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Welcome to the 3rd lesson on module 2 on analysis
of strain.
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.
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.This particular lesson we have designated
as analysis of strain 3. In the last lesson,
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we have
discussed certain aspects of strain. In this
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particular lesson we will be discussing some
more
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aspects of strain.
.
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It is expected that after this particular
lesson is completed, one will be able to understand
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the
concept of normal strain in a three dimensional
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stress body and thereby we can go for the
generalization of Hooke’s law. One should
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be able to understand the concept of determinate
and
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indeterminate systems, also one should be
able to understand the concept of compatibility
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which
will eventually arise out of this indeterminate
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system which we are going to discuss
subsequently.
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..
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This particular lesson includes the recapitulation
of previous lesson which we will be doing
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through the questions which we posed last
time. We will be discussing those questions
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and
eventually it will give you the recapitulation
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of the previous lesson, evaluation of axial
strain in a
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three dimensional stress body, concept of
determinate and indeterminate systems and
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thereby the
compatibility criteria, how it will arise
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out from that particular system and finally
the evaluation
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of forces for indeterminate systems.
.
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Here are some questions:
The first question is: What is Poisson’s
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ratio and what is the range of its values?
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..
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Now, if we remember, last time we discussed
that if a body which is subjected to axial
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pull (is
held up at this end and is being pulled by
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axial load p) eventually, this bar extends
or elongates
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and this elongation we had indicated by delta
and thereby there is contraction on the other
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sides.
The French scientist Poisson demonstrated
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that the ratio of this strain which is the
transverse
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strain to the axial strain is constant for
stresses within the proportional limit and
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this particular
constant is designated as Poisson’s ratio.
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So far as the definition of the Poisson ratio
is concerned, the ratio of the lateral strain
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to the
longitudinal strain and if you remember we
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had designated this Μ as the transverse strain
to the
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longitudinal strain and remember that we had
put this particular sign as negative because
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where
this is being pulled by an axially tensile
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pull, then there is contraction on other sides
and this
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negative sign indicates that there is contraction.
Similarly, if this particular bar is subjected
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to an
axial compressive force where the length in
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the axial direction is going to shorten thereby
the
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other two directions is going to increase
and eventually that shortening of the axial
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length is
being designated as negative to the expansion
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of the other two sides and that is why this
negative
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sign comes in.
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..
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We said that what is the range of its values;
now is what is its ratio for different material,
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where
there is very small range of values, generally
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the values vary between 0.25 to 0.35 for some
limiting materials, we get values in the range
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of around 0.1 or for some material we get
in the
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range of 0.5 which is that of rubber. And
in fact 0.5 is the limiting value at the other
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side in the
sense that the range of the values in general
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we can say between 0.1 to 0.5 and most of
the
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material lies between 0.25 to 0.35.5 is the
maximum value of the Poisson’s ratio.
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.
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What is shear modulus and what is its unit?
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..
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The shear strain in an elastic body is proportional
to the applied shear stress. Now this is the
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body which is acted on by
this is subjected to the surface force and
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there by the shear stress tau
equal to the applied load by the cross-sectional
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area is given as the shear stress and within
the
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limit of the proportionality, the shearing
strain is proportional to the shearing strain
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which we
have designated as gamma. So tau is proportional
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to gamma and removing this proportionality
sign, we say tau is equal to G into gamma
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where G is termed as shear modulus this is
called as
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shear modulus and this has the unit similar
to the unit of modulus of elasticity. Now
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which is in
Pa or in N by m square or MPa which is 10
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to the power 6 into N by m square or GPa,
thus the
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unit of the shear module G.
.
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..
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The 3rd question is; how you will evaluate
stain in a three dimensional stress body?
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Let us assume that these are the three directions
this is x, this is y and this is z and
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correspondingly as we have defined earlier
the stress components sigma x , sigma y and
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sigma z .
Now as we have seen earlier that when a bar
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is subjected to axial pull, the stress at
any cross
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section is P by a the cross sectional area
which we have designated as normal stress
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sigma x and
correspondingly the strain in the axial direction
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is epsilon x and as per Hooke’s law epsilon
x is
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equal to sigma x by e. Also, we have noted
from the concept of Poisson’s ratio, that
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if we have
axial load acting in the x direction only
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and thereby there is strain in the x direction
as epsilon x
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the strain in the other two direction y and
z direction epsilon y is equal to epsilon
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z is equal to -Μ
into epsilon x is as per the definition of
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Poisson’s ratio.
Also, if we consider that the body subjected
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to the axial load in the y direction instead
of in the x
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direction, that means we have a bar which
is subjected to load in the y direction it
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is p there by
the stress again at any cross-section is p
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by a and let us call that stress as sigma
y which is p is
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acting in the y direction so this is p by
a. Let us call this as p y this is p x . Now
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because of this
application of the load, there is strain in
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the y direction which we designate as epsilon
y and that
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is going to be equal to according to the Hooke’s
law is sigma y by e. Now due to Poisson’s
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effect,
then the strain in the x and z direction will
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be epsilon x is equal to epsilon z is equal
to minus Μ
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epsilon y and if we substitute the values
of epsilon x and epsilon y in this corresponding
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places this
is Μ sigma x by e and this is minus Μ sigma
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y by e.
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..
