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Welcome to the 4th lesson of module 2 which
is on analysis of strain part IV. In the
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previous lesson we discussed some aspects
of analysis of strain. In this particular
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lesson
we are going to discuss some more aspects
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of strain analysis. Let us look into those
aspects.
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.
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..
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After completing this particular lesson it
is expected that one should be able to
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understand the concept of strain due to change
in temperature. We have discussed about
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the aspects of strain and we have seen how
to compute strain at a point in a body and
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in a
member when it is subjected to axial pole.
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Now let us discuss some aspects of
development of strain due to change in temperature.
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Also, one should be able to
understand the concept of thermal stresses.
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Earlier we discussed about the determinate
and indeterminate systems where we find that
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we need additional criteria to evaluate the
internal forces which we termed as compatibility
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condition.
.
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.This particular lesson includes recapitulation
of the previous lesson. Let us look into
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some of the aspects. It is evaluation of strain
in a body due to variation in temperature.
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we will be looking into the concept of thermal
stresses and indeterminate system such as
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this change in temperature, the strain which
is caused due to change in temperature,
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whether this is determinate system or indeterminate
system and subsequently we will
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evaluate the stresses due to change in temperature
in different systems.
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.
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Questions:
What is meant by indeterminate system and
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how is it different from determinate ones?
.
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.In indeterminate system the number of unknown
forces is more than the number of
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equilibrium equations available and those
kinds of systems are called as statically
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indeterminate system.
We have seen what is meant by equilibrium
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equations and what the solution of such
systems is; we have seen that we need additional
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criteria to be brought in so that we can
evaluate the unknown forces and the way it
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differs from the determinate system. The
equilibrium equations are adequate to evaluate
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the reactive forces or the internal forces,
we call those kind of system as determinate
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system and these distinguishes between the
indeterminate system and the determinate system.
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.
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..
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The 2nd question was, what is meant by compatibility
equation?
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Let us look into the definition of compatibility.
As we had seen last time that statically
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indeterminate problem has geometric restrictions
imposed on its deformation and the
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mathematical expression of such constrains
of the restrictions are known as the
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compatibility equations.
.
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.Let us look into this particular example
which is quiet common. You must have seen
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that
the vertical members made out of concrete
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are provided with some steel bars if we cut
a
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cross section here and look from the top then
the section looks like this. Let us assume
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that the cross section of this vertical member
is a square one with the sizes of the sites
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as
A.
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This particular cross section of the member
is provided with four steel rods, if bearing
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plate is connected to the top of this member
and it is subjected to the load p onto top
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of
the plate. If we take a free body diagram,
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if we cut the member here and take the free
body of the top bar, the free body diagram
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looks like this, where the top force is p
and for
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the equilibrium of the whole system, we will
have the forces generated internal forces
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which will be exerted by the concrete, let
us call that as p c and the steel rods in
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a
combined form as P s so the equilibrium condition
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demands that P is equal to P c plus P s .
As you can see here, that we have two unknown
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parameters P c and P s and we have only
one equilibrium equation, hence the two unknowns
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from one equation cannot be
evaluated and hence we need an additional
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condition which you call as a compatibility
criteria. When we put the restrictions on
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the deformation and convert that in terms
of
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mathematical expression we call that particular
condition as the compatibility condition.
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.
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.Similarly, here when the load is applied
onto this member, the deformation of the
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concrete and steel, these two different materials
which are placed in this particular
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diagram or in this particular member, the
deformation of concrete and the deformation
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of
steel should be same. If we say that the deformation
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of concrete is deltac this should be
deformation in steel.
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In other words, we can say the strain in concrete
equals to strain in steel. Now, we
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convert this strain right in terms of stresses
using Hooke’s law which we call as the
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constitutive relationship between the stress
and the strain and that will give us another
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condition for forces between the concrete
and the steel. Now that we have one equation
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from here, we get another equation from here
and using these two equations, you can
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solve for P c and P s. So this particular
condition that P is equal to P c plus P s
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is termed as
equilibrium equation and another equation
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in terms of forces which are generated from
the second half equation where strain in concrete
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equals to strain in steel we call these as
compatibility equations. And as you can absorb
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here that the restrictions on the
deformation, we have said that the deformations
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of concrete and deformation of steel is
same and based on that this compatibility
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criteria has been derived.
.
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.The 3rd question was;
What will be the effect of temperature variation
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in a stressed system?
This is the aspect which we are going to discuss
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about.
.
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Strain due to temperature:
This is known that, if a body is subjected
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to temperature variation then it undergoes
changes in its dimensions. This volume expands
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if there is an increase in the temperature
or it contracts if there is decrease in the
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temperature.
In general, this change in temperature, if
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the temperature is uniformly increased or
decreased you get same amount of expansion
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or contraction in the body and this is tropic
in nature. The variation in temperature causes
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change in the length of the member which
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.you call as deformation and that deformation
leads to strain in the body. The strain due
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to
variation in temperature is given by this
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particular expression, epsilon T is the strain
due
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to change in temperature and deltaT is the
change in temperature.
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Let us say if the initial temperature is T
0 the final temperature is equal to T then
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deltaT is
equal to T minus T 0 and alpha is a term which
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is known as the coefficient of thermal
expansion. So the strain due to temperature
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variation is given as alpha(deltaT) for T
minus T 0 and deltaT is (positive) when we
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take this expansion or there is a (positive)
change in the temperature, where there is
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a raise in the temperature we call deltaT
as
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(positive), when there is a decrease in the
temperature we call that as (negative) or
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contraction of the body.
