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So, we are talking about energy method. Talking
about, means just we have started
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talking on this. The technique is energy method.
It is based on basically the concept of
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strain energy. The strain energy, we are trying
to explain with the form of a simple bar
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problem, under tension.
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So, we took a bar in that form and applied
a load P. Now, this is a case of a very simple
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stress problem, because entire bar will have
identical stress and strength. So, anywhere
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stress is P by A and strain will be your total
elongation divided by length of the member.
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So, it is a case of uniform strain, uniform
stress, whatever you can say.
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Now, the elongation, we wrote in the last
class it was Pl by AE, and we have drawn a
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curve like this; your load versus elongation,
straight line, because whole thing is a linear
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system. And for any load, if you start applying
load from 0, gradually increase, go up to
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a load P, that elongation will increase from
0 to delta, and the area under this load
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deflection curve, we have found that it is
basically the work carried out the force P.
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Now, it is not straight away P and delta,
because P is not constant throughout, because
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P
is starting from 0 to that P and it is delta
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is starting from 0 to delta. So, total work,
so that
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we have put in the form of half P into delta.
So, that work, it will be stored in the form
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of
some energy. We said this strain energy; that
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strain energy will be equal to this work
done half P into delta.
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Now, here, load why we are putting in a gradual
manner? Because we want to maintain
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the static condition. If we apply suddenly,
load whole thing will be dynamic problem.
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So, that will be much more difficult case.
So, we want to understand with a very simple
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type of problem. So, load in a static manner
if we want to apply, we have to apply
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gradually. So, always there will be balance
between internal-external force system.
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Basically, this strain energy, though it is
expressed in the form of P into delta, it
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is rather
the measure of some internal quantities; internal
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quantities means there will be
deformation that we have defined in the form
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of strain and due to the strain some stress
will be there. So, whole thing is a measure
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of the strain and stress inside. And you must
have remembered we have written this one as
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this sigma epsilon into v, because that P
we have divided by area, epsilon divided by
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l. So, it became stress; it became strain;
and
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this l and A, it becomes V, that is the volume,
or sometimes we write it is U 0 V. So, U 0
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is this part; this sometimes say it is strain
energy, density.
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So, half stress into strain is the strain
energy density multiplied by the volume will
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be the
total energy. Now, this is a case of uniform
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stress and strain. So, everywhere your sigma
and epsilon is constant. For a general case,
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we have written U equal to your half sigma
epsilon dv. Whereas, this was written for
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a case, for an arbitrary type of problem,
as
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stress strain will vary from point to point.
So, this is valid for this problem as stressstrain
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is uniform. So, if we take, again, where stress-strain
is not uniform, it is varying,
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so this concept is valid, for a small point,
for a small element. So, that concept applied,
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say, V become dV. So, all the element if we
combined, so we are supposed to get the
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expression of a strain energy is equal to
U equal to half integration sigma epsilon
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dV.
Now, it is basically strain energy. So, related
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to strain or related to stress. So, it is
basically internal thing. Though we have tried
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to explain at the beginning in terms of
external work, because external work, internal
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work in the form of strain energy we have
tried to equate, for applying a load in a
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gradual manner, in a static condition. But
in this
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strain energy concept is valid for dynamic
problem also. You apply load, your internal
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stress-strain it will gradually increase from
0 to a certain value. So, it will not be
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suddenly generated. So, this half term will
be coming there. Only thing, the internal
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system and external system will be not be
in equilibrium; naturally, it will be in a
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moving
condition, in a dynamic condition; otherwise,
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this concept will be there.
So, for a expression of energy, in a generalized
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form, we can say U equal to half sigma
epsilon dV. Now, here number of stress component
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is one, number of strain component
is one, because it is a one-dimensional problem.
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If we have a stress analysis problem
where number of stress component is more;
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say, if we take a plate it is under stretching;
so it will be a biaxial system of stress.
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So, it will be a plane stress type of problem.
So,
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there will be three stress components, two
normal stress components, one along x
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another y plus shear stress. Similarly, three
strain components will be there. So, in that
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case, we can define sigma x multiplied by
epsilon x plus sigma y multiplied by epsilon
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y
plus your tau xy into gamma xy. So, for a
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specific case, we can write in this form;
for a
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biaxial system of force, say, it is sigma
x, epsilon x, sigma y, epsilon y, tau xy,
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gamma
xy into dv. Say, if we take a thin plate,
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we can take it is dx dy; we can multiply it
by a
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thickness; t is the thickness of the plate.
