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So, today we will be discussing deflection
of beam.
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So, that is
deflection of beams; that is the general topic
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of today’s discussion. In our
previous class, we are talking about stresses,
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and we were tried to get the idea that stress
is a very important quantity. All the time
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we have to determine, calculate stress, to
have
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some idea regarding the structural safety,
because that stress should be within certain
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limit, and that limit, we have defined as
allowable stress.
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Now, deflection is not that important like
stress, but there are some situations where
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we
may require getting the information about
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deflection. In that case, the knowledge of
calculating deflection at some point is necessary;
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so, that is one aspect. Apart from that,
there is another requirement, which I should
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say it is much more important. In that
context, I can explain in that manner, say,
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a structure, we divide the type of structure
into
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two categories: one is called statically determinate
problem; and another type of
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problem, we say statically indeterminate problem.
So, if we say a structure in general - structure:
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so one type is statically determinate type;
another is indeterminate type. Now, what is
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statically determinate type and statically
indeterminate type? Last class, we were talking
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about equation of statics and we have
seen there are three equations, if we handle
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a two-dimensional problem or a plane
problem. Summation of all the forces along
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x direction we have taken 0; summation of
all the forces along y direction equal to
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0; plus we can take moment at any point of
all
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the forces it will be equal to 0. So, these
are basically equation of statics if we think
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in
terms of plane problem.
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Now, there are some structural problems, where
the unknowns, which are basically the
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reactions which we are getting from the supports;
those quantities we can determine with
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the help of equation of statics. So, it can
be determined by the equation of statics,
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which
is statically determinate. Already we have
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taken a problem - two supports; one is hinge;
another is roller; we got three unknown reactions;
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and there are three equations we have
applied; we obtained the support reactions.
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So that problem was absolutely a statically
determinate problem.
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Now, all the structures will not be statically
determinate. Say, there might be more
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supports, may be both the supports hinge.
So, you will get four reactions; you may get
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five reactions; you may get more number of
reactions; or not necessarily you cannot
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force that support should be such the reaction
number will be always within this number
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3 lk. Depending on the practical solution,
it may be any number. So, it cannot be solved
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by only the equation of statics. So, that
category of problem we say statically
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indeterminate problem. So, it cannot be determined
with the equation of statics. So, that
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is why we say it is statically indeterminate
problems.
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Now, the question is - how we will solve a
statically indeterminate problem, but real
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structure may be statically indeterminate;
it can be solved. So, we will require more
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number of equations. So, we have three equation
of statics. So far, the problem is within
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a plane. Say our unknowns are five. So we
will require another two equations; and those
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two additional equations, it will come from
the information regarding the deformation
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of
the structure. So, structural deformation
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is nothing but the deflection of the structure.
So,
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deflection of structure, we will try to utilize
to get some more information; that more
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information will be reflected in the form
of more equations.
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So, structural deformation or structural deflection
those relationships will be utilized in
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the form of additional equation in addition
with the equation of statics. And that will
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help
to solve the entire problem. So, ultimately
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what will be our problem? Our problem will
be, if we are interested only bending moment,
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shear force, or stresses, and problem is not
determinate, we have to go for some calculation
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of deflection of the structure, because
that will help to give some equation.
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So, that is, I think, one of the major requirement
for having some understanding
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regarding the deformation of a beam or deformation
of any structure. So we can start
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with a beam, because it is one of the very,
very elementary structural element. So, that
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concept can be extended to some other complex
type of structure like, say, frame
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problem or other type of structural problem.
So, here our starting point is deflection
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of the beam. And we will start with statically
determinate problem. Everything will be known;
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at least the forces, reactions. So, how to
find out deflection, that part we will study.
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Once that understanding will be developed,
with that we can go ahead for a statically
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indeterminate type of problem for finding
out
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the unknowns, so that we can determine internal
forces, internal moments;
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simultaneously we can find out the deflection
of the structures also. So, it will be a
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integrated process; deformation calculation,
stress calculation - both will be carried
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out
simultaneously.
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Now, if we try to find out the deflection
of beam, rather there are many methods
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available we can use, but here my attention
is to cover only very, very basic type of
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methods. At least I will try to cover two,
three, maximum four types of method; there
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is
no point of covering all the type of methods
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available in the market. What are very basic
type, very fundamental type, which can be
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used without much problem we can go ahead
with that.
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Now, I will start with a method; it is called
deferential equation technique. So, this is
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defined in a different manner; sometimes it
is defined differential equation method,
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differential equation technique or integration
of differential equation, integration of
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differential equation technique. So, basically
it is dealing with some differential equation
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of the beam problem. Here, when we will be
handling with differential equation, we have
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to start from the deformation of that beam,
with some bending moment, with shear force,
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with load, and all those quantities.
So, before coming to that exact treatment
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what is done in this particular method, let
us
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try to find out the relationship between the
different quantities. The different quantities
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just now I have mentioned: one is the load,
shear force, bending moment, slope,
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deflection. So, they are not separate quantities;
they are closely associated with each
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other; and some dependencies are there in
between. So, let us try to get the idea how
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they
are related.
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Now, we can take a beam like this; can take
any beam; there might be some loading here.
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So, it is just I have drawn a simply supported
beam; means beam one end is hinge
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support, another is a roller support, and
combination of that it will give a stable
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structure,
determinate structure, and some loading. Here
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hinge will provide two reactions; roller
will provide one reaction. So, three reactions.
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It can be determined with the equation of
statics.
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Now, if we try to make the hinge or the roller
the left support, what will happen?
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Number of unknown will be less; two your equation
will be more; in that case structure
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will not be stable. So, all the time we cannot
make it as minimum as possible. So, a
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minimum number of destiny is also necessary,
because any structure, basically, we want
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to fixed in its place; so, it will require
some support. So, at least three direction
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it should
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be resting; it should not move this way or
that way; it should not be stable, but rotate.
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So,
all the possible movements we have to restrain
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again.
So, these are the minimum thing, and if we
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go beyond minimum, then it will be statically
indeterminate. Now, here, I have drawn a line
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to represent the beam. So, basically, a
beam has a very small dimension along the
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depth and width compared to the length. So,
length is the predominant dimension. And normally,
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all the breadth and depth quantity,
we more or less represent in a compact form;
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we take high value. So, everything is
incorporated there for representing the bending.