.
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Uniform change in temperature throughout the
body causes uniform thermal strain. This
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is important to know that there is strain
which is developed in the body due to change
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in
temperature; we call that as thermal strain.
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Supposing this is a body which undergoes change
in temperature, if it is free to move,
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this will undergo expansion or contraction
depending on the change in the temperature.
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If
temperature rises it will expand, if temperature
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decreases then it will contract. If it is
not
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restricted or constricted by any actions from
outside it will undergo movement and it will
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undergo strain. If we take the characteristic
length of this particular member as L or for
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that matter any of this length which undergoes
deformation then the deformation deltaT
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of that length can be given as the strain(length).
And strain as we have seen due to
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temperature is equal to alpha(deltaT) so the
deformation deltaT because of temperature
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is
alpha(deltaT) L where L is the dimension of
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the body which we are concentrating on.
.
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.If the thermal deformation occurs that if
it is allowed to move freely then there is
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no
possibility of inducing any stress in the
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member as there are no internal forces induced
if
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thermal deformation is allowed to occur freely.
There will be strain induced but there will
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not be any stress in the member. As there
are no internal forces being developed naturally
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there will not be any stresses. But since
it is undergoing deformation, either this
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is
expanding because of the change in the temperature
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or raise in the temperature or it is
contracting because of decrease in the temperature.
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So there is movement, there is
deformation in turn there is strain but since
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the movement of the body is not restricted
there is no internal forces that are getting
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generated. Hence there will not be any stresses
in the member. That is why it says that strain
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will be induced but no stress will be
developed. Total or partial restriction of
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deformation induces internal forces, if this
particular body which is free to move due
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to change in the temperature.
Supposing if I hold this body from two sides
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and do not allow this to move since it will
try to expand it, will exert pressure on this
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support and as a result the body will be
subjected to some internal forces and thereby
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there will be stresses induced in the body.
If
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total or partial restrictions are imposed
on these deformations, then they are going
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to
resist these forces due to thermal expansion
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or contraction and thereby there will be
stresses caused by these internal forces and
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these stresses.
Generally designated as thermal stresses,
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you must have notice that in many times is
structures will provide a gap between two
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structural elements. We allow the structural
member to move expand or contract due to the
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variation in temperature, you must have
noticed in the railway tracks are not continuous,
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some gaps are maintained at certain
distance interval. Due to change in the temperature
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on the railway track, the track
undergoes expansion. If those gaps are not
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kept, then the rail track will be subjected
to
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tremendous amount of stresses.
.
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.The forces developed due to the changing
temperature, causes forces if they are restricted
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and these forces cannot be determined from
equilibrium equations as we have seen
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earlier. Since we cannot determine the internal
forces which are getting developed due to
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change in temperature, we call this kind of
systems as indeterminate system according
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to
the definition.
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As we have seen earlier, for the systems which
we can evaluate the internal forces based
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on the equations of equilibrium, we call those
systems as determinate system, but if we
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cannot evaluate the forces which are getting
developed due to change in the temperature
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from the equilibrium equations, then we call
this system as indeterminate system, hence
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statically these kinds of thermal stresses
are induced in the member. Basically, statically
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indeterminate and we need to use the equation
of compatibility for the solution of these
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forces for evaluating the internal forces.
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.
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.If we summarize the statically indeterminate
problems; we need the equilibrium
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conditions the equilibrium conditions has
to be satisfied and that is either in a local
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concept or in a global sense or the whole
structure.
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Secondly, the geometric compatibility has
to be satisfied if we get more number of
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unknowns than the number of equilibrium equations
available. They are known as
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compatibility conditions or correspondingly
compatibility equations which give us
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additional information based on which we can
solve the unknown forces.
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Thirdly, we have written down the compatibility
in terms of the geometric deformation
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and we have related the strain parameter to
the corresponding stress parameter using
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Hooke’s law. And this relationship between
the stress and the strain is generally referred
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as the constitutive relationship which in
this case is based on Hooks law we are arriving
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at the constitutive relationship. These three
are the essential parts for the solution of
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statically indeterminate problems.
First one is equilibrium equation, second
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one is the compatibility equation and Third
one
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is the constitutive relationship which gives
us the relationship between the strain and
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the
stress and thereby we can write down the equations
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in terms of forces. We have one
equation which you have written in terms of
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equilibrium criteria. We have another
equation which we have written in terms of
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compatibility criteria from which we can
solve for unknown internal forces. So, solution
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00:20:59,820 --> 00:21:07,250
of equilibrium and compatibility
equations for unknown forces using equilibrium
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and compatibility criteria.
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.
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00:21:12,539 --> 00:21:21,320
.Having known about this different kinds of
strains that is getting induced into a body
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either when they are subjected to external
loads or they are subjected to change in
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00:21:27,509 --> 00:21:47,879
temperature, let us look into some of the
examples of that how do we evaluate these
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forces?
The example which was given earlier was a
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00:21:54,830 --> 00:22:06,950
rigid bar A, B of negligible weight is
supported by a pin at B and two vertical rods.