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Likewise, if we assume that the body is acted
on by force in the z direction p then the
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stress at
any cross section, we call that as sigma z
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is equal to p z by a. Also the strain corresponding
to that
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in the z direction is epsilon z is equal to
sigma z by e epsilon z is equal to sigma z
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by e. As we have
seen in the previous cases that when it is
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acted on by the load p either in the x direction
or in the
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y direction, we could compute the strains
in the other two directions in the present
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case also
when it is subjected to axial pull in the
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z direction and is the strain which is acting
in the z
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direction is epsilon z the corresponding strain
in the epsilon in the x and y directions are
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epsilon x
and epsilon y is equal to minus Μ into epsilon
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z . So in this particular case epsilon x is
equal to
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epsilon y is is equal to minus Μ into epsilon
z which is Μ into sigma z by e.
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Hence, if we combine all three cases together,
if we propose them since we are considering
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the
body with in the elastic limit. Hence if this
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proposition is valid we can combine them together
with the individual results. If we join them
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together, we will get the same effect when
all these
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are acting simultaneously so in that particular
case when all the three are acting
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we can write
then the strain in the x direction epsilon
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x is equal to sigma x by e because of the
load which is
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acting in directly that and the strain which
we are getting in the x direction because
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of y and z
direction is equal to Μ into sigma y by e
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minus Μ sigma z by e or this is equal to
sigma x by e
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minus Μsigma y plus sigma z by e. Likewise,
the strain in the y direction epsilon y can
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be written
as sigma y by e minus Μ into sigma x plus
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sigma z by e
and epsilon z is equal to sigma z by e minus
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Μ into sigma x plus sigma y by e.
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..
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.
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Now if we reduce this three dimensional form
to a two dimensional one giving the x and
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y plane,
then the stresses which will be acting sigma
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x and sigma y and eventually sigma z is equal
to 0. In
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that particular case, we will get epsilon
x is equal to sigma x by e minus Μ sigma
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y by e and
epsilon y is equal to sigma y by e minus Μ
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sigma x by e. This results we have seen it
earlier so this
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is in fact in a generalized form of Hooke’s
law which is applicable in case of a three
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dimensional
body in three direction epsilon x epsilon
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y epsilon z . The strain in the three directions
can be
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represented in terms of the stresses sigma
x sigma y and sigma z and in terms of the
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modulus of
velocity e.
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..
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These are the aspects we discussed.
.
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Now let us look into the important aspects
which we do come across in many problems related
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to
different fields which we designate as statically
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indeterminate problems. Now when we talk
about statically indeterminate problem naturally
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we will first know what we really mean by
a
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determinate system we have seen earlier that
we have three states of equilibrium equations
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in
statics there is summation of horizontal forces
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are 0 summation vertical forces are 0, summation
moments are 0. These are the equations of
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equilibrium. If we consider any system from
which
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we can evaluate the reactive forces by using
these equilibrium equations we call those
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systems as
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.statically determinate you see if equilibrium
equations are adequate this is important that
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equilibrium equations are adequate to evaluate
the reactive forces then the system is called
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as
determinate. Let us look into some example.
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.
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00:17:09,549 --> 00:17:22,819
We have a rigid bar which is pinned at this
particular end and is held up by a rod and
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is subjected
to a vertical load p, now if we draw the free
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00:17:29,860 --> 00:17:37,350
body diagram of this particular configuration
removing this supports the free body diagram
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00:17:37,350 --> 00:17:53,380
it will look like this. This bar being pin
connected
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is subjected to axial loads only hence there
is axial pull P b and these are the reactive
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forces R and
H and this the external load P. This is the
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free body diagram of this particular system.
Now in this particular free body diagram the
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unknown forces are the reactive forces R H
and the
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axial pull in the member P b . So these three
parameters can be evaluated from the equations
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of
static that means if we take summation horizontal
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forces is equal to 0 summation vertical force
is
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equal to 0, summation moment is equal to 0
and if we apply these three equations for
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this; we
can evaluate R H and P radically from these
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00:18:46,169 --> 00:18:51,499
equations. Hence this particular system is
statically
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determinate. Now let us change the configurations
of this particular system later.
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..
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This is connected by two wires of this particular
form. Now these rods will undergo extension
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because they are subjected to axial pull when
this particular load will be acting.
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.
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Now if we try to plot the free body diagram
of this particular system, this is the rigid
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00:20:03,549 --> 00:20:11,649
bar pinned
over here, this is subjected to the reactive
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force vertical and horizontal as before here
we have the
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00:20:15,809 --> 00:20:29,159
applied load P, now two rods which are supporting
this bar will have axial pull in the rods.
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00:20:29,159 --> 00:20:32,909
So
there are forces which are reactive forces
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00:20:32,909 --> 00:20:39,580
R, H and let us call this as P 1 and P 2 which
are under
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00:20:39,580 --> 00:20:48,860
the action of the axial load P or the transverse
load P. Now in this particular case, if we
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00:20:48,860 --> 00:20:55,471
look into
we have 4 unknown reactive forces R, H, P
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00:20:55,471 --> 00:21:00,590
1 and P 2 and since we have only three equations
of
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00:21:00,590 --> 00:21:09,350
.statics this particular system cannot be
solved using these equations of equilibrium
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00:21:09,350 --> 00:21:15,749
alone. Hence
there are systems like this where you cannot
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00:21:15,749 --> 00:21:20,369
solve or evaluate the unknown forces based
on the
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00:21:20,369 --> 00:21:27,499
equations of equilibrium alone or equations
of statics alone. Now you need an additional
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00:21:27,499 --> 00:21:41,880
equation here over and above these three equations
to solve these unknown reactive forces.
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00:21:41,880 --> 00:21:42,880
.
