Newton so these divided
346
00:41:09,660 --> 00:41:18,650
by the cross sectional area, 5 Kilo Newton
divided by 100 mm square will give us the
347
00:41:18,650 --> 00:41:26,069
stress in FAC and correspondingly 4 Kilo
Newton by 100 mm square will give me the stress
348
00:41:26,069 --> 00:41:32,999
in FAB.
Likewise we have to compute the forces in
349
00:41:32,999 --> 00:41:41,900
this bar, in this bar, this bar and this bar.
Supposing if we take or cut the structure
350
00:41:41,900 --> 00:41:52,999
from here and take the free body of the upper
part of it then we get a configuration which
351
00:41:52,999 --> 00:42:07,349
is like this here; we have 3 Kilo Newton,
here
352
00:42:07,349 --> 00:42:23,440
we have 6 Kilo Newton, his is force call FBD,
this is A, this is B, this is C and let us
353
00:42:23,440 --> 00:42:29,980
call
this as D and this point here we call as E.
354
00:42:29,980 --> 00:42:46,019
So this is ABD member in this direction, which
is FCD and here we have force FCE. We
355
00:42:46,019 --> 00:42:53,269
can compute forces if we take the moment of
all the forces with respect to this then,
356
00:42:53,269 --> 00:42:55,730
3
Kilo Newton is going to distribute the forces
357
00:42:55,730 --> 00:43:01,839
three times the moment and FBD will have
the moment this times this distance which
358
00:43:01,839 --> 00:43:06,880
is equal to 3. So 3 Kilo Newton into 4 which
is
359
00:43:06,880 --> 00:43:15,430
in anticlockwise moment, if we take moment
about C is equal to 0, so 3 into 4 is equal
360
00:43:15,430 --> 00:43:22,569
to
FBD into 3. This is going to be in anticlockwise
361
00:43:22,569 --> 00:43:32,950
direction, so thereby FBD is equal to 4
Kilo Newton. And again this is at a joint
362
00:43:32,950 --> 00:43:36,059
at this direction, so the member the force
will be
363
00:43:36,059 --> 00:43:41,470
in this particular direction so this gives
us again a tensile pull, so this is a tensile
364
00:43:41,470 --> 00:43:46,049
force. So
this force divided by the cross sectional
365
00:43:46,049 --> 00:43:50,880
area will give me the stress in the member.
366
00:43:50,880 --> 00:43:59,380
.Now this particular angle we have assumed
as theta and this is also theta and again
367
00:43:59,380 --> 00:44:03,839
we
can take the horizontal and the equilibrium
368
00:44:03,839 --> 00:44:12,890
forces of all the forces and we can compute
the value of FCE and FCD and thereby the value
369
00:44:12,890 --> 00:44:24,099
of FCD we will get as equals 5 Kilo
Newton and this is a tensile pull and the
370
00:44:24,099 --> 00:44:34,850
force FCE we will get as 10 Kilo Newton which
is compressive force. And so we have seen
371
00:44:34,850 --> 00:44:43,210
AB, BC, BD, CD, CE and the member which
is BC.
372
00:44:43,210 --> 00:44:49,309
If we take the equilibrium of this particular
joint and we have 6 Kilo Newton here and 6
373
00:44:49,309 --> 00:44:56,359
Kilo Newton at joint B, we have member AB,
we have member BD and we have member
374
00:44:56,359 --> 00:45:10,029
BC where force will be FBC and value from
this will come as 6 Kilo Newton, which is
375
00:45:10,029 --> 00:45:12,660
a
compressive force.
376
00:45:12,660 --> 00:45:18,799
Once we get force in each of this member,
force divided by cross section area will give
377
00:45:18,799 --> 00:45:26,089
me the stress in each member. That is how
we compute the stresses in the member. Now
378
00:45:26,089 --> 00:45:35,539
by stating that the truss joints are pinned
means that the forces transmitted in each
379
00:45:35,539 --> 00:45:45,040
of
these members are purely axial in nature.
380
00:45:45,040 --> 00:45:46,040
.
381
00:45:46,040 --> 00:45:51,329
Having discussed that particular problem related
to the evaluation of the stress as we had
382
00:45:51,329 --> 00:45:59,779
discussed in the previous module, how to compute
stresses at a point in a stress body,
383
00:45:59,779 --> 00:46:05,630
now in this particular lesson we have seen
how to compute the deformation in a member
384
00:46:05,630 --> 00:46:13,010
for an axially loaded member or there is variation
in the load of the cross sectional area,
385
00:46:13,010 --> 00:46:17,799
what is going to be deformation in the member
along with the length of the member.
386
00:46:17,799 --> 00:46:26,730
Now let us look into some examples; how do
you compute these deformations if the
387
00:46:26,730 --> 00:46:31,950
forces in the member is known, we know the
cross sectional area of the member and we
388
00:46:31,950 --> 00:46:37,880
know the material property which is of the
elasticity of the material. Now in this
389
00:46:37,880 --> 00:46:47,200
example, we have steel rod, which is having
cross sectional area of three hundred
390
00:46:47,200 --> 00:46:54,739
millimeter square and the length of the bar
is 150m, is suspended vertically from one
391
00:46:54,739 --> 00:46:55,749
end.