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00:22:06,950 --> 00:22:15,240
They are hung from the top and connected
to this bar, find the vertical displacement
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00:22:15,240 --> 00:22:25,000
of the 50 kN weight which is placed at this
particular point which is at a distance of
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00:22:25,000 --> 00:22:29,539
1 meter from end A. So we will have to find
out
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00:22:29,539 --> 00:22:35,539
the displacement of this particular point
where this 50 kN weight is hung, the crosssectional
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00:22:35,539 --> 00:22:44,299
area of steel bar is 300 mm square, this is
the steel bar of length 2 meter and this
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00:22:44,299 --> 00:22:50,850
is the aluminum bar of length 1 meter.
The cross-sectional area of steel bar is 300
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mm square and that of aluminum is 1,000 mm
square. The modulus of elasticity of steel
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00:22:58,700 --> 00:23:03,519
E s is equal to 2 into 10 to the power 5 MPa,
and
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00:23:03,519 --> 00:23:10,161
modulus of elasticity of aluminum is 0.7 into
10 to the power 5 MPa. We will have to
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00:23:10,161 --> 00:23:23,309
find out the vertical displacement of this
particular point from where the 50 kN were
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00:23:23,309 --> 00:23:32,590
designed. The first thing is that we will
have to write down the equations of equilibrium.
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If we take the free body of the whole system
this is the rigid bar A, B and this is
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connected by a pin at end B.
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.
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00:23:54,109 --> 00:24:05,690
.We have one vertical rod connected, this
end we have another vertical rod connected
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at
this end, this is made up of steel, this is
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made of aluminum. Let us call the force in
the
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aluminum rod as p aluminum and the force of
the steel as P s and the 50 kN load is hung
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at this point which is at a distance of 1m,
this is 1m, this is 50 kN. Now if we take
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of even
this B end part of it, we will have the reactive
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component, the vertical force and the
horizontal force, let us call this as R and
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00:24:59,909 --> 00:25:08,241
H. Since we do not like to evaluate R and
H, we
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00:25:08,241 --> 00:25:13,789
are interested in p s and p al because we
would like to find out how much force these
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00:25:13,789 --> 00:25:16,080
rigid
rods will be subjected to.
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Let us take the moment of all these forces
about point B, so summation of all the forces
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with B is equal to 0, this is one of the equilibrium
equations. And this gives as the p
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aluminum into 1 plus p steel into 3 is equal
to 50 into 2 or p aluminum plus 3p steel is
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equal to 100, this is equation 1 which we
call as the equilibrium equation. We are not
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00:26:15,710 --> 00:26:20,210
looking into other equations which are summation
of vertical forces as 0 and summation
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00:26:20,210 --> 00:26:26,370
of horizontal forces as 0 from which we will
get R and H since we are interested to
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00:26:26,370 --> 00:26:33,269
evaluate under this. Because we have four
unknown parameters p s , p al, R and H, these
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00:26:33,269 --> 00:26:36,570
are
the four unknown forces and we have three
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00:26:36,570 --> 00:26:42,010
equations which are summation of horizontal
force as 0, summation of vertical force as
215
00:26:42,010 --> 00:26:48,140
0 and summation of moment at this point
equals to 0.
216
00:26:48,140 --> 00:26:56,649
In this particular equation we see that we
have two unknowns and one equation from
217
00:26:56,649 --> 00:27:02,710
which we cannot really solve these unknown
forces. Therefore we need an additional
218
00:27:02,710 --> 00:27:09,549
criteria or additional condition from which
we can evaluate these forces. If we look into
219
00:27:09,549 --> 00:27:20,630
the deformation of this member, this particular
bar A B is a rigid one. Since this is
220
00:27:20,630 --> 00:27:29,629
subjected to a load 50 kN this will undergo
a movement in a circular path considering
221
00:27:29,629 --> 00:27:37,769
point B as a center and since the deformation
is small this circular approximate as a
222
00:27:37,769 --> 00:27:48,480
straight one and we join this line, this gives
us an excorticated form of a deformation.
223
00:27:48,480 --> 00:27:54,809
This point undergoes deformation here; this
point undergoes deformation here, also this
224
00:27:54,809 --> 00:27:58,500
point has deformed to the step.
225
00:27:58,500 --> 00:28:08,039
.From this triangular configuration, if we
call this deformation as deltaal and this
226
00:28:08,039 --> 00:28:19,629
deformation as deltas then from this triangular
configuration we can write deltaal by 1 is
227
00:28:19,629 --> 00:28:38,369
equal to delta s by 3 because this claim is
3 or deltaal is equal to 1 by 3(deltas). Now
228
00:28:38,369 --> 00:28:43,419
we
have one equation of equilibrium. We had obtained
229
00:28:43,419 --> 00:28:49,370
another equation which is
corresponding to the compatibility of the
230
00:28:49,370 --> 00:28:54,809
deformation which we call as compatibility
equation. So, equation two is the compatibility
231
00:28:54,809 --> 00:29:01,149
equation.
Next thing which we will have to do is impose
232
00:29:01,149 --> 00:29:06,639
the constitutive relationship. That means
first you write this deformation in terms
233
00:29:06,639 --> 00:29:11,740
of strain and that we relate to the stress.
This if
234
00:29:11,740 --> 00:29:28,330
you write in terms of load or rather the stresses
we can say that p al length al by a al into
235
00:29:28,330 --> 00:29:38,381
e al, this is the deformation of the al bar,
this equal to 1 by 3 (p s) is the load which
236
00:29:38,381 --> 00:29:42,100
is
acting on the steel rod times length of the
237
00:29:42,100 --> 00:29:51,259
steel rod divided by area of steel rod times
modulus of elasticity of the steel rod.