So, this is a special case of strain energy
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for a
stress problem, where number of stress components
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are more. So, this idea can be
generalized.
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Now, at this moment, we are interested with
simple problems. So, it is only one
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component of stress. Now, this expression
of strain energy, we want to apply directly
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to
a bending problem, because we are handling
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a problem having some bending of the
beam or bending of the frame member. So, bending
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part we are mostly dealing at this
moment. So, this strain energy if we put in
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a bending mode, so what will be happening?
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So, if we write
strain energy in bending, we know any member,
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if it is under bending,
will get bending stress. And, that bending
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stress, it will vary from lower fiber to upper
fiber, in a gradual manner; it follows a linear
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manner; a neutral axis, it is basically 0;
and
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it will increase, it will go away from the
neutral axis. So, along that depth variation
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it
will vary. And along the width, normally,
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we assume it is uniform. So, depth wise
variation is there; width wise variation is
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not there; plus along the length it is minimum;
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it will vary; naturally it will vary. So,
if we think in terms xyz along length it will
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vary,
along depth it will vary, along width it will
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be constant. So, it is not a uniform state
of
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stress for the bending stress component.
Now, in bending, there is only one stress
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component; that is a bending stress. Now due
to
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shear or shear force there will be a shear
stress, but due to shear stress, we do not
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consider any shear strain. In elementary bending
theory, the effect of shear deformation
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normally we do not consider; this is one of
the assumptions. Because the beam theory
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what we have the understanding, we assume
a plane perpendicular to the neutral axis
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before bending, remain straight and normal
to the deflected neutral plane or neutral
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line
after bending. So, that assumption is based
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on that idea that we are not considering the
effect of shear deformation. Because shear
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stress, it varies in a parabolic manner. If
you
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take the effect of shear deformation, so it
will be not a normal line; it will not be
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a
straight line; and it will not be normal.
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So, the normal it will be violated plus it
will have
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a working of the whole section.
Normally, what happens, if we take a beam,
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its length is quite… the value of the length;
normally, it is higher compared to the depth
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and width of the beam. Now, that is mostly
common in our structural member. Now in that
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mode, so bending is predominant. So,
bending deformations is predominant; the effect
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of shear deformation is little less.
Just if we take a case, say, there is a structure
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like this, there is a support, there is a
support, and if we put some load here. So,
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along member, under load, so bending will
be
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predominant. Now, if we try to shift the supports;
so, if you shift the supports; if support
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is here; support is here. So, this support
will be shifted here; this support will be
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shifted
here. Your bending deformation will be reduced.
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In one case, say, the supports - both the
supports - will be tending towards the load.
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So, there are two supports; there is a load;
what will be the mode of failure? It will
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be just, it is shear mode of failure.
Just we can think there is a cantilever beam,
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or same this problem, we can think in terms
of cantilever. If we put a load here, it will
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try to bend. Now if you try to bring it here,
if
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you bring it here, what will be happening?
So, there will be certain shear of the whole
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thing, but if you put the load here, it will
undergo bending. So, fiber will start failing
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gradually, form top or bottom, but here once
the length will be reduced; so it depends,
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basically, the depth versus this length. So,
once this length is more compared to this,
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bending will be predominant, if this length
is very small compared to this. So, it will
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be
in a shear mode.
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We cut a piece of paper by scissors; means
two forces we are putting more or less on
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the
same point with the slight shifting. So, if
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we put two blades, it will never cut, it will
be
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under bending. So, when length is more, your
bending aspect will be predominant. When
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length will be small, it will be in shear
mode. This shearing of the rivet; two plates
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with
the rivet, if you try to pull, practically
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there is no gap, so it will shear. But if
you place
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here, there is a long neck, two plate; so,
whole thing will try to bend.
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So, it depends what is the size of our structure.
So, normally, our structure member will
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have a large span compared to the depth. So,
bending will be predominant, and normally,
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we take the deformation due to the bending
action. So, bending stress will produce
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bending strain and shear stress will be there;
but shear strain will be very small. And if
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you count, if it is a 1 percent or less than
1 percent, there is no point of making the
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analysis very complicated, because we have
to take the effect of shear, some other
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expression will be coming in to our equation;
we have to handle that; and maximum
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00:15:20,579 --> 00:15:26,150
benefit we will get, say, 1 percent or half
percent, if the span is quite large, so we
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do not
consider that.