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00:21:42,880 --> 00:21:49,879
This is what is indicated here; if the number
of unknown forces in a system exceeds the
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00:21:49,879 --> 00:21:55,269
number
of equilibrium equations then the system is
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00:21:55,269 --> 00:21:59,990
called statically indeterminate system. The
system is
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00:21:59,990 --> 00:22:08,429
indeterminate because we cannot solve the
unknown forces using the equations of equilibrium
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00:22:08,429 --> 00:22:15,070
but then it does not mean we cannot evaluate
the unknown forces for those indeterminate
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00:22:15,070 --> 00:22:23,809
systems so for the solution of this indeterminate
systems, will have to find out ways by which
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00:22:23,809 --> 00:22:27,710
we
can generate additional equations from which
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00:22:27,710 --> 00:22:35,850
we can solve this particular unknown reactive
forces.
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00:22:35,850 --> 00:22:36,850
..
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00:22:36,850 --> 00:22:45,710
So the indeterminate problems are the ones
in which a statically indeterminate problem
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00:22:45,710 --> 00:22:49,679
always
has geometric restrictions imposed on its
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00:22:49,679 --> 00:22:56,169
deformation. In fact from this, the compatibility
criteria
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00:22:56,169 --> 00:23:05,120
comes in, the indeterminate system always
gives rise to geometric restrictions and these
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00:23:05,120 --> 00:23:15,029
geometric restrictions when expressed mathematically
gives rise to the equation of compatibility.
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00:23:15,029 --> 00:23:21,400
We have seen that we have three equations
of equilibrium which has statical equations
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00:23:21,400 --> 00:23:26,100
of
equilibrium, over and above if we write down
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00:23:26,100 --> 00:23:30,070
the equations of compatibility which do arise
from
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00:23:30,070 --> 00:23:37,240
this geometric restrictions for the deformation,
this equation of compatibility along with
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00:23:37,240 --> 00:23:40,940
the
equations of equilibrium will lead us to the
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00:23:40,940 --> 00:23:44,539
system of equations from which you can solve
the
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00:23:44,539 --> 00:23:49,040
unknown reactive forces in the indeterminate
system.
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00:23:49,040 --> 00:23:55,309
Now you can distinguish between a determinate
system and indeterminate system. Once again
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00:23:55,309 --> 00:23:58,820
a
determinate system is the one for which the
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00:23:58,820 --> 00:24:02,309
unknown reactive forces can be evaluated based
on
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00:24:02,309 --> 00:24:10,249
the equations of equilibrium. And for indeterminate
systems we cannot evaluate the unknown
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00:24:10,249 --> 00:24:16,629
reactive forces based on the equations of
equilibrium alone and for that we need to
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00:24:16,629 --> 00:24:19,539
have
additional equations depending on the number
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00:24:19,539 --> 00:24:27,299
of unknown reactive forces you have and those
additional equations do generate from the
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00:24:27,299 --> 00:24:30,970
equations of compatibility. And the equations
of
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00:24:30,970 --> 00:24:38,909
compatibility can be written down in terms
of the geometrical constraints of the restrictions
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00:24:38,909 --> 00:24:47,179
which we impose in terms of deformation.
So, compatibility conditions provide additional
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00:24:47,179 --> 00:24:55,750
equations for evaluating unknown forces and
thereafter we can adopt this Hooke’s law
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00:24:55,750 --> 00:24:59,830
to express the deformation in terms of forces.
Because
223
00:24:59,830 --> 00:25:07,620
we first write down the compatibility criteria
based on the deformation and then we adopt
224
00:25:07,620 --> 00:25:14,129
Hooke’s law to relate the deformation to
the forces and finally we get equations from
225
00:25:14,129 --> 00:25:18,600
these
compatibility criteria which help us to solve
226
00:25:18,600 --> 00:25:23,989
the unknown reactive forces.
227
00:25:23,989 --> 00:25:24,989
..
228
00:25:24,989 --> 00:25:38,809
Now let us
look into the problems where such indeterminacy
229
00:25:38,809 --> 00:25:49,820
comes in but before that let us look
into some example problems which we already
230
00:25:49,820 --> 00:26:10,460
discussed. Determine the elongation of
231
00:26:10,460 --> 00:26:15,970
the
tapered cylindrical aluminum bar caused by
232
00:26:15,970 --> 00:26:20,169
the 30 kilo Newton axial load and the value
of e is
233
00:26:20,169 --> 00:26:26,850
given over here and in this particular problem,
it is given that the diameter of this bar
234
00:26:26,850 --> 00:26:30,769
at this
place is 20 mm and the diameter at this end
235
00:26:30,769 --> 00:26:37,249
is 30 mm and over the length of 400 mm it
is varying
236
00:26:37,249 --> 00:26:43,190
gradually.
Now our job is to find out the deformation
237
00:26:43,190 --> 00:26:48,049
to find out the elongation of the gradually
varying
238
00:26:48,049 --> 00:26:55,840
member. And if you remember we have discussed
how to compute the elongation in such
239
00:26:55,840 --> 00:27:04,629
members since the cross section is varying
at every position. Hence the cross sectional
240
00:27:04,629 --> 00:27:09,759
area is not
constant and we cannot compute directly delta
241
00:27:09,759 --> 00:27:15,440
from the expression P l by ae, where a is
no longer
242
00:27:15,440 --> 00:27:22,010
a constant parameter over the length l. Hence
what we need to do is that we got to compute
243
00:27:22,010 --> 00:27:27,320
delta
is equal to ∫over the length of the member
244
00:27:27,320 --> 00:27:33,169
as P d x over the small segment. We say the
area which
245
00:27:33,169 --> 00:27:42,309
we write as a function of x into e and from
this expression we compute the value of delta.