238
00:29:51,259 --> 00:29:52,259
.
239
00:29:52,259 --> 00:30:09,529
If we substitute the values
we get P al into by length of al is 1,000
240
00:30:09,529 --> 00:30:23,159
mm or 1m by crosssectional area of a al rod
is 1,000 mm square into e which is 0.7 into
241
00:30:23,159 --> 00:30:37,119
10 to the power 5
MPa is equal to 1 by 3(P s ) length is equal
242
00:30:37,119 --> 00:30:42,119
to 2m by cross-sectional area of steel is
300
243
00:30:42,119 --> 00:30:56,730
into 2 into 10 to the power 5. This gives
us the relationship that P al is equal to
244
00:30:56,730 --> 00:31:09,259
7 by 9 P s .
This is another relationship we got from between
245
00:31:09,259 --> 00:31:13,080
p in the aluminum and p in the steel.
246
00:31:13,080 --> 00:31:14,080
..
247
00:31:14,080 --> 00:31:23,549
Earlier we had one equation which is equation
of equilibrium, this P al plus 3 into P s
248
00:31:23,549 --> 00:31:27,279
is
equal to 100.
249
00:31:27,279 --> 00:31:28,279
.
250
00:31:28,279 --> 00:31:40,090
This is, let us call the equation 2 so from
equations 1 and 2, we have then P s (7 by
251
00:31:40,090 --> 00:31:51,340
9 plus
1) or plus we have 3 of this is 3 is equal
252
00:31:51,340 --> 00:32:03,909
to 100. We have P al plus 3(P s ).
253
00:32:03,909 --> 00:32:04,909
..
254
00:32:04,909 --> 00:32:05,909
.
255
00:32:05,909 --> 00:32:29,259
P al (7 by 9 plus 3) is equal to 100 which
gives us the value of P s as 26.45 kN, and
256
00:32:29,259 --> 00:32:34,039
from
this, the value of al load in al is equal
257
00:32:34,039 --> 00:32:50,909
to 7 by 9 into 26.45 and this comes as 20.63
kN and
258
00:32:50,909 --> 00:32:57,249
once we get the load we can compute the deformation.
259
00:32:57,249 --> 00:32:58,249
..
260
00:32:58,249 --> 00:33:04,009
Once we note the steel rod and the aluminum
rod they are subjected to the load P s and
261
00:33:04,009 --> 00:33:07,659
P al
which we have evaluated. Hence we can compute
262
00:33:07,659 --> 00:33:12,779
that how much deformation this
particular bar will undergo and how much deformation
263
00:33:12,779 --> 00:33:16,929
this bar will undergo. Since we
are interested to evaluate the deformation
264
00:33:16,929 --> 00:33:23,130
of this particular point, if we know the
deformation of this particular bar from this
265
00:33:23,130 --> 00:33:27,860
triangular configuration we can compute how
much deformation this particular point will
266
00:33:27,860 --> 00:33:31,720
undergo.
.
267
00:33:31,720 --> 00:33:50,909
.Hence the deformation of the steel bar deltas
is equal to P sl by A S E S is
268
00:33:50,909 --> 00:34:01,220
equal to P S you
got 26.45 into 10 to the power 3 so much of
269
00:34:01,220 --> 00:34:16,960
N into A S which is 2,000 mm and A s is equal
to 300 into E s which is 2 into 10 to the
270
00:34:16,960 --> 00:34:26,000
power 5 hence this gives as a value, is equal
to
271
00:34:26,000 --> 00:34:35,870
0.882 mm so this is the deformation of the
steel rod which is shown in this particular
272
00:34:35,870 --> 00:34:40,570
diagram.
.
273
00:34:40,570 --> 00:34:51,250
The deformation which we have obtained here
deltas is equal to 0.882 mm. From this
274
00:34:51,250 --> 00:35:00,319
triangular configuration
we can compute the deformation of this particular
275
00:35:00,319 --> 00:35:04,690
point which is
at a distance of 1m from here, the deformation
276
00:35:04,690 --> 00:35:14,430
of this is equal to 2 by 3 of this particular
deformation.
277
00:35:14,430 --> 00:35:15,430
.
278
00:35:15,430 --> 00:35:31,540
.The deformation of the point where the load
is acting say delta at load point, where 50
279
00:35:31,540 --> 00:35:39,490
kN
load is acting this is equal to 2 by 3 into
280
00:35:39,490 --> 00:35:51,369
0.882 mm is equal to 0.888 mm so this gives
us
281
00:35:51,369 --> 00:36:02,620
the value of the deformation of the load point
and this is what we were interested to know
282
00:36:02,620 --> 00:36:08,030
that this much deformation this load point
and according to this calculation, we find
283
00:36:08,030 --> 00:36:13,839
this
is equal to 0.588 mm.
284
00:36:13,839 --> 00:36:14,839
.
285
00:36:14,839 --> 00:36:15,839
.
286
00:36:15,839 --> 00:36:16,839
..
287
00:36:16,839 --> 00:36:26,059
I am sure that you could calculate the values,
you can check the values whether the
288
00:36:26,059 --> 00:36:33,619
matching with this solution or not, let us
look into another example based on the aspects
289
00:36:33,619 --> 00:36:39,900
which you have discussed today that due to
the change in the temperature there will be
290
00:36:39,900 --> 00:36:45,530
deformation in the member and that causes
stresses in the member which we call as a
291
00:36:45,530 --> 00:36:47,720
thermal stress.