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So, in bending mode, basically, though there
are two stresses, but the strain due to the
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00:15:34,060 --> 00:15:38,240
bending part will be predominant, that is
why we take the bending strain only. We do
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00:15:38,240 --> 00:15:41,059
not
consider the shear strain; shear strain means
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strain produced by the shear stress. So, two
stresses, and two strains. So, one of the
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00:15:48,640 --> 00:15:52,170
strain is basically 0, and energy is stress
into
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00:15:52,170 --> 00:15:59,080
strain. So, energy due to the shear force,
shear stress, shear strain - this part we
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can
simply neglect. So, it will be a one-dimensional
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problem; means, it is a problem with
only one stress component; that is a bending
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stress and its corresponding bending strain.
Now, the second aspect is the stress components
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will vary. As I am mentioned, it will
vary along the depth, along the length; only
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width wise we are keeping it more or less
uniform. Now, if we write the energy, so U
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00:16:32,419 --> 00:16:43,120
already we have written; it is sigma epsilon
dv, we can write dx dy dz. So, dv is dx dy
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dz. So, we are putting it in a separate form.
And here this sigma, for a bending problem,
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we know it is MY by I.
So, it is depending on y. So, if x is the
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length; x is the coordinate along the length
of the
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00:17:08,620 --> 00:17:13,439
member, along the axis of the member; y is
perpendicular to that along the depth; and
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00:17:13,439 --> 00:17:14,439
z
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00:17:14,439 --> 00:17:24,220
is along the width. So, sigma we can write
MY by I. Now, here, if I put the expression
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of
sigma, now what we will get? Now, strain is
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00:17:30,020 --> 00:17:41,940
also depended on stress. So, strain is stress
by E. So, this expression can be defined in
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00:17:41,940 --> 00:17:48,600
terms of stress entirely or it can be defined
entirely in terms of strain. So, this sigma
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00:17:48,600 --> 00:17:50,789
we can write as E into strain. So, it will
be
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00:17:50,789 --> 00:17:55,980
totally strain dependent or this we can express
in the form of sigma YE. So, it will be
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entirely stress dependent, but here we know
the expression of stress. So, it is better
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00:18:00,750 --> 00:18:04,790
to
write in the form of stress. So, we can write
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00:18:04,790 --> 00:18:20,250
in the second step half it is epsilon square
divided by E dx dy dz. So, it will be sigma
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sigma sigma square divided by E into dx dy
dz. Now, here we will put the value of sigma
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- that is MY by I.
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So, if I go to the next page. So, we can write
U equal to half integral. So, it is sigma
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00:18:42,350 --> 00:18:58,120
square; means, it will be M square, Y square,
I square, denominator E is there. So, dx dy
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00:18:58,120 --> 00:19:19,780
dz. Now, this half, this M square by I square
E, here you just observe. IE more or less
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00:19:19,780 --> 00:19:27,350
this quantity - will be a constant quantity.
Here M it will be fixed at a station; for
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00:19:27,350 --> 00:19:31,910
a
particular value of x it is constant.
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So, if we integrate about y and z. So, yz
means along the cross section of the member,
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00:19:38,370 --> 00:19:44,790
because this is x; it is y; it is z. So, along
the cross section if we integrate, along the
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00:19:44,790 --> 00:19:47,970
cross
section means at a particular station M it
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00:19:47,970 --> 00:19:50,790
will be constant. So, M will vary from from
one
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00:19:50,790 --> 00:19:57,240
station to another station. So, here we can
integrate. So, if we integrate about dy dz
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00:19:57,240 --> 00:20:02,200
about
yz, so here y square will be coming into the
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equation. So, it is what? dy dz, if we write,
it
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00:20:04,530 --> 00:20:14,230
is da. So, integrant y square da is what?
Second moment of area. So, we can write this
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one as I dx. So, this one and this one, if
we handle y square dy dz. So, that will lead
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00:20:26,180 --> 00:20:36,020
to
your I. Now, this I and this I will cancel.
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00:20:36,020 --> 00:20:51,260
So, it will be your M square by EI dx; rather
this is a very common expression for the energy
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in bending.
So, we have started with U equal to half sigma
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00:21:01,940 --> 00:21:08,930
epsilon dx dy dz. Epsilon we have
expressed in the form of sigma ye. And the
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00:21:08,930 --> 00:21:18,070
sigma we have written MY by I. So, this Y
square dy dz. So, that part is simply I. So,
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00:21:18,070 --> 00:21:28,020
ultimately it will lead to M square EI dx
integral half; it will be the energy.