246
00:27:42,309 --> 00:27:52,450
Now if we take a section which is at a distance
of x from the left end, then the diameter
247
00:27:52,450 --> 00:28:02,850
at this
particular section d is equal to 20 plus 30
248
00:28:02,850 --> 00:28:10,230
so 30 minus 20 is the value at the end. So
at this
249
00:28:10,230 --> 00:28:19,900
particular point it will be in the ratio of
x to 400 so this is 30 minus 20 into x by
250
00:28:19,900 --> 00:28:28,179
40 and 30 minus
20 by 2 and twice of that since on this side
251
00:28:28,179 --> 00:28:30,159
we have an extra and on this side we have
an extra
252
00:28:30,159 --> 00:28:39,350
about 20. This is 2, so 20 plus 30 minus 20
into x by 400. This is nothing but is equal
253
00:28:39,350 --> 00:28:45,669
to 20 plus
x by 40. This is the diameter at this cross
254
00:28:45,669 --> 00:28:49,840
section which is at a distance of x from the
left end. Let
255
00:28:49,840 --> 00:28:58,549
us compute delta for this particular case.
256
00:28:58,549 --> 00:28:59,549
..
257
00:28:59,549 --> 00:29:06,289
So we have the bar which is tapered and we
have just seen that at a cross section which
258
00:29:06,289 --> 00:29:11,989
is at a
distance of x, the diameter here d is equal
259
00:29:11,989 --> 00:29:21,809
to 20 plus x by 40 mm hence the cross sectional
area a
260
00:29:21,809 --> 00:29:39,129
is equal to pi by 4 into 20 plus x by 40 square
and delta is equal to ∫02400 the load is
261
00:29:39,129 --> 00:29:44,520
30 kilo
Newton. So 30 into 10 cube so much of Newton
262
00:29:44,520 --> 00:29:55,409
dx by a which is pi 20 plus x by 40 square
into
263
00:29:55,409 --> 00:30:07,999
4pi by 2 is equal to 4 into e which is 72
Gpa so 70 into 10 to the power 9 so much of
264
00:30:07,999 --> 00:30:16,299
mm and this
if we compute this will give rise to is equal
265
00:30:16,299 --> 00:30:32,809
to 120 by 72pi(10 to the power minus 6) ∫0
to 400.
266
00:30:32,809 --> 00:30:50,710
This is dx by 20 plus x by 40 square. If we
substitute 20 plus x by 40 as I said, we get
267
00:30:50,710 --> 00:31:00,570
a dz as 1
by 40 dx and when x is 0, z is equal to 20
268
00:31:00,570 --> 00:31:04,289
and when x is equal to 400 z is equal to 30
so it varies
269
00:31:04,289 --> 00:31:18,080
from 20 to 30. This ∫gets transformed to
this particular part as ∫20 to 30 dz 40
270
00:31:18,080 --> 00:31:30,220
into dz by z square.
This is nothing but minus 1 by z∫ which
271
00:31:30,220 --> 00:31:39,929
is 20 to 30 eventually. Then this comes as
is equal to
272
00:31:39,929 --> 00:31:40,929
delta.
273
00:31:40,929 --> 00:31:41,929
..
274
00:31:41,929 --> 00:31:58,259
If we compute the whole thing it is 4800(10
to the power minus 6) by 72pi (1 by 20 minus
275
00:31:58,259 --> 00:32:12,980
1 by
30 is equal to 3.53(10 to the power minus
276
00:32:12,980 --> 00:32:20,919
7). This is the value of the delta which we
get from the
277
00:32:20,919 --> 00:32:27,509
competition for the value of the delta for
the varying diameter.
278
00:32:27,509 --> 00:32:28,509
.
279
00:32:28,509 --> 00:32:39,129
So, for a member which is tapered and if we
know the diameters at its two ends then we
280
00:32:39,129 --> 00:32:53,519
can
compute the amount of elongation the member
281
00:32:53,519 --> 00:32:54,570
will be undergoing.
282
00:32:54,570 --> 00:32:55,570
..
283
00:32:55,570 --> 00:33:12,190
The next example is in terms of the indeterminate
problem. Now here a bearing plate connected
284
00:33:12,190 --> 00:33:18,870
with a copper rod is placed inside an aluminum
sleeve as shown in figure so here this is
285
00:33:18,870 --> 00:33:26,869
the
bearing plate and this bearing plate is connected
286
00:33:26,869 --> 00:33:32,109
with a copper rod going inside so this is
the
287
00:33:32,109 --> 00:33:46,399
copper rod and this is the aluminum sleeve.
This is placed inside an aluminum sleeve as
288
00:33:46,399 --> 00:33:50,090
shown
in figure the copper rod is 0.05 mm longer
289
00:33:50,090 --> 00:34:03,019
than the sleeve so this gap is 0.05 mm so
the copper
290
00:34:03,019 --> 00:34:11,200
rod is longer than this sleeve by 0.05 mm.
Now we will have to find the maximum safe
291
00:34:11,200 --> 00:34:17,649
load that can be applied to the bearing plate.
On this
292
00:34:17,649 --> 00:34:24,660
plate the maximum load P that can be applied
has to be computed. The cross-sectional areas
293
00:34:24,660 --> 00:34:27,310
of
copper rod and aluminum sleeve are 1000 mm
294
00:34:27,310 --> 00:34:34,540
square and 2000 mm square e values of copper
and aluminum are 1.4(10 to the power 5) MPa
295
00:34:34,540 --> 00:34:37,859
and 0.7(10 to the power 5) MPa. Allowable
stress
296
00:34:37,859 --> 00:34:44,820
in copper is 100 MPa and that in aluminum
is 50 MPa. The maximum permissible stress
297
00:34:44,820 --> 00:34:48,030
in
aluminum and copper are given. The cross-sectional
298
00:34:48,030 --> 00:34:59,390
area for the copper and the aluminum sleeve
are given. Now our job is to compute that
299
00:34:59,390 --> 00:35:09,660
what will be the value of P which can be safely
transferred on top of this bearing plate.