.
292
00:36:47,720 --> 00:36:59,510
.In this particular example we stated that
a bronze bar which is a 3m long bar, let us
293
00:36:59,510 --> 00:37:06,170
call
this as A B this bronze bar 3m with the cross-sectional
294
00:37:06,170 --> 00:37:12,369
area of 320 mm square is placed
between two rigid walls. This is one wall,
295
00:37:12,369 --> 00:37:19,970
this is another wall placed in between these
two at a temperature of minus 20 degree C.
296
00:37:19,970 --> 00:37:22,160
There is a gap between the age of the bar
and
297
00:37:22,160 --> 00:37:29,500
the wall the age of this bar and the wall,
there is a gap of 2.5 mm.
298
00:37:29,500 --> 00:37:36,860
Now what we will have to do is, we will have
to find out the temperature at which the
299
00:37:36,860 --> 00:37:51,420
compressive stress
in the bar will be 35 MPa given the value
300
00:37:51,420 --> 00:37:56,600
of alpha which is coefficient
of thermal expansion as is equal to 18 into
301
00:37:56,600 --> 00:37:59,839
10 to the power 8 by degree C and the value
of
302
00:37:59,839 --> 00:38:14,040
modulus of elasticity given is 80 GPa when
the temperature goes off from minus 20
303
00:38:14,040 --> 00:38:24,810
degrees. As there is an increase in the temperature
the bar will expand since there is a gap
304
00:38:24,810 --> 00:38:34,050
between the wall and the bar first due to
expansion, the bar will touch the wall till
305
00:38:34,050 --> 00:38:35,240
that
particular time.
306
00:38:35,240 --> 00:38:44,349
Since it is free to move, it will not experience
any stress but further extension beyond the
307
00:38:44,349 --> 00:38:51,260
touching of the wall, the wall will not allow
the bar to move so naturally, there will be
308
00:38:51,260 --> 00:38:55,510
the
stress induced in the bar. What we will have
309
00:38:55,510 --> 00:39:05,589
to find out is that, we can allow maximum
value of this stress to go up to 35 MPa and
310
00:39:05,589 --> 00:39:13,420
we will have to find out that temperature
which will allow this expansion of this bar
311
00:39:13,420 --> 00:39:18,330
to cause the stress within this body as 35
MPa.
312
00:39:18,330 --> 00:39:19,330
.
313
00:39:19,330 --> 00:39:27,640
.So if we draw a free body diagram then it
will be easier to visualize that how the whole
314
00:39:27,640 --> 00:39:35,700
process is undertaken. In such problems where
we try to evaluate the stresses due to
315
00:39:35,700 --> 00:39:43,470
change in the temperature and where such restrictions
are imposed like the bar is fixed
316
00:39:43,470 --> 00:39:49,960
between the walls or the bar at this point
is not allowed to go beyond a certain value.
317
00:39:49,960 --> 00:39:54,690
We
try to first remove either personally at the
318
00:39:54,690 --> 00:40:03,220
whole the restrictions allow the bar to move
freely due to change in the temperature.
319
00:40:03,220 --> 00:40:08,230
We calculate how much deformation it undergoes
because of the change in the
320
00:40:08,230 --> 00:40:16,430
temperature and then if there would have been
restrictions then to bring back to its
321
00:40:16,430 --> 00:40:22,230
particular position, how much force we need
to impose on that we try to evaluate that
322
00:40:22,230 --> 00:40:25,349
and
that is what is the compatibility criteria.
323
00:40:25,349 --> 00:40:30,170
We allow it to move freely thereby there will
be
324
00:40:30,170 --> 00:40:36,280
deformation because of the change in the temperature
and now we impose some external
325
00:40:36,280 --> 00:40:42,200
force to bring back to deformation to its
original position, thereby, how much force
326
00:40:42,200 --> 00:40:45,299
we
need to put to bring back the member to its
327
00:40:45,299 --> 00:40:49,880
initial form?
We evaluate that and if we compare these two
328
00:40:49,880 --> 00:40:56,950
from these we can drive at an expression
which we generally call as compatibility equation.
329
00:40:56,950 --> 00:41:03,390
That is what we need to do in this
particular case.
330
00:41:03,390 --> 00:41:04,390
.
331
00:41:04,390 --> 00:41:22,490
.We have the bar and on this side it is fixed
to the wall, here we have one restriction
332
00:41:22,490 --> 00:41:26,250
which
is another wall but there is gap between the
333
00:41:26,250 --> 00:41:34,559
wall and the bar. When the temperature goes
up from minus 20 degrees to some value T which
334
00:41:34,559 --> 00:41:41,280
we do not know which we need to
evaluate, initially when it starts expanding
335
00:41:41,280 --> 00:41:44,829
it goes up to the wall without any problem
and
336
00:41:44,829 --> 00:41:52,299
till that time since it is not experiencing
any abstraction the bar we will expanding
337
00:41:52,299 --> 00:41:56,460
freely
and thereby there will be strain but there
338
00:41:56,460 --> 00:42:11,730
will not be any stress in the bar.