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00:21:28,020 --> 00:21:33,920
Now, for bending mode. So, whatever bending
moment is there, square of that, divided
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00:21:33,920 --> 00:21:43,440
by EI into dx half, that will be the U. Now,
if you try to compare with your basic
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00:21:43,440 --> 00:21:54,600
expression of strain energy, that was half
sigma into epsilon into dx; here, sigma, you
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00:21:54,600 --> 00:22:03,340
can think as the analogous term of sigma is
bending moment M. And M square is M into
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00:22:03,340 --> 00:22:11,350
M, you can write this M into M by EI, M by
EI is what? d 2 y by dx square. So, d 2 y
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00:22:11,350 --> 00:22:16,770
by
dx square is what? Curvature. And d 2 y by
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00:22:16,770 --> 00:22:22,410
dx square curvature multiplied by EI is equal
to moment. So, we can say moment curvature
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00:22:22,410 --> 00:22:27,030
with a factor EI. So, EI is just like your
E;
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00:22:27,030 --> 00:22:31,200
and d 2 y by dx square curvature is strain;
and this is stress. So, it is not stress;
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00:22:31,200 --> 00:22:34,760
it is stress
resultant. And curvature is not the strain;
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00:22:34,760 --> 00:22:39,140
it is a generalized curvature. So, that thing,
in
198
00:22:39,140 --> 00:22:52,010
this form we are trying to express.
Once that is there, the energy, after beam
199
00:22:52,010 --> 00:23:00,680
under bending, now there is a theorem of
energy in terms of calculating the deflection.
200
00:23:00,680 --> 00:23:11,310
And that theorem was defined as a energy
theorem, and that was first proposed by Castigliano;
201
00:23:11,310 --> 00:23:26,260
we will say it is Castigliano’s
theorem. So, Castigliano. So, the energy principle
202
00:23:26,260 --> 00:23:33,170
is based on the theory proposed by
Castigliano. We say it is Castigliano’s
203
00:23:33,170 --> 00:23:43,630
theorem. So, what is that theorem? Basically
Castigliano has two theorems or whatever he
204
00:23:43,630 --> 00:23:48,020
has proposed it has two components. So,
one of the components, it will be utilized
205
00:23:48,020 --> 00:23:57,600
here. So, if we say U, if we write the
Castigliano’s theorem, so U you can express
206
00:23:57,600 --> 00:24:10,890
the energy of the system, we can express in
the form of the forces, because the energy
207
00:24:10,890 --> 00:24:20,530
here, energy is half M square EI dx.
So, if we take a beam problem, M we can find
208
00:24:20,530 --> 00:24:26,170
P into X omega X square by 2 something.
So, it will be expressed in the form of some
209
00:24:26,170 --> 00:24:39,320
force. Now, it can be also expressed in the
form of displacement. Now we are interested
210
00:24:39,320 --> 00:24:44,340
for the force part. So, U if it is a function,
211
00:24:44,340 --> 00:24:50,900
and it is defined in terms of some force,
it may be one force; it may be a number of
212
00:24:50,900 --> 00:24:53,160
forces.
213
00:24:53,160 --> 00:25:07,500
So, there is a beam problem; it is subjected
to P 1, P 2, P 3, P 4 like this. So, the bending
214
00:25:07,500 --> 00:25:10,800
for the different region, you can express
in the form of… because reaction would be
215
00:25:10,800 --> 00:25:15,309
in
terms of P 1, P 2, P 3, P 4, Pn, and that
216
00:25:15,309 --> 00:25:18,830
reaction into that distance, then the force
into
217
00:25:18,830 --> 00:25:24,960
some distance, automatically the moment expression
will be in the form of P 1, P 2, P 3,
218
00:25:24,960 --> 00:25:29,721
P 4, Pn. And energy expression is nothing
but your M square EI dx dy. So, whole thing
219
00:25:29,721 --> 00:25:35,830
will be express in the form these forces.
Now one of that expression will be available.
220
00:25:35,830 --> 00:25:46,590
Now, if we take partial derivative about U
for P 1, we will get delta 1; or if we just
221
00:25:46,590 --> 00:25:59,070
P 2, it
will be delta 2; or like this Pn, it will
222
00:25:59,070 --> 00:26:03,630
be delta n. Now, if the force is only 1 P.