300
00:35:09,660 --> 00:35:14,490
Let us look into the free body diagram of
this particular
301
00:35:14,490 --> 00:35:15,490
system.
302
00:35:15,490 --> 00:35:16,490
..
303
00:35:16,490 --> 00:35:27,160
If we cut off the sleeve at some distance
from the top, then this is the bearing plate
304
00:35:27,160 --> 00:35:35,300
the copper
rod is attached with this. And this is acted
305
00:35:35,300 --> 00:35:46,740
on by load p now when the rod will get compressed
and cover up the distance 0.05 mm. Eventually
306
00:35:46,740 --> 00:36:00,539
it will touch the sleeve so this bearing plate
along
307
00:36:00,539 --> 00:36:11,140
with the copper rod it has compressed cover
up covered up that 0.05 mm gap between the
308
00:36:11,140 --> 00:36:15,190
sleeve
and the copper rod. That much of compression
309
00:36:15,190 --> 00:36:21,099
as or the deformation has undergone into the
copper plate and eventually the bearing plate
310
00:36:21,099 --> 00:36:27,750
is going to touch on the sleeve and when the
bearing plate will be touching the sleeve,
311
00:36:27,750 --> 00:36:31,380
then the load which is acting on the bearing
plate will
312
00:36:31,380 --> 00:36:37,849
be transmitted to the sleeve.
So at this free body diagram, if we now transfer
313
00:36:37,849 --> 00:36:41,809
the reactive forces, the reactive forces go
as
314
00:36:41,809 --> 00:36:51,460
acting on this. They are the force in this
copper bar and the forces in the combined
315
00:36:51,460 --> 00:36:56,010
form in the
sleeve. These two together, let us call that
316
00:36:56,010 --> 00:37:02,559
the load in the aluminum. We have these two
forces P c
317
00:37:02,559 --> 00:37:12,859
and P aluminum which are being generated because
of the externally applied load P. So if we
318
00:37:12,859 --> 00:37:16,440
write
down the equilibrium equation so equilibrium
319
00:37:16,440 --> 00:37:27,470
of this particular system is that P is equal
to P c
320
00:37:27,470 --> 00:37:39,130
plus P aluminum . This is equation 1.
In this particular system we cannot have any
321
00:37:39,130 --> 00:37:43,119
other equilibrium equation. So we have only
one
322
00:37:43,119 --> 00:37:52,430
equation of equilibrium and there are two
unknown forces and they are P c and P a the
323
00:37:52,430 --> 00:37:55,369
two
unknown forces and one single equation. Hence
324
00:37:55,369 --> 00:38:02,760
we cannot solve this particular system unless,
we have an additional equation and this additional
325
00:38:02,760 --> 00:38:09,559
equation can be generated from the
comparative criteria in terms of the deformation
326
00:38:09,559 --> 00:38:15,230
and then we can solve for these two unknown
forces. So now let us look into what is going
327
00:38:15,230 --> 00:38:23,390
to be the comparative equation in this particular
case. Let me look into another free body diagram
328
00:38:23,390 --> 00:38:32,410
of the top part of the system.
This is the bearing plate along with a copper
329
00:38:32,410 --> 00:38:49,099
rod and the sleeve is in this location and
this is the
330
00:38:49,099 --> 00:39:00,079
gap since copper rod is 0.05 mm greater than
the sleeve. This is an excess rated from and
331
00:39:00,079 --> 00:39:03,470
let say
after deformation, the bearing comes in this
332
00:39:03,470 --> 00:39:09,440
particular position hence the deformation
that copper
333
00:39:09,440 --> 00:39:18,240
rod undergoes is this much this we call as
delta c and in the process when copper rod
334
00:39:18,240 --> 00:39:19,240
deform to
335
00:39:19,240 --> 00:39:25,220
.this extend and if this is the position where
the bearing plate has come, then the deformation
336
00:39:25,220 --> 00:39:33,410
which is existing in the or which has occurred
in the aluminum is this much this we call
337
00:39:33,410 --> 00:39:36,440
as
delta aluminum .
338
00:39:36,440 --> 00:39:44,940
From this particular configuration, we can
write delta c is equal to delta aluminum plus
339
00:39:44,940 --> 00:39:53,630
0.05 mm. So
this is the criterion which is getting developed
340
00:39:53,630 --> 00:39:56,839
from the compatibility of the two systems
that
341
00:39:56,839 --> 00:40:01,930
initially when the load is applied on this
bearing plate, first the bearing plate along
342
00:40:01,930 --> 00:40:05,560
with the
copper rod has to be deformed to the extent
343
00:40:05,560 --> 00:40:11,290
of 0.05 mm. Then the load gets transferred
on to this
344
00:40:11,290 --> 00:40:18,599
sleeve. Then the sleeve deforms and this is
the final level of the deformation and if
345
00:40:18,599 --> 00:40:21,920
these total
deformation we call as delta c then delta
346
00:40:21,920 --> 00:40:29,619
c consist of two parts; one is this 0.05 mm
gap which has
347
00:40:29,619 --> 00:40:37,089
to be made up by the copper tube at the rod
along with the deformation of this sleeve.