Let us remove this wall
339
00:42:11,730 --> 00:42:18,930
and allow this bar to move freely and because
of the change in
340
00:42:18,930 --> 00:42:32,630
the temperature it will move. This particular
expansion is delta. This delta which is total
341
00:42:32,630 --> 00:42:44,160
deformation has two parts: one is this initial
2.5 mm which was unrestricted movement
342
00:42:44,160 --> 00:42:50,490
and then subsequently it will be encountered
in the wall. Since I have removed the wall
343
00:42:50,490 --> 00:42:58,000
so this is also moving freely and let us call
this particular distance as delta prime so
344
00:42:58,000 --> 00:43:03,730
delta
the total deformation is equal to delta plus
345
00:43:03,730 --> 00:43:15,910
2.5 mm.
Our job is to find out the temperature. Since
346
00:43:15,910 --> 00:43:20,450
the increase in the temperature is going to
cause the deformation of the bar we will have
347
00:43:20,450 --> 00:43:30,030
to find out how much temperature we can
allow so that when it will be encountering
348
00:43:30,030 --> 00:43:37,970
this wall it will be subjected to a compressive
force, it will be subjected to a stress and
349
00:43:37,970 --> 00:43:42,569
the restrictions imposed on this particular
case is
350
00:43:42,569 --> 00:43:51,500
that the stress level can go up to 32 MPa.
Therefore till the time it reaches 35 MPa
351
00:43:51,500 --> 00:43:54,150
the
temperature is allowed to increase, as soon
352
00:43:54,150 --> 00:44:00,369
as it goes to 35 Mpa we cannot increase the
temperature further because then the material
353
00:44:00,369 --> 00:44:09,299
may not be able to withstand that amount of
increasing temperature. That means when I
354
00:44:09,299 --> 00:44:16,350
have removed this particular wall and
allowed the bar to move freely naturally the
355
00:44:16,350 --> 00:44:23,140
second consideration which is a realistic
situation is that I will have to apply some
356
00:44:23,140 --> 00:44:29,349
force on this bar and bring back this
deformation to this particular stress so that
357
00:44:29,349 --> 00:44:35,630
the member is between these two walls. that
means these additional deformation delta prime
358
00:44:35,630 --> 00:44:43,089
which is getting caused from the wall to
this end has to be brought back to this particular
359
00:44:43,089 --> 00:44:50,650
position and for that I will have to apply
a load p and this load is going to cause a
360
00:44:50,650 --> 00:45:01,039
stress in the bar is equal to P by cross-sectional
area. And p by cross-sectional area the stress
361
00:45:01,039 --> 00:45:05,069
is to be limited to 35 MPa and under that
362
00:45:05,069 --> 00:45:16,910
.consideration we will have to find out what
is the value of T so that the stress level
363
00:45:16,910 --> 00:45:22,690
is
exactly 35 MPa.
364
00:45:22,690 --> 00:45:30,430
If we calculate the compatibility equation
from this particular concept from the concept
365
00:45:30,430 --> 00:45:41,349
of deformation if we compute the compatibility
criteria, then what we can do is the delta
366
00:45:41,349 --> 00:45:48,810
is equal to deformation which you are getting
in the bar which you have designated as
367
00:45:48,810 --> 00:45:54,160
delta because of the change in the temperature.
The initial temperature was minus 20
368
00:45:54,160 --> 00:46:01,210
degrees, the final temperature is T so the
deltaT the change in the temperature is equal
369
00:46:01,210 --> 00:46:07,950
to
T minus (minus 20) is equal to T plus 20,
370
00:46:07,950 --> 00:46:18,240
this is the value of deltaT. The
strain ΕT is
371
00:46:18,240 --> 00:46:26,460
equal to alpha into deltaT or deformation
due to change in temperature which we have
372
00:46:26,460 --> 00:46:33,869
called as deltaT is equal to alpha into deltaT
into length l.
373
00:46:33,869 --> 00:46:39,940
So in this particular case the coefficient
of thermal expansion which is given as 18
374
00:46:39,940 --> 00:46:47,789
into 10
to the power minus 6 by degree C into change
375
00:46:47,789 --> 00:47:01,020
in temperature is T plus 20 into l is 3,000
the original length 3 m so this is equal to
376
00:47:01,020 --> 00:47:07,329
delta which is delta prime plus 2.5 so this
is 2.5
377
00:47:07,329 --> 00:47:25,619
plus delta prime. Now from these we get the
value of delta prime is equal to 54 into 10
378
00:47:25,619 --> 00:47:35,000
to
the power minus 3 into T plus 20 minus 2.5.
379
00:47:35,000 --> 00:47:45,180
Now this delta prime which is being caused
from the wall to this end and this particular
380
00:47:45,180 --> 00:47:51,650
part has to be brought back to this particular
wall level by applying this force P so this
381
00:47:51,650 --> 00:48:00,859
delta can be written as Pl by AE by the
application of external force p, we are bringing
382
00:48:00,859 --> 00:48:09,540
back this delta to the wall position and p
by a p by cross-sectional area, we can write
383
00:48:09,540 --> 00:48:13,190
this as stress.
In this particular problem since the stress
384
00:48:13,190 --> 00:48:16,260
is limited we write this expression in terms
of
385
00:48:16,260 --> 00:48:26,539
stress. So p by a is the parameter sigma into
l by e equals to this particular quantity
386
00:48:26,539 --> 00:48:30,930
which
is 54 into 10 to the power minus 3 into T
387
00:48:30,930 --> 00:48:36,690
plus 20 minus 2.5.
388
00:48:36,690 --> 00:48:37,690
.
389
00:48:37,690 --> 00:48:57,400
.From this we can compute the value of T.