392
00:46:55,749 --> 00:47:04,359
.The rod supports the tensile load of 20 Kilo
Newton at the free end we will have to find
393
00:47:04,359 --> 00:47:10,630
out the elongation of the rod. Or first we
will have to find out the deformation. The
394
00:47:10,630 --> 00:47:15,029
value
of E is given as 2 into 10 to the power 5
395
00:47:15,029 --> 00:47:24,519
MPa. So it us like you have a bar of length
150m
396
00:47:24,519 --> 00:47:32,999
and it is supported at one end, it is hung
from the top and at this free end this bar
397
00:47:32,999 --> 00:47:40,740
is
subjected to a load of 20 Kilo Newton, the
398
00:47:40,740 --> 00:47:50,519
length is 150m.
So we are interested to compute that what
399
00:47:50,519 --> 00:47:57,519
is going to be the elongation of the bar because
of the axial pull? Bar is going to be stretched
400
00:47:57,519 --> 00:48:04,460
or it is going to deform, so we are interested
to know how much this bar under go deformation
401
00:48:04,460 --> 00:48:08,259
because of the axial pull which is
acting, one end is fixed and the other end
402
00:48:08,259 --> 00:48:20,279
is pulled. Now here as we have seen
deformation delta is equal to PL by AE where
403
00:48:20,279 --> 00:48:24,259
P is the axial pull acting in the bar, L is
the
404
00:48:24,259 --> 00:48:29,279
length of the bar, A is the cross sectional
area of the member, which is uniform here
405
00:48:29,279 --> 00:48:33,269
and
E is the modulus of elasticity of the material.
406
00:48:33,269 --> 00:48:46,140
So this is equal to P is given as 20 Kilo
Newton, so 20 into 10 cube Newton, L is 150m,
407
00:48:46,140 --> 00:48:52,349
so 150 into 10 cube so much of
millimeter divided by cross sectional area
408
00:48:52,349 --> 00:49:01,930
A which is 300 mm square into E which is 2
into 10 to the power 5 which is N by mm square
409
00:49:01,930 --> 00:49:13,229
MPa.
So this 10 cube, 10 cube cancels with 10 to
410
00:49:13,229 --> 00:49:22,670
the power 5 and 0, 6. These two 0s cancel.
This 2 and this 2 gets canceled. So we have
411
00:49:22,670 --> 00:49:31,650
15 is equal to 3 into 15 is equal to 50 mm.
So
412
00:49:31,650 --> 00:49:39,460
the amount of elongation, this particular
bar will be under going because of this load.
413
00:49:39,460 --> 00:49:40,460
.
414
00:49:40,460 --> 00:49:48,749
Now let us look into another example problem,
this is interesting, because here we have
415
00:49:48,749 --> 00:49:54,269
a
bar, a plastic bar is in the elastic limit,
416
00:49:54,269 --> 00:49:59,660
and in this we have variable cross section.
The bar
417
00:49:59,660 --> 00:50:07,849
cross section is not uniform as in the previous
case. Here for this particular stretch of
418
00:50:07,849 --> 00:50:12,839
the
bar, here one cross sectional area, let us
419
00:50:12,839 --> 00:50:16,799
call this as A and this as B and this is C
and this
420
00:50:16,799 --> 00:50:25,470
is D, here part AB, the segment AB is having
area of 1000m square, segment BC is
421
00:50:25,470 --> 00:50:35,680
having 1000m square and segment CD is of 1000m
square and this particular bar is
422
00:50:35,680 --> 00:50:42,299
.subjected to pull here of 200 Kilo Newton,
here it is 400 Kilo Newton, here it is 100
423
00:50:42,299 --> 00:50:46,569
Kilo
Newton and here it is 100 Kilo Newton.
424
00:50:46,569 --> 00:50:55,049
Now for such kind of system, we need to look
into whether the whole bar is under the
425
00:50:55,049 --> 00:51:04,499
action of equilibrium of the forces or not.
We look into a bar which is subjected to the
426
00:51:04,499 --> 00:51:10,979
action of axial force of 400 Kilo Newton in
the positive x-direction. In the negative
427
00:51:10,979 --> 00:51:19,059
xdirection which is 100 plus 100 plus 200
is equal to 400 so this is in equilibrium.
428
00:51:19,059 --> 00:51:22,619
Now if
we take the free body of different part of
429
00:51:22,619 --> 00:51:25,749
this particular bar, then we can find out
that
430
00:51:25,749 --> 00:51:30,630
what is the amount of forces that is each
member is subjected to.