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So, the variable here is only along x and
this line is basically the neutral axis or
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neutral
line of the beam. Now, under the load it will
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undergo some bending. So, we are trying to
plot the deformation of the beam, with its
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neutral axis; we sometimes we say it is a
elastic line, because whole thing is elastic
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analysis and it is a line; we say it is a
elastic
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line. It will undergo some deformation in
an arbitrary manner depending on the type
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of
load. Now, if I put here some axis system,
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say x along the axis of the member, and y
is
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perpendicular to that. So, we can say the
deformation of the beam or deflection of the
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beam, it can be represented as, say, y equal
to function of x. So, y at any point, say,
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at a
distance x, so this is x that will be the
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y. Now, it is different x, it will be different
y; and
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it is another x, it will be another y. So,
x is a variable quantity; depending on x,
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you will
get the value of y. So, y is a function of
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x. Now what is y? y is the deflection.
Now, if I take the derivative, say, y dash
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it is nothing but your dy by dx or f dash
x. If we
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have the expression of the function, we can
take derivative. So, it will give the slope
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of
the line. Slope of the elastic curve that
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we say it is sometimes, we write as theta;
theta is
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the slope of the elastic curve. Now, the y
is the deflection; its derivative is theta.
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Now, if
we take further derivative, we will get d
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2 y dx square or y double dash x or f double
dash x; that quantity will have some relationship
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with moment.
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So, if we take y double dash or this one;
this quantity will be equal to M by EI . So,
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for
slope in first derivative, second derivative
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is, this term, basically the curvature and
that
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curvature will be equal to the bending moment
divided by EI. EI term already we have
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defined, which is basically the flexile rigidity
of the beam. So, M is the moment. And
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that expression is very, very important. Sometimes
we write as minus EI d 2 y dx square
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equal to M. So, that equation in a different
fashion we have written.
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So, this part d 2 y by dx square - this part
- is called basically curvature. It is basically
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curvature. Curvature is inverse of radius
of curvature. The curve will have some radius
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of curvature. So, inverse of that will be
curvature. So, if we take a straight line,
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radius of
curvature will be infinite and curvature will
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be 0. So, straight line no curvature, but
if we
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have some curvature, so we will have some
finite value of radius of curvature. So, this
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d
2 y by dx square will be the curvature and
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M is the bending moment. A is the flexile
rigidity, and that relationship is called
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moment curvature relationship.
Now, just now, I have mentioned d 2 y by dx
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square is curvature; curvature of the
deformed shape of the beam or the elastic
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line. So, if we say it is a deformed shape
of the
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beam, now anywhere if that is the radius R.
So, d 2 y by dx square it will be 1 by d 2
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y
by dx square. So, this is the radius of curvature
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and this is d 2 y by dx square is
curvature. So, the reverse term of that is
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your radius of curvature.
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In that context, I should say the expression
of radius of curvature and curvature is not
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exactly the same. For a very arbitrary curve,
say, any arbitrary curve, the R if we try
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to
represent, so R is basically your d 2 y dx
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square divided by 1 plus dy dx. So, that is
the
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general expression for that; that will be
1 by R. So, 1 by R is the curvature, it is
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d 2 y by
dx square 1 plus dy dx whole square whole
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to the power 3 by 2 or 1.5 whatever you can
say. But here, the denominator part we are
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not taking; that reason very simple, because
we will be dealing with structural problem,
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and structural problem deformation will not
be very, very large; it will be very small
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deformations; and when deformation is small,
its rate of change is basically slope dy by
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dx. Slope will be much more small and square
of that, it will be much more small. So, this
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quantity will be practically going to be 0;
we
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can neglect that.
So, this part if you draw, it will be 1 to
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dy 3 by 2, it will be basically your 1. So,
approximately, we can write 1 by R is your
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d 2 y dx square for small deformation
problem; and in our case it will be a small
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deformation problem. So, we can simply write
1 by R equal to d 2 y by dx square; and this
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d 2 y by dx square, they are related to
basically bending moment, with a factor of
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EI, and some minus sign will come.
Here I want to tell you we can compare that
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equation with your normal stress-strain
relationship. Say normal stress-strain relationship
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is, say, here I am writing in the side
sigma equal to your E into epsilon. Now, sigma
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you can compare with M, and strain you
can compare with your curvature, and E you
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can compare with EI. So, it is basically
something similar or analogous to your stress-strain
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relationship. So, it is not stress; it is
moment; moment is nothing but one of the stress
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resultant, because throughout the depth
of the beam, your bending stress will vary;
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it will be maximum, 0, minimum. So, total
effect if you take, that is basically the
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bending moment. So, M is the resultant of
the
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stresses. So, it is stress; one of the stress
resultant. So, that stress resultant we can
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compare with stress. And here strain, we can
compare with the curvature; and E here
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something related to EI. Now, we have written
minus. Why we are writing minus here?
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It is basically a matter of sign convention.
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Sometimes we write here, say, d 2 y dx square
EI. We say somewhere you will find the
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relation is something like this. So, you can
take plus; you can take minus; where we will
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take plus, where we will take minus, that
will depend absolutely on the sign convention.
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Say, I am taking a axis system, whatever we
have taken earlier. So, it is x; it is y.
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Now,
there might be a bending phenomena like this;
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there might be a bending phenomena like
that; so, earlier we are trying to define
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a sagging/hogging type. Now, here there will
be a
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type of moment, and here there will be a type
of moment. Now, xy d 2 y by dx square is
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what? Your d 2 y dx square we can say it is
d dy dx of dy by dx; and dy by dx is what?
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It
is basically theta. So, we can write it is
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d theta by bx.
Now, what is theta? Theta is basically if
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we draw a line here that will be the theta.
So, it
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should be theta. And theta will be positive
when the way we have written or drawn it is
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basically theta. If we increase x, your y
value is increasing, but if you draw like
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this with
the increase of x, y is decreasing; y it is
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going towards the negative side. So, that
will be
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a negative slope; that will be a positive
slope.