Here sigma is given as 35 MPa and
390
00:48:57,400 --> 00:49:00,809
the length
initially was 3m and then we have allowed
391
00:49:00,809 --> 00:49:06,260
it to move because of the increase in the
temperature and since there was a gap between
392
00:49:06,260 --> 00:49:14,880
the wall and the bar of 2.5 mm this 2.5
mm movement is unrestricted so the stress
393
00:49:14,880 --> 00:49:18,501
start getting the generator only when the
bar
394
00:49:18,501 --> 00:49:26,000
starts the wall. So length which is going
to cause the stress or which is going to cause
395
00:49:26,000 --> 00:49:33,289
this
additional deformation is equal to 3002.5,
396
00:49:33,289 --> 00:49:42,710
this 3,000 plus 2.5 is the length and this
divided by e which is 80 GPa area so 80 into
397
00:49:42,710 --> 00:49:48,920
10 to the power 3 MPa is equal to 54 into
10
398
00:49:48,920 --> 00:49:57,359
to the power minus 3 into T plus 20 minus
2.5.
399
00:49:57,359 --> 00:50:11,289
Or, if you evaluate this it comes to 1.3136
plus 2.5 is equal to 54 into 10 to the power
400
00:50:11,289 --> 00:50:24,350
minus 3 into T plus 20 and then T plus 20
from this expression is equal to 70.62, if
401
00:50:24,350 --> 00:50:28,220
we
compute this you get this as 70.62 and hence
402
00:50:28,220 --> 00:50:42,299
the T is equal to 50.62. So, if the
temperature goes up to 50.6 degree C from
403
00:50:42,299 --> 00:50:47,000
minus 20 then the bar will be experiencing
a
404
00:50:47,000 --> 00:50:54,330
stress of 35 MPa and that is the problem.
If you allow the bar to undergo a change in
405
00:50:54,330 --> 00:51:00,059
the
temperature from minus 20 to 50 plus 50 to
406
00:51:00,059 --> 00:51:04,470
62 the bar will be experiencing a stress of
35
407
00:51:04,470 --> 00:51:05,470
MPa.
408
00:51:05,470 --> 00:51:06,470
.
409
00:51:06,470 --> 00:51:13,650
.But the interesting part of this to be noted
here is that this value of T we have arrived
410
00:51:13,650 --> 00:51:18,450
at
from the criteria where we have equated the
411
00:51:18,450 --> 00:51:26,300
deformation part of the bar. Initially, allowed
the bar to deform in such away that undergoes
412
00:51:26,300 --> 00:51:37,819
movement freely over the 2.5 mm and
then it hits wall and then keeps on increasing
413
00:51:37,819 --> 00:51:44,290
because that there is a change in the
temperature but when it is giving the thrust
414
00:51:44,290 --> 00:51:47,780
on the wall is giving a reactive force which
is
415
00:51:47,780 --> 00:51:56,880
causing as internal stress in the member.
We have derived these equation based on the
416
00:51:56,880 --> 00:52:03,420
deformation compatibility which we call as
the compatibility equation and so the change
417
00:52:03,420 --> 00:52:10,850
in the temperature is basically an indeterminate
system and we solve the problem in terms
418
00:52:10,850 --> 00:52:15,960
of indeterminate form.
.
419
00:52:15,960 --> 00:52:26,890
.Let us look into another example where a
rigid block of mass M is supported by the
420
00:52:26,890 --> 00:52:31,079
three
symmetrically placed rods and the ends of
421
00:52:31,079 --> 00:52:41,510
the rods were leveled before the block was
attached. They are in the same level, we will
422
00:52:41,510 --> 00:52:51,839
have to determine the largest allowable
value of M that can be this bar, can carry
423
00:52:51,839 --> 00:52:54,270
cross-sectional areas of steel and copper
rods
424
00:52:54,270 --> 00:53:03,190
are 1200 and 900 mm square then modulus of
elasticity of steel is 200 Gpa modulus of
425
00:53:03,190 --> 00:53:09,589
elasticity of copper is 120 MPa and permissible
stress for copper is 70 MPa and that of
426
00:53:09,589 --> 00:53:16,549
steel is 140 MPa. These are the two copper
bars and this is the steel bar of length 240
427
00:53:16,549 --> 00:53:21,970
mm
and length of the copper bar is 160 mm and
428
00:53:21,970 --> 00:53:26,890
they are placed at 1 meter interval
symmetrically placed. Before these blocks
429
00:53:26,890 --> 00:53:34,089
were placed they were on the same level.
After the block is placed we will undergo
430
00:53:34,089 --> 00:53:37,160
deformation. What we will have to find out
is
431
00:53:37,160 --> 00:53:45,900
that how much weight we can put so that they
are within their stresses, their stresses
432
00:53:45,900 --> 00:53:54,690
do
not go beyond these values. Let us write down
433
00:53:54,690 --> 00:53:58,890
the equilibrium equation.
.
434
00:53:58,890 --> 00:54:08,460
This is the bar or the block whose mass is
M so the load that it transmits on those three
435
00:54:08,460 --> 00:54:19,339
bars equals the weight of it which is M into
g. We have the three bars placed, central
436
00:54:19,339 --> 00:54:25,220
is
the steel one and two copper rods, let us
437
00:54:25,220 --> 00:54:29,210
call the load that will be carried by steel
rod as
438
00:54:29,210 --> 00:54:37,750
P s and those of copper by P c.
So we can write down the summation of vertical
439
00:54:37,750 --> 00:54:47,940
forces as 0, we will give us w is equal to
P s plus 2P c so this is the equation of equilibrium.