348
00:40:37,089 --> 00:40:40,609
This is the
compatibility equation.
349
00:40:40,609 --> 00:40:48,240
Now if we apply Hooke’s law to this deformation
equation we can get additional equation from
350
00:40:48,240 --> 00:40:59,490
which you can evaluate as P c and P aluminum
. Now delta c , we can write in terms of the
351
00:40:59,490 --> 00:41:04,740
load and
cross sectional area which is PL by ae. Let
352
00:41:04,740 --> 00:41:08,660
us call the P c is the force which acting
in the copper
353
00:41:08,660 --> 00:41:21,970
rod times length of the copper is L c by area
ac into ec is equal to P in the aluminum times
354
00:41:21,970 --> 00:41:30,640
l of
aluminum by a of aluminum into e of aluminum
355
00:41:30,640 --> 00:41:38,529
plus 0.05. Now in this particular case since
the
356
00:41:38,529 --> 00:41:45,440
values of limiting stresses are given, let
us compute the values in terms of stresses
357
00:41:45,440 --> 00:41:50,470
now as we
know P by the cross sectional area is the
358
00:41:50,470 --> 00:41:55,579
stress. This particular part P c by a c we
can write as the
359
00:41:55,579 --> 00:42:04,390
sigma c the normal stress in the copper rod
and P l by a l , we can write as sigma aluminum
360
00:42:04,390 --> 00:42:09,470
which is
the normal stress in the aluminum sleeve.
361
00:42:09,470 --> 00:42:16,589
Hence this compatibility equation in terms
of stresses we can write as sigma c into L
362
00:42:16,589 --> 00:42:25,990
c by e c , this
is equal to sigma aluminum into length of
363
00:42:25,990 --> 00:42:37,190
the aluminum by e of aluminum plus 0.05. Now
if we
364
00:42:37,190 --> 00:42:44,440
substitute the values the value of L c the
length of the copper tube the length of the
365
00:42:44,440 --> 00:42:49,539
aluminum
sleeve is given as 250 mm and since the copper
366
00:42:49,539 --> 00:42:59,130
rod is 0.05 mm longer than the aluminum sleeve,
the length of the copper rod is 250 plus 0.05
367
00:42:59,130 --> 00:43:07,839
so 250.05. sigma c (250.05) is the length
of the
368
00:43:07,839 --> 00:43:18,640
copper rod by e of copper rod which is 1.4(10
to the power 5 ); this is equal to sigma aluminum
369
00:43:18,640 --> 00:43:25,940
into
the length of the aluminum is 250 by e which
370
00:43:25,940 --> 00:43:46,359
0.7 into 10 to the power 5 plus 0.05.
371
00:43:46,359 --> 00:43:47,359
..
372
00:43:47,359 --> 00:44:06,640
Now this if we compute this comes as 0.00179
into sigma c is equal to 0.00357 sigma aluminum
373
00:44:06,640 --> 00:44:18,760
plus
0.05 or sigma c is equal to 1.995 into sigma
374
00:44:18,760 --> 00:44:31,920
aluminum plus 27.93. Now this is the relationship
between the stresses in the copper rod to
375
00:44:31,920 --> 00:44:40,060
the stress in the aluminum rod. Here the limiting
values
376
00:44:40,060 --> 00:44:47,819
of the stresses in copper and aluminum is
given now the maximum allowable stress in
377
00:44:47,819 --> 00:44:56,020
copper rod
is 100 MPa sigma cmax is 100 MPa as sigma
378
00:44:56,020 --> 00:45:02,980
aluminum allowable or max as we call it is
equal to 15
379
00:45:02,980 --> 00:45:15,099
MPa. Hence if we take the value of allowable
stress in aluminum as 50 MPa and if you substitute
380
00:45:15,099 --> 00:45:30,990
it here then we get the value of sigma c is
equal to 1.99 into 50 plus 27.93 eventually
381
00:45:30,990 --> 00:45:49,289
and the
value of sigma c comes as 127.68 MPa. Since
382
00:45:49,289 --> 00:45:53,059
the maximum allowable stress in copper is
100
383
00:45:53,059 --> 00:46:00,010
MPa, we cannot afford to go for stress up
to this level.
384
00:46:00,010 --> 00:46:06,180
We cannot apply the stress in the aluminum
up to 50 MPa, because if we go up to a stress
385
00:46:06,180 --> 00:46:09,890
level
of 50 MPa in aluminum and the stress level
386
00:46:09,890 --> 00:46:13,010
that is expected in the copper rod is going
beyond
387
00:46:13,010 --> 00:46:19,309
the allowable stress of 100 MPa. Hence we
cannot allow the stress in aluminum sleeve
388
00:46:19,309 --> 00:46:23,240
to go up
to its limiting value which is 50 MPa. That
389
00:46:23,240 --> 00:46:26,599
means we will have to limit the stress in
the copper
390
00:46:26,599 --> 00:46:33,829
rod up to 100 MPa and check on how much stress
it can generate in the aluminum sleeve from
391
00:46:33,829 --> 00:46:42,150
which we can evaluate the safe load. Hence
if we go the other way limiting the value
392
00:46:42,150 --> 00:46:47,930
of sigma c
to 100 MPa we need to see how much we can
393
00:46:47,930 --> 00:46:51,170
get as the value of sigma aluminum.
394
00:46:51,170 --> 00:46:52,170
..