431
00:51:30,630 --> 00:51:38,059
Now let us take a section somewhere here,
let us call section one-one and if you take
432
00:51:38,059 --> 00:51:44,150
free
body of this, this particular segment is subjected
433
00:51:44,150 --> 00:51:52,749
to pull here, 200 Kilo Newton, and as
we have seen the free body in our module one,
434
00:51:52,749 --> 00:52:00,230
the opposing force will also be 200 Kilo
Newton. This particular member is subjected
435
00:52:00,230 --> 00:52:08,700
to pull of 200 Kilo Newton. If we take
section here, section two-two and if you take
436
00:52:08,700 --> 00:52:22,769
the free body part of it, then here we have
force 200 Kilo Newton, here we have a force
437
00:52:22,769 --> 00:52:28,499
which is 400 Kilo Newton acting in the
other direction, so the resulting force 200
438
00:52:28,499 --> 00:52:32,240
Kilo Newton acting in the opposite direction,
if
439
00:52:32,240 --> 00:52:43,499
we balance this has to be 200 Kilo Newton.
So this is being pulled which has compressive
440
00:52:43,499 --> 00:52:53,450
force of two hundred Kilo Newton and if
we take the free body of this part, then we
441
00:52:53,450 --> 00:53:10,890
have the free body part is like this, this
is 200,
442
00:53:10,890 --> 00:53:18,920
this is 400 and this is 100, so 100 plus 200
on this side and 400 on this side, so it has
443
00:53:18,920 --> 00:53:21,900
to
be balanced by hundred on this side. So this
444
00:53:21,900 --> 00:53:28,730
particular member is subjected to 100 here
and 100 here which is under compression. So
445
00:53:28,730 --> 00:53:31,089
the first part, this part is under tensile
pull
446
00:53:31,089 --> 00:53:35,829
of 100 Kilo Newton and central part is under
compression of two hundred Kilo Newton,
447
00:53:35,829 --> 00:53:40,890
the third part is in the compression of 100
Kilo Newton.
448
00:53:40,890 --> 00:53:53,759
Now for these if we compute, if we calculate
the deformation for the three segments we
449
00:53:53,759 --> 00:54:02,349
can find out the total delta as the sum of
summation of P L by A E. As in each of these
450
00:54:02,349 --> 00:54:09,119
three segments the cross section area over
these stretch, over this cross sectional area
451
00:54:09,119 --> 00:54:11,030
of
particular section remains same. For this
452
00:54:11,030 --> 00:54:15,849
particular cross sectional area the cross
sectional
453
00:54:15,849 --> 00:54:27,349
area is same so we can write this equation
as, for the segment AB as the P 1 L 1 by A
454
00:54:27,349 --> 00:54:38,920
1 E
plus P 2 L 2 by A 2 E plus P 3 L 3 by A 3
455
00:54:38,920 --> 00:54:45,400
E for the three segments. Now for the second
segment this P 2 is negative, this is compressive,
456
00:54:45,400 --> 00:54:48,280
if we call P 1 as tensile as positive, for
the
457
00:54:48,280 --> 00:54:57,769
third segment P 3 is compressive again negative.
Now if we substitute the values as we know
458
00:54:57,769 --> 00:55:04,140
P, we know the value of L 1 , we know the
value of L 1 , correspondingly P 2 , L 2 and
459
00:55:04,140 --> 00:55:09,299
A 2 and we know the value of P 3 , L 3 and
A 3 and
460
00:55:09,299 --> 00:55:22,729
if we calculate that we can compute the vale
of the deformation and this delta is equal
461
00:55:22,729 --> 00:55:32,069
to
200 and 10 cube into 2000 by 1000(2 into 10
462
00:55:32,069 --> 00:55:50,290
to the power 5), second one will be minus
is equal to 100 minus this 100 into 10 cube
463
00:55:50,290 --> 00:55:59,400
into 1000 by 1000 (10 to the power 5) is equal
to 1 mm.
464
00:55:59,400 --> 00:56:00,400
..
465
00:56:00,400 --> 00:56:12,099
Here is another problem which is aluminium
bar with cross sectional area of 160 mm
466
00:56:12,099 --> 00:56:19,039
square carries the axial loads, you have to
compute the values of deformation to compute.
467
00:56:19,039 --> 00:56:20,039
.
468
00:56:20,039 --> 00:56:26,279
Now, to summarize; this particular lesson
included concept of strain at a point and
469
00:56:26,279 --> 00:56:30,359
axial
and normal strain, stress-strain relationship
470
00:56:30,359 --> 00:56:35,719
and the relevance of different point in the
stress strain curve, example to demonstrate
471
00:56:35,719 --> 00:56:45,089
how to evaluate the strain and
the
472
00:56:45,089 --> 00:56:46,469
deformation stress body.
473
00:56:46,469 --> 00:56:47,469
..
474
00:56:47,469 --> 00:56:50,729
Here are some questions:
What is meant by elastic limit?
475
00:56:50,729 --> 00:56:54,400
What is the difference between nominal stress
and true stress?
476
00:56:54,400 --> 00:58:47,819
How will you evaluate strain in a bar with
gradually varying cross section?
477
00:58:47,819 --> 00:58:47,819
.