So, energy
223
00:26:03,630 --> 00:26:09,280
will be expressed in terms of P. So, partial
derivative will be full derivative; it will
224
00:26:09,280 --> 00:26:13,070
be dU
by dP will be delta. So, what is delta? Delta
225
00:26:13,070 --> 00:26:18,850
is the deflection under the load P 1. So,
delta
226
00:26:18,850 --> 00:26:27,020
1 is deflection under P 1; delta 2 is deflection
under P 2.
227
00:26:27,020 --> 00:26:41,570
I think, if I draw a figure it will be better.
Say it is P 1; if it is P 2; if it is P 3;
228
00:26:41,570 --> 00:26:49,570
this will be
your delta 1; this will be delta 2; this will
229
00:26:49,570 --> 00:26:54,630
be delta 3; like this. Say, here I have just
drawn
230
00:26:54,630 --> 00:26:59,930
any arbitrary beam subjected to P 1, P 2,
P 3. So, reactions can be evaluated in the
231
00:26:59,930 --> 00:27:02,790
form
of P 1, P 2, P 3. So, moment expression you
232
00:27:02,790 --> 00:27:06,660
can find out in terms of P 1, P 2, P 3. So,
if
233
00:27:06,660 --> 00:27:11,860
moment is in terms of P 1, P 2, P 3 ultimately
your energy will be a function of P 1, P 2,
234
00:27:11,860 --> 00:27:12,860
P 3.
235
00:27:12,860 --> 00:27:17,670
Now, if you take derivative of U in terms
of P 1, you will get delta 1; delta 1 is the
236
00:27:17,670 --> 00:27:23,940
deflection under the load; and delta 2 is
deflection under the load P 2; delta 3 is
237
00:27:23,940 --> 00:27:26,590
the
deflection under the load P 3; like that you
238
00:27:26,590 --> 00:27:30,820
will get. So, if there is only one load P,
so if
239
00:27:30,820 --> 00:27:47,011
you take that derivative above that P, you
will get the deflection under that load.
240
00:27:47,011 --> 00:27:51,940
Now, say there might be a question - at the
middle there is no load, and we want to get
241
00:27:51,940 --> 00:28:02,350
the deflection here - is the theory valid
there? Should I apply this? Definitely, we
242
00:28:02,350 --> 00:28:06,950
can
apply. Here we have to put a load, say, P
243
00:28:06,950 --> 00:28:16,780
at the center. And we will apply this theorem;
at the end, we will put Pc is equal to 0.
244
00:28:16,780 --> 00:28:22,710
So, we will put PC as a unknown and apply
this
245
00:28:22,710 --> 00:28:28,990
theorem. Actually Pc, there is a no force,
so this value will be 0. At the end, we have
246
00:28:28,990 --> 00:28:31,890
to
apply. But initially we have to take it; otherwise
247
00:28:31,890 --> 00:28:37,860
you cannot proceed. I think those part, if
you take some example, it will be much more
248
00:28:37,860 --> 00:28:41,480
clearer.
249
00:28:41,480 --> 00:28:49,610
Say you start with the problem, which already
we have solved by other method. Now we
250
00:28:49,610 --> 00:29:02,820
know the result. Say it is a cantilever beam;
it is subjected to load P. It has some length
251
00:29:02,820 --> 00:29:09,240
l,
some E, some I, something is there. So, length
252
00:29:09,240 --> 00:29:16,510
is l, E the material property, I is the
second moment of area. Now, due to that load,
253
00:29:16,510 --> 00:29:19,970
say, deflection under that load we are
interested.
254
00:29:19,970 --> 00:29:27,580
So, first job is we have to find out the expression
of the bending moment. Now, you can
255
00:29:27,580 --> 00:29:32,460
start x from this way or x from that side.
So, it is better to start from this side,
256
00:29:32,460 --> 00:29:34,740
because
expression will be much more easier, P into
257
00:29:34,740 --> 00:29:38,559
X. This side, it will be P into l minus x.
So,
258
00:29:38,559 --> 00:30:00,130
if we start x from here, so M will be your
P into x, and our energy is half EI dx. What
259
00:30:00,130 --> 00:30:09,170
is
M square EI by dx? It will be half P square
260
00:30:09,170 --> 00:30:17,650
x square EI dx, limit will be 0 to l. So,
entire
261
00:30:17,650 --> 00:30:29,490
span, it is the expression. Now, this advantage
we may not get in other cases; if we put a
262
00:30:29,490 --> 00:30:33,760
separate load moment expression, it will be
different. So, the integration will be from,
263
00:30:33,760 --> 00:30:42,540
say, 0 to l by 2, l by 2 to l like this. Now,
this integration if we perform. So, it will
264
00:30:42,540 --> 00:30:55,370
be...