395
00:46:52,170 --> 00:47:00,020
So we limit the value of sigma c as is equal
to 100 MPa and from the relationship which
396
00:47:00,020 --> 00:47:08,000
we have
sigma c is equal to 1.995 sigma aluminum plus
397
00:47:08,000 --> 00:47:16,000
27.93. The value of sigma aluminum comes as
100 minus
398
00:47:16,000 --> 00:47:35,500
27.93 by 1.995 and this is is equal to 36.13
MPa. So by allowing the copper rod to go up
399
00:47:35,500 --> 00:47:39,579
to its
limiting value of 100 MPa, the maximum stress
400
00:47:39,579 --> 00:47:42,740
that can be generated in the aluminum sleeve
is
401
00:47:42,740 --> 00:47:51,809
36 MPa which is well within the allowable
limit of the 50 MPa stress. Hence the safe
402
00:47:51,809 --> 00:47:56,279
load that
can be transfer on the bearing without causing
403
00:47:56,279 --> 00:48:02,650
any distress to the member is equal to P c
plus
404
00:48:02,650 --> 00:48:09,289
P aluminum safe load in the copper and the
aluminum copper rod and the aluminum sleeve,
405
00:48:09,289 --> 00:48:13,490
this is
equal to sigma c into area of the copper rod
406
00:48:13,490 --> 00:48:18,210
plus sigma aluminum which is the maximum we
can
407
00:48:18,210 --> 00:48:26,140
allow multiplied by the area of the aluminum.
And this is is equal to 100 multiplied by
408
00:48:26,140 --> 00:48:31,349
the cross
sectional area of the copper which is 1000
409
00:48:31,349 --> 00:48:44,650
mm square plus sigma allowable is 36.13 for
aluminum into 2000 mm square and these eventually
410
00:48:44,650 --> 00:48:55,240
gives 172.26 kilo Newton. So the load
which can be applied on the bearing must be
411
00:48:55,240 --> 00:49:02,099
lesser than or is equal to 172.26 kilo Newton
and
412
00:49:02,099 --> 00:49:11,029
thereby the maximum stress that can be generated
in the copper rod is 100 MPa but the
413
00:49:11,029 --> 00:49:17,339
aluminum sleeve will not go up to 50 MPa.
But it will be well within 50 MPa and thereby
414
00:49:17,339 --> 00:49:24,789
the
whole system will be safe.
415
00:49:24,789 --> 00:49:25,789
..
416
00:49:25,789 --> 00:49:30,380
Let us look into another example of such indeterminate
system.
417
00:49:30,380 --> 00:49:31,380
.
418
00:49:31,380 --> 00:49:45,480
Here AB is a rigid bar, AB is supported by
A pinned at B and by two rods. These bars
419
00:49:45,480 --> 00:49:50,940
is pinned
at B is supported by two rods which are at
420
00:49:50,940 --> 00:49:55,640
a distance of 0.8 meter and 1.8 meter from
B. This rod
421
00:49:55,640 --> 00:50:07,190
is a steel rod of length 1m and this is aluminum
rod of length 2 meter and this bar is subjected
422
00:50:07,190 --> 00:50:12,430
to
a load of 100 kilo Newton at the point A.
423
00:50:12,430 --> 00:50:17,660
Now we will have to find out the stress in
each bar or
424
00:50:17,660 --> 00:50:24,039
each rod and the cross sectional area of those
bars are given steel bar is 500 mm square
425
00:50:24,039 --> 00:50:29,780
and that
of aluminum bar is 300 mm square and the value
426
00:50:29,780 --> 00:50:33,200
of e are given as 2(10 to the power 5) and
as
427
00:50:33,200 --> 00:50:40,589
0.7(10 to the power 7) MPa for the steel and
the aluminum and let us assume that the weight
428
00:50:40,589 --> 00:50:41,589
of
429
00:50:41,589 --> 00:50:54,369
.this bar is negligible. Let us write down
the free body of this system and write down
430
00:50:54,369 --> 00:51:02,160
the
equations of equilibrium and absorb what happens.
431
00:51:02,160 --> 00:51:03,160
.
432
00:51:03,160 --> 00:51:18,250
If we take the free body this is the rigid
bar pinned at this end acted on by two rods
433
00:51:18,250 --> 00:51:26,000
they are steel
and aluminum and at these end we have the
434
00:51:26,000 --> 00:51:30,250
reactive forces for the pin and at these end
the load
435
00:51:30,250 --> 00:51:49,470
100 kilo Newton is acting. Now if we like
to find out the forces let us take the moment
436
00:51:49,470 --> 00:51:54,050
of now
here we have 1, 2, 3, 4 unknown parameters
437
00:51:54,050 --> 00:51:57,279
and we have three equations of equilibrium.
Hence
438
00:51:57,279 --> 00:52:04,980
we cannot solve it means statically determinate
one or using the equations of equilibrium.
439
00:52:04,980 --> 00:52:09,890
Hence
we will have to have an additional equation
440
00:52:09,890 --> 00:52:15,920
which are generating from the equation of
compatibility and the equations of equilibrium
441
00:52:15,920 --> 00:52:23,160
if we take the moment about B, then we have
100
442
00:52:23,160 --> 00:52:39,970
moment about B is equal to 0 so we have 100(2.5)
is equal to P s (0.8) plus P aluminum (1.8).
443
00:52:39,970 --> 00:52:46,710
So the
equilibrium equation is 0.8P s plus 1.8 P
444
00:52:46,710 --> 00:53:01,580
aluminum is equal to 250. So this is the equilibrium
equation.