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Now, if I take that type of problem, here
slope will be positive and here slope will
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be
negative. So, d 2 y by dx square is rate of
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00:24:25,490 --> 00:24:31,960
change of slope. So, it will be… it is changing
from your positive slope to negative slope.
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00:24:31,960 --> 00:24:42,299
So, curvature will be in that case negative,
because here slope is positive, here slope
202
00:24:42,299 --> 00:24:49,760
is negative. So, whatever negative value minus
that positive quantity will be negative divided
203
00:24:49,760 --> 00:24:53,370
by dx. So, here curvature is a negative
204
00:24:53,370 --> 00:24:57,600
curvature. But here this is a negative curvature,
negative slope, and that is a positive
205
00:24:57,600 --> 00:25:06,240
slope; so, positive slope means with increase
of x the value is increasing in the positive
206
00:25:06,240 --> 00:25:16,490
direction of y, and here if you increase x,
y part is decreasing. So, this is a negative
207
00:25:16,490 --> 00:25:21,809
slope; that is a positive slope. So, rate
of change of slope - this minus this divided
208
00:25:21,809 --> 00:25:25,450
by
length dx - that will be basically d theta
209
00:25:25,450 --> 00:25:30,020
by dx or d 2 by dx square. So, this will be
a
210
00:25:30,020 --> 00:25:36,049
positive curvature.
So, this should be a positive curvature, and
211
00:25:36,049 --> 00:25:53,799
this should be a negative curvature, because
if it is a sagging one in terms of xy system,
212
00:25:53,799 --> 00:25:56,950
your curvature is becoming negative - d 2
y
213
00:25:56,950 --> 00:26:02,450
dx square is becoming negative - because a
slope we have defined, which is plus and
214
00:26:02,450 --> 00:26:09,110
minus, and change of - rate of change of slope
- that is basically d 2 y by dx square. That
215
00:26:09,110 --> 00:26:13,919
is becoming negative here; this is becoming
positive here for the case of hogging.
216
00:26:13,919 --> 00:26:22,460
Now, it is a matter of sign convention. If
we take sagging moment is positive, then here
217
00:26:22,460 --> 00:26:31,130
our this side sign will be minus. If we take
as hogging is positive, we will take the sign
218
00:26:31,130 --> 00:26:35,740
of
M will be plus. So, it is again a sign conventional.
219
00:26:35,740 --> 00:26:41,410
So, normally, last day we are trying to
take sagging as positive; we can go on with
220
00:26:41,410 --> 00:26:48,049
that - sagging as positive. So, here at that
level we will write moment with a minus sign.
221
00:26:48,049 --> 00:26:52,990
So, we have started y, its first derivative
is slope, second derivative with EI will be
222
00:26:52,990 --> 00:27:02,090
minus moment.
Now, next part is your dM by dx. If you take
223
00:27:02,090 --> 00:27:09,490
the derivative of moment, you will get
shear force or if we put here, here it will
224
00:27:09,490 --> 00:27:19,090
be minus EI d 3 y dx cube. So, shear force
is
225
00:27:19,090 --> 00:27:25,440
what? Derivative of moment. What is moment?
Moment is your minus EI d 2 y by dx
226
00:27:25,440 --> 00:27:31,860
square. So, further derivative of that and
minus sign we have to incorporate.
227
00:27:31,860 --> 00:27:43,960
And the last term is the loading Q is equal
to minus dv by dx or minus minus it will be
228
00:27:43,960 --> 00:27:56,970
plus; it will be EI d 4 y dx to the power
of 4. So, what is Q? Q is basically the intensity
229
00:27:56,970 --> 00:28:02,490
of
the load acting on the member; member is here;
230
00:28:02,490 --> 00:28:16,299
every problem it is a beam member. So,
we are putting a load; we are getting a displacement.
231
00:28:16,299 --> 00:28:22,950
In between, your slope is there,
moment is there, shear is shear is there,
232
00:28:22,950 --> 00:28:26,470
and they are related like this.
So, if we know the equation of the deflected
233
00:28:26,470 --> 00:28:32,200
curve, we take the first derivative, we will
get slope; second derivative multiplied by
234
00:28:32,200 --> 00:28:36,820
minus shear, you will get moment; an another
derivative you will get shear force; another
235
00:28:36,820 --> 00:28:43,620
derivative you will get how much is the
intensity of the load. So, all the three,
236
00:28:43,620 --> 00:28:48,789
four quantities: deflection, slope, moment,
shear,
237
00:28:48,789 --> 00:29:00,830
and load - rather five quantities; they are
related like this.
238
00:29:00,830 --> 00:29:07,280
Now, sometimes we use some of the equations
for finding some of the important
239
00:29:07,280 --> 00:29:14,460
quantities, say, moment and shear force - relationship
is just one order derivative
240
00:29:14,460 --> 00:29:23,210
difference. Say we want to get the maximum
bending moment within a particular region.
241
00:29:23,210 --> 00:29:30,090
We have the equation and equation is not very
symmetric type. So, from the physical
242
00:29:30,090 --> 00:29:34,940
look, you can get the idea it will be at the
center, it may be somewhere intermediate,
243
00:29:34,940 --> 00:29:37,520
but
you have to determine what will be the maximum
244
00:29:37,520 --> 00:29:42,510
value and where it will be located.
You can find out the value of moment, because
245
00:29:42,510 --> 00:29:47,030
if you substitute x equal to how much,
from the equation you can get it, but you
246
00:29:47,030 --> 00:29:51,590
require the information where will be the
location - means what will be the value of
247
00:29:51,590 --> 00:30:01,679
x? So, in that case, that will be very useful
where moment will be maximum, so shear force
248
00:30:01,679 --> 00:30:05,919
would be 0. So, any maximum or
minimum quantity it is derived. So, it will
249
00:30:05,919 --> 00:30:13,010
go to a peak and come, come to the lower
side; means there the slope means derivative
250
00:30:13,010 --> 00:30:19,190
will be equal to 0. So, shear force equation
if we equate 0, you will get the location,
251
00:30:19,190 --> 00:30:22,600
the value of x; at what x value shear force
is
252
00:30:22,600 --> 00:30:28,020
becoming 0; corresponding to that value, your
bending moment will be maximum. So,
253
00:30:28,020 --> 00:30:31,940
that x value, you can substitute in the expression
of bending moment and get the
254
00:30:31,940 --> 00:30:37,070
maximum value of the bending moment there.