440
00:54:47,940 --> 00:54:58,102
The
compatibility equation, next
441
00:54:58,102 --> 00:55:04,920
equation which is here, also you see unknown
parameters are P s and P c and of course we
442
00:55:04,920 --> 00:55:13,930
have to find out W in terms of these. We need
another equation which is in terms of its
443
00:55:13,930 --> 00:55:19,880
deformation compatibility. And since the block
is placed when the bars are uniform and
444
00:55:19,880 --> 00:55:28,130
is expected that it will be in the same level
the strain in the copper bar or copper rods
445
00:55:28,130 --> 00:55:31,230
will
be equal to the strain in steel or the deformation
446
00:55:31,230 --> 00:55:41,410
that the bars will undergo such as the
strain in the steel bar will have the same
447
00:55:41,410 --> 00:55:45,670
value of the strain in the copper bars. These
lead
448
00:55:45,670 --> 00:55:54,430
us to the criteria that, if we substitute
the values of this strain which is P c into
449
00:55:54,430 --> 00:55:56,329
L c by A c .
450
00:55:56,329 --> 00:56:10,349
.into E c the suffix stands for copper is
equal to P s L s by A s E s this gives us
451
00:56:10,349 --> 00:56:18,819
a relationship
from which we get P c is equal to 27 by 40
452
00:56:18,819 --> 00:56:29,500
into P s if we compute the value of W in terms
of P c and P s .
453
00:56:29,500 --> 00:56:38,549
Here another criteria is to be noted that
the maximum value of P c and P s can be evaluated
454
00:56:38,549 --> 00:56:44,829
from the given stress. Since the allowable
stresses on these rods are given, the maximum
455
00:56:44,829 --> 00:56:52,390
load that the rods can carry equals the stress
in the rod multiplied by the cross-sectional
456
00:56:52,390 --> 00:56:53,390
area.
.
457
00:56:53,390 --> 00:57:00,910
So the value of P s is equal to the stress
in the steel rod multiplied by the cross-sectional
458
00:57:00,910 --> 00:57:06,180
area of the steel rod and the load maximum
load that the copper rods can carry is equal
459
00:57:06,180 --> 00:57:11,410
to
sigma c into A c . So from these we compute
460
00:57:11,410 --> 00:57:17,420
the value of W and from the criteria that
from
461
00:57:17,420 --> 00:57:27,420
the maximum value of steel rod if we compute
we will get W is equal to 789.6 kN or
462
00:57:27,420 --> 00:57:36,650
mass we get around 80.6 into 10 cube kilograms
whereas from the other criteria from the
463
00:57:36,650 --> 00:57:46,029
limiting criteria for P c we get mass is equal
to 22.6 into 10 cube kilograms. Since we are
464
00:57:46,029 --> 00:57:49,819
getting two values and out of these this is
the lowest one, therefore this is the maximum
465
00:57:49,819 --> 00:57:56,789
value we can apply beyond which if we apply
that the value of the stresses in the bar
466
00:57:56,789 --> 00:58:04,679
will
not exceed the permissible limits.
467
00:58:04,679 --> 00:58:05,679
..
468
00:58:05,679 --> 00:58:13,559
We have another problem similar to the previous
problem. We got to determine the
469
00:58:13,559 --> 00:58:19,430
lengths of the two copper rods so that stresses
in all the three reach their allowable limits
470
00:58:19,430 --> 00:58:24,970
simultaneously. In this case, we are allowing
the stresses in the bars to reach their
471
00:58:24,970 --> 00:58:33,430
limiting values simultaneously then what should
be the length of the copper rods?
472
00:58:33,430 --> 00:58:34,430
.
473
00:58:34,430 --> 00:58:42,250
Also, we have another problem which are related
to this thermal or temperature changes
474
00:58:42,250 --> 00:58:48,720
in the member that all members of the steel
truss shown in this figure have the same
475
00:58:48,720 --> 00:58:54,859
.cross-sectional area. If the truss is stress
free at 10 degree C, determine the stresses
476
00:58:54,859 --> 00:58:59,519
in the
members at 90 degree C, for steel alpha is
477
00:58:59,519 --> 00:59:02,550
equal to 11.7 into 10 to the power minus 6
by
478
00:59:02,550 --> 00:59:06,400
degree C and E is equal to 200 GPa.
.
479
00:59:06,400 --> 00:59:11,690
To summarize; in this particular lesson we
have included the concept of strains due to
480
00:59:11,690 --> 00:59:18,609
change in temperature, we have included the
concept of thermal stresses, we have seen
481
00:59:18,609 --> 00:59:24,309
the steps to evaluate unknown forces for indeterminate
systems and then we have looked
482
00:59:24,309 --> 00:59:28,730
into some examples to demonstrate the evaluation
of stresses due to change in
483
00:59:28,730 --> 00:59:29,730
temperature.
.
484
00:59:29,730 --> 00:59:32,770
.We have some questions:
(1)
485
00:59:32,770 --> 00:59:37,970
What are the steps for the evaluation of unknown
forces for indeterminate
486
00:59:37,970 --> 00:59:39,490
systems?
(2)
487
00:59:39,490 --> 00:59:42,609
What is the expression for thermal stress?
(3)
488
00:59:42,609 --> 00:59:56,210
Is the evaluation of thermal stress in a stressed
body a determinate one?
489
00:59:56,210 --> 00:59:56,210
.