445
00:53:01,580 --> 00:53:08,780
Now from the compatibility if this load is
acting at this end and this is the pinned
446
00:53:08,780 --> 00:53:12,940
end this will
undergo; the bar will move in this particular
447
00:53:12,940 --> 00:53:16,501
form and this will move in the form of a circle
with
448
00:53:16,501 --> 00:53:22,130
the center of this particular point. Now this
being a smaller deformation, we consider this
449
00:53:22,130 --> 00:53:25,840
arc as a
straight length; hence this is a triangular
450
00:53:25,840 --> 00:53:33,060
form. At these locations this bar will undergo
deformation and let us call this deformation
451
00:53:33,060 --> 00:53:38,869
as delta s and the deformation here as delta
aluminum .
452
00:53:38,869 --> 00:53:44,609
Hence the compatibility is that delta from
this particular triangular configuration.
453
00:53:44,609 --> 00:53:54,910
We can write deltas by 0.8 is equal to delta
aluminum by 1.8. So this is the compatibility
454
00:53:54,910 --> 00:54:00,869
relation and
delta if we write in terms of P in terms of
455
00:54:00,869 --> 00:54:06,020
using the Hooke’s law which is P l by a
e. So this is P s
456
00:54:06,020 --> 00:54:20,559
L s by A s e s . That is for steel is equal
to 0.8 by 1.8 into P aluminum length aluminum
457
00:54:20,559 --> 00:54:27,269
divided by
A aluminum into e aluminum and if we substitute
458
00:54:27,269 --> 00:54:32,579
this values we get the relationship between
the load in
459
00:54:32,579 --> 00:54:44,319
P s and the force in P aluminum. Eventually
we get P s is equal to 4.233 P aluminum , this
460
00:54:44,319 --> 00:54:47,559
is the second
equation. Now from these two equations one
461
00:54:47,559 --> 00:54:55,070
and two we can solve for P s and P aluminum
and if we
462
00:54:55,070 --> 00:55:02,809
substitute the values of P s over here, we
get the values of P aluminum as is equal to
463
00:55:02,809 --> 00:55:07,720
48.2 kilo
464
00:55:07,720 --> 00:55:26,060
.Newton and thereby P s is equal to 204.04
kilo Newton. These are the forces acting in
465
00:55:26,060 --> 00:55:29,650
these two
rods and once we know the forces that are
466
00:55:29,650 --> 00:55:31,509
acting then the stresses can be known.
.
467
00:55:31,509 --> 00:55:49,099
The stress in the aluminum bar stress in aluminum
rod is equal to P aluminum by A aluminum which
468
00:55:49,099 --> 00:55:51,369
is
cross sectional area of aluminum which is
469
00:55:51,369 --> 00:55:57,410
equal to 48.2 is the load 10 cube so much
of Newton
470
00:55:57,410 --> 00:56:06,710
by 300 is the cross sectional area and these
eventually comes as 160.7 MPa. And stress
471
00:56:06,710 --> 00:56:14,500
in steel
bar or steel rod this is equal to P steel
472
00:56:14,500 --> 00:56:25,410
by A steel is equal to 204.40(10 cube) so
much of Newton by
473
00:56:25,410 --> 00:56:39,829
500 and this comes as 408.08 MPa. Here, though
the stresses level of stresses appear to be
474
00:56:39,829 --> 00:56:43,760
little
higher may be beyond the limiting capacity
475
00:56:43,760 --> 00:56:46,900
of this material. However, theoretically we
are
476
00:56:46,900 --> 00:56:51,819
computing what are the values of the stresses
if the limiting values are given, we can always
477
00:56:51,819 --> 00:56:56,680
compare and we can say whether this is safe
or unsafe, whether the system will be able
478
00:56:56,680 --> 00:57:03,690
to
withstand this load or not that we can always
479
00:57:03,690 --> 00:57:06,010
compute.
480
00:57:06,010 --> 00:57:07,010
..
481
00:57:07,010 --> 00:57:13,609
Now we have another problem which is a similar
type but of little variation that the rigid
482
00:57:13,609 --> 00:57:19,039
bar AB
of negligible weight is supported by a pin
483
00:57:19,039 --> 00:57:23,970
at B and by two vertical rods of aluminum
and steel;
484
00:57:23,970 --> 00:57:29,470
we will have to find out the vertical displacement
of the 50 kilo Newton weight at this location,
485
00:57:29,470 --> 00:57:40,259
how much displacement it undergoes etc, the
parameters of these members
486
00:57:40,259 --> 00:57:49,190
are given.
To summarize the whole thing we can say; in
487
00:57:49,190 --> 00:57:52,589
this particular lesson we included the concept
of
488
00:57:52,589 --> 00:57:59,779
strains in a three dimensional stress body
and concept of determinate and indeterminate
489
00:57:59,779 --> 00:58:02,130
systems.
Now, we clearly know what we really mean by
490
00:58:02,130 --> 00:58:07,660
determinate system and indeterminate system
and thereby how to arrive at the compatibility
491
00:58:07,660 --> 00:58:13,619
criteria to solve the unknown reactions in
indeterminate systems. Also, we know some
492
00:58:13,619 --> 00:58:16,299
examples to demonstrate the evaluation of
forces in
493
00:58:16,299 --> 00:58:17,809
indeterminate system.
494
00:58:17,809 --> 00:58:18,809
..
495
00:58:18,809 --> 00:58:19,809
.
496
00:58:19,809 --> 00:58:23,819
Here are some questions to be answered:
What is meant by indeterminate system and
497
00:58:23,819 --> 00:58:27,410
how is it different from determinate ones?
What is
498
00:58:27,410 --> 00:58:38,119
meant by compatibility equation and what will
be the effect of temperature variation in
499
00:58:38,119 --> 00:59:49,339
a stressed
system?
500
00:59:49,339 --> 00:59:49,339
.