So, this information sometimes you will be
255
00:30:37,070 --> 00:30:42,650
using for finding out some specific problem.
So, these are very, very important relationship:
256
00:30:42,650 --> 00:30:48,860
deflection, slope, bending moment, shear
force, and the load distribution. So, they
257
00:30:48,860 --> 00:30:56,970
are related with that.
Now, in connection with our primary theme
258
00:30:56,970 --> 00:31:01,909
of today’s discussion is - deflection of
beam.
259
00:31:01,909 --> 00:31:07,600
The first line on this page, that is moment
curvature relation EI d 2 y dx square equal
260
00:31:07,600 --> 00:31:12,539
to
plus or minus M, with that we will try to
261
00:31:12,539 --> 00:31:19,430
solve deflection of beam using differential
equation technique, because this is the differential
262
00:31:19,430 --> 00:31:25,220
equation, and the differential equation
we will utilize for finding out the deflection
263
00:31:25,220 --> 00:31:34,100
of the beam. Now, we shall take some
simple cases and try to explain how the method
264
00:31:34,100 --> 00:31:38,330
works.
265
00:31:38,330 --> 00:31:54,900
I can take a cantilever beam problem, which
is quite simple type of problem. Say, load
266
00:31:54,900 --> 00:32:05,730
is
acting at the free end, so it is P, and it
267
00:32:05,730 --> 00:32:13,820
has some length, some E values, some I value.
So,
268
00:32:13,820 --> 00:32:22,760
this is a beam, left support is fixed or clamped;
right side is free; at the free end there
269
00:32:22,760 --> 00:32:26,460
is a
vertical load P; the total length of the beam
270
00:32:26,460 --> 00:32:33,660
is L; it has a material property E; cross
sectional property is I, I means it is second
271
00:32:33,660 --> 00:32:37,640
moment of area or moment of inertia of the
cross section. It is a uniform cross section;
272
00:32:37,640 --> 00:32:43,870
everywhere it is EI.
Now under the load, the deformation will take
273
00:32:43,870 --> 00:32:53,320
place like this. At at this support, there
will be no deflection; there will be no slope.
274
00:32:53,320 --> 00:32:54,540
So, it will go; gradually it will deform;
at
275
00:32:54,540 --> 00:33:00,700
the tip there will be a maximum, deflection
maximum slope. And here, if we take some
276
00:33:00,700 --> 00:33:11,380
axis system like this x, and it is y. We can
start with our differential equation; differential
277
00:33:11,380 --> 00:33:20,460
equation is nothing but your moment-curvature
relationship. So, we can write EI d 2 y dx
278
00:33:20,460 --> 00:33:27,960
square equal to minus m.
If we say sagging is positive, it will be
279
00:33:27,960 --> 00:33:33,760
minus M. Now, what will be the expression
of M?
280
00:33:33,760 --> 00:33:44,330
So, M part; M is bending moment at any cross
section; say this is xy. So, if we take any
281
00:33:44,330 --> 00:33:51,600
value of x, what will be the moment? If it
is x, this part will… be total length is
282
00:33:51,600 --> 00:33:57,350
L; L
minus x. So, P into L minus x. And P L minus
283
00:33:57,350 --> 00:34:07,980
x will be your hogging moment. And
sagging moment will be minus of that. So,
284
00:34:07,980 --> 00:34:30,800
it will be P L minus x minus. If sagging is
considered to be positive, say, here, any,
285
00:34:30,800 --> 00:34:35,300
any station. So, P is the force and distance
is L
286
00:34:35,300 --> 00:34:37,770
minus x. So, P L minus x.
287
00:34:37,770 --> 00:34:44,159
What will be the type of moment? The moment
will be in that direction. So, that moment
288
00:34:44,159 --> 00:34:49,829
will generate stress - tensile stress - at
the top. So, it is something like this. So,
289
00:34:49,829 --> 00:34:53,779
hogging
moment, tensile stress will develop at the
290
00:34:53,779 --> 00:34:58,869
top. So, if we take sagging as positive, we
have
291
00:34:58,869 --> 00:35:04,770
to get a moment. So, the moment we are getting,
it is basically negative moment. So, it is
292
00:35:04,770 --> 00:35:12,440
minus; minus is, it is not sagging; it is
hogging; and the value will be PL minus x.
293
00:35:12,440 --> 00:35:23,420
Now, here, if I substitute that value, this
will be equal to EI d 2 y dx square; this
294
00:35:23,420 --> 00:35:29,240
minus
minus, it will be plus; L minus x; we could
295
00:35:29,240 --> 00:35:36,360
take hogging as positive, in that case we
could write this is plus, and this we could
296
00:35:36,360 --> 00:35:39,520
write plus. If hogging is positive, this would
be
297
00:35:39,520 --> 00:35:46,080
plus. And here also it should be… it is
basically hogging moment, so it will be plus.
298
00:35:46,080 --> 00:35:48,490
And
if we take sagging also, in both the case
299
00:35:48,490 --> 00:35:52,480
it is negative.
In any case, if you start sagging as positive
300
00:35:52,480 --> 00:35:55,140
or hogging as positive, this equation will
be
301
00:35:55,140 --> 00:36:04,030
automatically coming to the same equation.
Now, this is our differential equation. So,
302
00:36:04,030 --> 00:36:08,060
EI
d 2 y by dx square equal to PL minus x. So,
303
00:36:08,060 --> 00:36:12,119
this equation we have to integrate. So, if
you
304
00:36:12,119 --> 00:36:19,370
integrate, this is a very straight forward
equation; it will be dy by dx. And it will
305
00:36:19,370 --> 00:36:26,890
be P L
minus x whole square divided by 2, and there
306
00:36:26,890 --> 00:36:33,650
will bea minus sign this side, because with
x there is a minus quantity, plus there will
307
00:36:33,650 --> 00:36:43,170
be some constant, say C 1.
Now, next step we can take a further derivative.
308
00:36:43,170 --> 00:36:57,099
It will be EI y; it will be minus Pl minus
x. So, 2 will be now 3, and below 3 will come.
309
00:36:57,099 --> 00:37:01,870
So, 3 into 2 it will be 6, and from x
another minus will come, and this minus and
310
00:37:01,870 --> 00:37:07,990
minus will be plus, and here it will be C
1 x
311
00:37:07,990 --> 00:37:21,720
plus another constant you will get, say C
2.
312
00:37:21,720 --> 00:37:26,580
So, integration of the differential equation
already obtained. In first integration, we
313
00:37:26,580 --> 00:37:29,599
have
got first derivative; from second derivative,
314
00:37:29,599 --> 00:37:34,880
in the second level, we have got without
derivative, direct y we got. So, the last
315
00:37:34,880 --> 00:37:42,401
line is the expression for y deflection; previous
line is the slope; previous to previous, it
316
00:37:42,401 --> 00:37:56,340
is basically the curvature relations here.
Now, in that process we have got one C 1 and
317
00:37:56,340 --> 00:38:03,349
C 2. These are the constants. And these
constants are unknown here. These constants
318
00:38:03,349 --> 00:38:07,150
we have to determine, in terms of, say, L
E
319
00:38:07,150 --> 00:38:11,660
P, something, whatever information we have
related to the structure. So, here it is a
320
00:38:11,660 --> 00:38:15,660
beam
problem. So, EI, l, P these are the quantities;
321
00:38:15,660 --> 00:38:22,369
so, in terms of those we have to determine.
And those quantities can be determined using
322
00:38:22,369 --> 00:38:30,670
the boundary conditions, where the two
ends there are some conditions, and those
323
00:38:30,670 --> 00:38:33,130
condition will be utilized, those are, we
say, it
324
00:38:33,130 --> 00:38:34,130
is boundary conditions.
325
00:38:34,130 --> 00:38:43,520
Say, here, this is a fixed support. So, y
or the deflection is 0; plus this point itself,
326
00:38:43,520 --> 00:38:47,790
slope
is equal to 0; not necessarily it will be
327
00:38:47,790 --> 00:38:49,830
all the time at the same point. If we have
a simple
328
00:38:49,830 --> 00:38:58,170
supported beam, deflection will be 0 here,
deflection will be 0 here. Now, we have to
329
00:38:58,170 --> 00:39:08,000
get
C 1 and C 2 from your boundary conditions.
330
00:39:08,000 --> 00:39:16,530
So, BC is boundary conditions. Now, we
have the equation.
331
00:39:16,530 --> 00:39:29,380
Let us put the condition, say, first of all,
at this level we put say at x equal to 0,
332
00:39:29,380 --> 00:39:34,599
your d 2
y by dx equal to 0; and the second level we
333
00:39:34,599 --> 00:39:38,500
will put at x equal to 0, y is equal to 0.
So,
334
00:39:38,500 --> 00:39:44,109
first we will get C 1. C 1 value you can substitute
here. And the second level, once you
335
00:39:44,109 --> 00:39:51,609
will put that requirement for deflection is
equal to 0, we will get C 2.
336
00:39:51,609 --> 00:40:08,840
Now, if I take at x equal to 0, theta equal
to dy dx 0. So, that condition will give your
337
00:40:08,840 --> 00:40:15,130
EI
into… it will be 0, because dy dx 0, minus
338
00:40:15,130 --> 00:40:27,200
P l minus 0 square divided by 2 plus C 1.
So,
339
00:40:27,200 --> 00:40:44,270
left part will be entirely 0; and C 1 we can
write in terms of minus Pl square divided
340
00:40:44,270 --> 00:40:53,450
by
2. Once C 1 is determined, we will take the
341
00:40:53,450 --> 00:41:03,710
last equation. So, at x equal to 0, your y
equal to 0. So, that condition will give your
342
00:41:03,710 --> 00:41:14,550
EI into 0; that will be equal to your P l
minus
343
00:41:14,550 --> 00:41:26,680
0 cube divided by 6 plus, C 1 part you can
substitute or you can keep it C 1 multiplied
344
00:41:26,680 --> 00:41:32,270
by
x, x will be equal to 0 plus C 2. So, these
345
00:41:32,270 --> 00:41:46,740
two terms are now 0.
So, C 2 will be minus Pl cube divided 6. C
346
00:41:46,740 --> 00:41:59,829
1 will be positive. C 1 will be positive,
because this is a minus sign it will be plus.
347
00:41:59,829 --> 00:42:09,369
C 2 will be negative. C 2 will be negative.
348
00:42:09,369 --> 00:42:16,609
Now, once C 1 and C 2 we have, we can write
the expression of slope; we can write the
349
00:42:16,609 --> 00:42:23,840
expression of deflection. Say if we write
the expression of slope, it will be theta
350
00:42:23,840 --> 00:42:36,040
equal to
dy by dx; that will be equal to 1 by EI minus
351
00:42:36,040 --> 00:42:52,250
P l minus x square divided by 2 plus C 1;
just now we have determined it is Pl square
352
00:42:52,250 --> 00:43:00,670
divided by 2. So, in last page we have made
minus P l minus x square divided by 2 plus
353
00:43:00,670 --> 00:43:06,630
C 1 is PL square by 2. EI part we have put
right side with a one-way form. So, that will
354
00:43:06,630 --> 00:43:12,950
be the expression of that. Anyhow, this
equation can be rearranged, because if you
355
00:43:12,950 --> 00:43:14,820
break Pl square by 2, that part will cancel.
So,
356
00:43:14,820 --> 00:43:21,910
you can put it like this or in a different
form. Even different book, you may get
357
00:43:21,910 --> 00:43:28,580
expression in a different manner, but ultimately
it will be the basic terms; it can be
358
00:43:28,580 --> 00:43:36,050
generated from there.
Now, if we want to find the maximum slope,
359
00:43:36,050 --> 00:43:45,460
maximum slope will occur at the free end.
So, we can get theta max; theta max will be
360
00:43:45,460 --> 00:43:56,940
1 by EI. We can put x is equal to L. So, this
part will become 0, and that will be Pl square
361
00:43:56,940 --> 00:44:11,400
divided by 2, or we can say Pl square
divided by 2 EI, and this will be at x equal
362
00:44:11,400 --> 00:44:18,339
to l. So, at the free end we will get the
theta
363
00:44:18,339 --> 00:44:38,030
will be maximum and that value will be P into
l square divided by 2 EI . So, this term is
364
00:44:38,030 --> 00:44:58,230
becoming 0, say x. Theta at x equal to L.
N slope, maximum slope that will be Pl square
365
00:44:58,230 --> 00:45:03,859
by 2 EI.
Now, this is one quantity Pl square by 2 EI;
366
00:45:03,859 --> 00:45:06,950
that is a cantilever subjected to a load at
the
367
00:45:06,950 --> 00:45:12,700
free endc we are getting the maximum slope
there, Pl square by 2 EI. This is one of the
368
00:45:12,700 --> 00:45:19,700
very important terms. There are some very
standard. This is one of the standard cases
369
00:45:19,700 --> 00:45:25,829
and the value is one of the standard quantity.
So, if we try to solve a relatively difficult
370
00:45:25,829 --> 00:45:28,839
problem, some of the quantities like this
PL
371
00:45:28,839 --> 00:45:34,109
square by 2 EI - that type of quantity, we
will derive some of the quantities here. Those
372
00:45:34,109 --> 00:45:43,790
quantities it will be utilized later on. Say,
a beam subjected to fully uniform redistributed
373
00:45:43,790 --> 00:45:51,300
load - what will be maximum moment at the
center? Omega l square by 8 and that part
374
00:45:51,300 --> 00:45:57,070
frequently we use that. We need not take a
piece of paper, try to derive what will be
375
00:45:57,070 --> 00:46:00,250
that
maximum. So, it is just instantaneously we
376
00:46:00,250 --> 00:46:06,770
say it is omega L square by 8.
So, if we try to evaluate deformation of a
377
00:46:06,770 --> 00:46:12,740
structure, deflection of beam, some of the
quantities at least we have to keep into our
378
00:46:12,740 --> 00:46:17,040
mind; otherwise, all the time we have to keep
a book, and take the help of that - what is
379
00:46:17,040 --> 00:46:22,240
the value of maximum slope here, maximum
deflection here.
380
00:46:22,240 --> 00:46:27,680
Now, in that process, I am trying to find
out some standard relationship plus the
381
00:46:27,680 --> 00:46:33,710
procedure I have explained, by that time I
think you got the feeling.
382
00:46:33,710 --> 00:46:37,950
So, the starting point is the differential
equation, and differential equation is nothing
383
00:46:37,950 --> 00:46:39,300
but
your moment curvature relationship; that is
384
00:46:39,300 --> 00:46:47,390
EI d 2 y dx square equal to M; and M that
will be the expression of the bending moment
385
00:46:47,390 --> 00:46:51,280
diagram - the expression of the bending
moment at any arbitrary section. So, it is
386
00:46:51,280 --> 00:47:00,089
a function of x. Once you get that expression,
you have to integrate twice, you will get
387
00:47:00,089 --> 00:47:02,609
two constants; and these two constants you
have
388
00:47:02,609 --> 00:47:07,390
determine with the help of boundary condition.
Say slope 0, deflection 0, some
389
00:47:07,390 --> 00:47:11,760
information you will have for a particular
structure.
390
00:47:11,760 --> 00:47:16,810
Now a slope expression already we have obtained
on the maximum slope. So, next step
391
00:47:16,810 --> 00:47:24,440
should be the deflection. C 1, C 2 we can
put in the equation of your expression of
392
00:47:24,440 --> 00:47:28,510
y. It
will give the equation of the deflection of
393
00:47:28,510 --> 00:47:34,030
the beam and we can find out the maximum
value. Normally that will occur at the same
394
00:47:34,030 --> 00:47:38,119
point of free end of the beam. So, let us
come
395
00:47:38,119 --> 00:47:40,370
to that part.
396
00:47:40,370 --> 00:48:07,020
So, it will be y equal to 1 by EI. And here
we got Pl minus x cube divided by 6 plus C
397
00:48:07,020 --> 00:48:18,740
1.
We got Pl square divided by 2 into x and C
398
00:48:18,740 --> 00:48:32,191
2 we got Pl cube divided by 6. So, we have
just substitute C 1 and C 2 in the expression
399
00:48:32,191 --> 00:48:37,430
of y. There EI part was in the left side,
that
400
00:48:37,430 --> 00:48:43,950
part we have put on the right side; in the
one-way form. So, that will be the expression.
401
00:48:43,950 --> 00:48:53,080
This can be arranged in a different form,
in a way you like; it may be given in different
402
00:48:53,080 --> 00:48:56,730
form in different places.
But if we are interested for finding out the
403
00:48:56,730 --> 00:49:11,891
maximum value of y, so at x equal to l y equal
to y max. So, y max it will be 1 by EI. The
404
00:49:11,891 --> 00:49:16,290
first term will be equal to 0, because x will
be
405
00:49:16,290 --> 00:49:26,680
l, l minus l, it will be 0. So, there it will
be Pl cube divided by 2 minus Pl cube divided
406
00:49:26,680 --> 00:49:37,109
by 6. So, it is basically both are Pl cube.
So, it is a half minus 1 by 6. So, half 1
407
00:49:37,109 --> 00:49:41,310
by 6. So,
if I just make a small calculation. So, it
408
00:49:41,310 --> 00:49:46,460
will be 6, means 3 minus 1. So, it will be
2 by 6
409
00:49:46,460 --> 00:49:55,810
or it we can say one-third. So, we can write
Pl cube divided by 3 EI; that will be the
410
00:49:55,810 --> 00:49:59,780
y
max. So, this is another important expression.
411
00:49:59,780 --> 00:50:08,680
So, we got the expression for the elastic
line. So, that is the line; that is the expression
412
00:50:08,680 --> 00:50:21,819
for that elastic line, and maximum value
we got Pl cube by 3 EI.
413
00:50:21,819 --> 00:50:28,430
Sometimes, we get confused whether it is l
square or l cube, l 4; that form is the all
414
00:50:28,430 --> 00:50:34,260
the
time you will find EI below, here P, l, the
415
00:50:34,260 --> 00:50:41,290
unit you can always check at least. So, EI
means unit will be in SI unit Newton per meter
416
00:50:41,290 --> 00:50:45,720
square and I will be meter to the power 4.
So, below it will be Newton meter square.
417
00:50:45,720 --> 00:50:49,330
So, here P will be Newton, it will cancel.
So,
418
00:50:49,330 --> 00:50:55,800
Newton meter square means it will be L cube.
So, meter cube by meter square it will be
419
00:50:55,800 --> 00:50:59,260
meter.
Say slope you got Pl square. So, it should
420
00:50:59,260 --> 00:51:04,090
be a dimensionless quantity. There should
be
421
00:51:04,090 --> 00:51:19,079
no unit there. So, we can apply those ideas
to check the order of the l, l cube, or l
422
00:51:19,079 --> 00:51:23,690
square,
or l 4. If you are little bit confused, try
423
00:51:23,690 --> 00:51:31,569
to recall from your memory.
So, it is basically a simple cantilever beam
424
00:51:31,569 --> 00:51:37,950
problem. We have tried to find out the
deflection using differential equation technique.
425
00:51:37,950 --> 00:51:51,680
And tried to get some maximum values
of bending moment and shear force. Now, up
426
00:51:51,680 --> 00:52:00,600
to this we can keep in this class. In the
next
427
00:52:00,600 --> 00:52:08,390
class, we will go ahead with similar type
of problem and try to find out some standard
428
00:52:08,390 --> 00:52:21,050
relationships.
Preview of Next Lecture
429
00:52:21,050 --> 00:52:35,280
So, we shall continue whatever we are trying
to cover in the last class, that is deflection
430
00:52:35,280 --> 00:52:42,440
of beam, and we are trying to solve the problem
with differential equation technique.
431
00:52:42,440 --> 00:52:49,869
We have taken a simple beam problem. It was
a cantilever, with some point loaded at the
432
00:52:49,869 --> 00:52:55,349
end, with that I have tried to explain you
how to find out the final equations, how to
433
00:52:55,349 --> 00:52:59,960
find
out the constants through boundary conditions,
434
00:52:59,960 --> 00:53:04,980
and from there, we try to obtain some
standard values of maximum deflection, maximum
435
00:53:04,980 --> 00:53:15,869
slope at one of the end.
Now, we will try some other standard cases,
436
00:53:15,869 --> 00:53:21,350
and try to get some important relationship.
Technique is same; problem will be little
437
00:53:21,350 --> 00:53:41,200
different. Now, I can take, again, cantilever
problem with a little different type of loader.
438
00:53:41,200 --> 00:54:13,660
So, structure is same; it has a length l,
EI, support identical, it is fixed, that is
439
00:54:13,660 --> 00:54:23,250
free, only
the loading is different. The idea is to handle
440
00:54:23,250 --> 00:54:30,241
a different type of loader. Also, I think,
before changing the type of structure; next
441
00:54:30,241 --> 00:54:37,470
step we can change the type of structure.
So, our first job is we have to find out the
442
00:54:37,470 --> 00:54:49,980
expression for the bending moment. Now, at
any distance x, so there is one cross section.
443
00:54:49,980 --> 00:54:57,300
What will be the expression of bending
moment? We can calculate from the right end,
444
00:54:57,300 --> 00:55:02,680
because that is a free end it is easier to
calculate. So, beyond that section right side,
445
00:55:02,680 --> 00:55:05,490
the loading is first of all the length of
that
446
00:55:05,490 --> 00:55:17,260
will be L minus x. And the loading is say
w omega; that is intensity of the distributed
447
00:55:17,260 --> 00:55:26,710
load; the load per unit length. So, length
is l minus x and w is per unit length - that
448
00:55:26,710 --> 00:55:30,790
is the
load. So, total load will be W into l minus
449
00:55:30,790 --> 00:55:34,060
x and that is any distributed manner.
So, moment will be, it is a uniform uniformly
450
00:55:34,060 --> 00:55:38,760
distributed load, means we can assume that
load is acting at the centroid of that part
451
00:55:38,760 --> 00:55:57,720
again. So, it is it will be acting half distance
of
452
00:55:57,720 --> 00:56:07,309
lyx. So, we can write omega is the intensity,
l minus x is the length, that multiplied by
453
00:56:07,309 --> 00:56:11,440
l
minus x divided by 2. So, that is one of the
454
00:56:11,440 --> 00:56:12,440
very familiar type of expression. If we start
from this side, well at omega into x that
455
00:56:12,440 --> 00:56:19,630
will be the load into x by 2, or straightaway
we
456
00:56:19,630 --> 00:56:23,950
will write omega x square by 2; in that case
x is not x, it is l minus x. So, it will be
457
00:56:23,950 --> 00:56:25,869
omega l minus x square divided by 2.
Now, we have to put some sign here again.
458
00:56:25,869 --> 00:56:26,869
At this moment, we have to decide which
sign convention you will take positively.
459
00:56:26,869 --> 00:56:27,869
If you take sign as positive, it should be
460
00:56:27,869 --> 00:56:30,260
negative, because it will give a falling type
of moment. So, if we follow the same thing,
461
00:56:30,260 --> 00:56:33,359
it will be minus and minus.
I think this part is clear - bending moment
462
00:56:33,359 --> 00:56:34,660
expression. Once that expression is ready,
we
463
00:56:34,660 --> 00:56:36,780
can use that moment-curvature relationship;
that directly we will put, the moment
464
00:56:36,780 --> 00:56:38,579
expression. So, there it will be EI d 2 y
dx square equal to minus M. So, it will be,
465
00:56:38,579 --> 00:56:39,579
we
can say it is minus M, and we can put here
466
00:56:39,579 --> 00:56:40,579
omega l minus x square divided by 2. So,
minus, minus; it will be plus. We are sagging;
467
00:56:40,579 --> 00:56:41,579
sagging we have taken positive, but here
moment is hogging, so we have put minus. And
468
00:56:41,579 --> 00:56:42,579
if it sagging as the positive moment,
expression will be with a minus M. So, this
469
00:56:42,579 --> 00:56:43,579
minus, minus it will be plus.
So, in a similar manner, we will integrate
470
00:56:43,579 --> 00:56:44,579
as we did in our previous problem. Iit will
be
471
00:56:44,579 --> 00:56:45,579
omega l minus x cube divided by 6 minus C
1. And EIY equal to omega l minus x 4. So,
472
00:56:45,579 --> 00:56:46,579
4 into 6 24, and minus and minus it will be
plus, C 1 x plus C 2. Absolutely same. Only
473
00:56:46,579 --> 00:56:47,579
the right side part is different. Once you
have that expression, our job is to determine
474
00:56:47,579 --> 00:56:48,579
C 1
and C 2 - these unknowns - through boundary
475
00:56:48,579 --> 00:56:49,579
condition. So, it is in a same manner at x
is
476
00:56:49,579 --> 00:56:50,579
equal to 0, so theta or dy by dx is equal
to 0. So, this is the first expression or
477
00:56:50,579 --> 00:56:51,579
first
condition. So, if we put here, I think. So,
478
00:56:51,579 --> 00:56:52,579
it will be EI into 0. It will be minus omega
l
479
00:56:52,579 --> 00:56:53,579
cube divided by 6 plus C 1 or we can say C
1 equal to omega l cube divided by 6. So,
480
00:56:53,579 --> 00:56:54,579
your C 1 part is available.
Next part is the C 2. So, next requirement
481
00:56:54,579 --> 00:56:55,579
is to find out C 2. And that we can find out
with the second boundary condition. That is
482
00:56:55,579 --> 00:56:56,579
deflection at this point - at x is equal to
0, it
483
00:56:56,579 --> 00:56:57,579
will be equal to 0.
484
00:56:57,579 --> 00:56:58,579
So, substitute that boundary condition - at
x equal to 0, y equal to 0. So, we will get
485
00:56:58,579 --> 00:56:59,579
EI
into 0. That will be omega l 4 24 plus C 1
486
00:56:59,579 --> 00:57:00,579
into 0 plus C 2 or we can get from here C
2
487
00:57:00,579 --> 00:57:01,579
equal to minus omega l 4 24 or these values
we can put in the expression of slope and
488
00:57:01,579 --> 00:57:02,579
deflection. So it will be 1 by EI minus.
So, C 1 and C 2 we have substituted the values
489
00:57:02,579 --> 00:57:03,579
in the expressions we obtained earlier for
slope and deflection, and it will be something
490
00:57:03,579 --> 00:57:04,579
like this - that is the expression of theta;
that is the expression of your deflection.
491
00:57:04,579 --> 00:57:05,579
Now, we will get the maximum deflection,
maximum slope. So, you here at x equal to
492
00:57:05,579 --> 00:57:06,579
l, theta will be theta max, and Y will be
Y
493
00:57:06,579 --> 00:57:07,579
max. So that part if we find out, so theta
x it will be 1 by EI, and here, this term
494
00:57:07,579 --> 00:57:08,579
will be 0;
it will be omega l cube by 6 or we can say
495
00:57:08,579 --> 00:57:09,579
omega l cube divided by 6 EI.
Now, if I find out Y max, 1 by EI; this part
496
00:57:09,579 --> 00:57:10,579
- it will be 0; and this part will be omega
l to
497
00:57:10,579 --> 00:57:11,579
the power 4 divided by 6, omega l to the power
4 divided by 24. So it is omega l to the
498
00:57:11,579 --> 00:57:12,579
power 4. So, 1 by 6 minus 1 by 24. So, if
you operate, it will be 24; this side it will
499
00:57:12,579 --> 00:57:13,579
be 4
minus 1. So 3. So, 3 24 means it will be 8.
500
00:57:13,579 --> 00:57:14,579
So, it will be omega l to the power 4 divided
by 8 EI. So, this is Y max; this is theta
501
00:57:14,579 --> 00:57:15,579
max. So, this is one of the quantity; that
is another
502
00:57:15,579 --> 00:57:16,579
important quantity. So, same problem with
a distributed load; we have tried to repeat
503
00:57:16,579 --> 00:57:17,579
that
process with a different moment expression,
504
00:57:17,579 --> 00:57:18,579
and got the values theta max, Y max.
Now, I will take a problem, where your type
505
00:57:18,579 --> 00:57:19,579
of structure will be different. So, beam will
not be a cantilever beam, say, we will take
506
00:57:19,579 --> 00:57:20,579
a simply supported beam. So, two supports
at
507
00:57:20,579 --> 00:57:21,579
the two end, and take some load, say, we can
take uniformly distributed load, whatever
508
00:57:21,579 --> 00:57:22,579
we have taken, at least load will be same
like this, but support condition will be different.
509
00:57:22,579 --> 00:57:23,579
So, this is a distributed load of intensity
omega or w; and the member is some EI; the
510
00:57:23,579 --> 00:57:44,369
flexible rigidity; this is the length; this
is x this is Y. So, our first job is to determine
511
00:57:44,369 --> 00:57:47,579
the
expression of bending moment. Now, if we start
512
00:57:47,579 --> 00:57:54,390
from this end or the other end, both the
sides supports are there. So, we have to know
513
00:57:54,390 --> 00:58:02,380
the value of those reactions.
Earlier case, support was on the left side
514
00:58:02,380 --> 00:58:04,910
and one free end was there. We were trying
to
515
00:58:04,910 --> 00:58:09,280
proceed from the free end; we need not bother
how much will be the reaction at the
516
00:58:09,280 --> 00:58:15,670
supports, but here we cannot start from this
end or other end; both end are supported.
517
00:58:15,670 --> 00:58:18,151
So,
first job is we have to find out the reaction.
518
00:58:18,151 --> 00:58:22,109
And that will be utilized for finding out
the
519
00:58:22,109 --> 00:58:33,050
bending moment expression. And that will go
to the differential equation. Now, here
520
00:58:33,050 --> 00:58:51,920
there will be one reaction and here. So, this
part we can say y; and this part we can say
521
00:58:51,920 --> 00:59:12,349
Ry; it is say Rx; let us say A, that A; this
is B; say left end if we say A, right end
522
00:59:12,349 --> 00:59:19,599
if we
say B; it is R along x at A; Ry y at A; this
523
00:59:19,599 --> 00:59:19,849
is RB.