There are two options. We can perform the
265
00:30:55,370 --> 00:31:01,100
integration, after that we can take the
derivative; other option is we can take the
266
00:31:01,100 --> 00:31:04,760
derivative - so, there are two options. Now,
it
267
00:31:04,760 --> 00:31:13,801
will give the same result.
Say, we put it in that manner, say, half M
268
00:31:13,801 --> 00:31:21,790
square part we can write as a M into M, or
let
269
00:31:21,790 --> 00:31:37,480
us put the derivative first: dU by dP that
will be the delta. Now, if you take the
270
00:31:37,480 --> 00:31:49,830
derivative, so here what will be happening?
Here P square will be there. P square will
271
00:31:49,830 --> 00:31:57,130
be
2 P. Or this expression we can little bit
272
00:31:57,130 --> 00:32:00,600
change, that will be easier to understand.
P square
273
00:32:00,600 --> 00:32:14,540
x square we can write as P into x. So, here
moment is P into x P into x whole square of
274
00:32:14,540 --> 00:32:24,740
that EI dx. Now, delta dU by dP, it will be
this square term will be 2 into that Px. So,
275
00:32:24,740 --> 00:32:34,210
here this 2 if you bring outside, so it will
be Px plus derivative of Px will be in terms
276
00:32:34,210 --> 00:32:35,580
of P
will be x.
277
00:32:35,580 --> 00:32:43,200
We are taking derivative about P not about
x. So, that divided by EI into dx and
278
00:32:43,200 --> 00:33:00,130
definitely it will be 0 to l. So, our energy
half M square EI dx. So, half here M we just
279
00:33:00,130 --> 00:33:09,840
put Px whole square. Now, delta is derivative
of U about P. So, this part 2 we have taken
280
00:33:09,840 --> 00:33:18,059
out. So, 2 and 2 it will cancel, and this
will be the Px, and it will be x. Now, what
281
00:33:18,059 --> 00:33:25,710
is this?
This Px is basically M and x is derivative
282
00:33:25,710 --> 00:33:33,470
of M with respect to P. So, it is derivative
of M
283
00:33:33,470 --> 00:33:44,760
with respect to P. So, that part we can explain
later also. So, that will be the expression.
284
00:33:44,760 --> 00:33:54,320
Now, if that is the expression, so it is Px
x, Px square divided by EI into dx. So, it
285
00:33:54,320 --> 00:34:00,380
will be
P if we take out, EI if we take out, it is
286
00:34:00,380 --> 00:34:09,440
basically integral dx 0 to l. So, integral
0 to lx
287
00:34:09,440 --> 00:34:16,280
square dx would be your x cube by 3, and limit
if you put, it will be l cube by 3; other
288
00:34:16,280 --> 00:34:27,059
part 0 it will be cancelled. So, whole thing
will be leading to Pl cube 3 EI. And that
289
00:34:27,059 --> 00:34:35,129
expression we have to determine by moment-area
theorem or differential equation
290
00:34:35,129 --> 00:34:41,720
technique; it is a standard expression. So,
here we obtain through your energy principle,
291
00:34:41,720 --> 00:34:52,030
we have written the expression of M, under
U we have put there, and taken derivative.
292
00:34:52,030 --> 00:35:10,190
Now, I shall take a different case. What I
was trying to tell little bit earlier, say,
293
00:35:10,190 --> 00:35:15,840
there is a
similar type of cantilever beam, what it is
294
00:35:15,840 --> 00:35:19,660
subjected to a distributed load omega, and
we
295
00:35:19,660 --> 00:35:27,760
are interested to finding out the deflection
here; under the free end there is no load.
296
00:35:27,760 --> 00:35:31,110
In
that case, there is a load; under the load,
297
00:35:31,110 --> 00:35:34,320
we have calculated the deflection. Now, I
am
298
00:35:34,320 --> 00:35:39,790
just talking about the simply supported beam,
span, deflection, no load. So, more or less
299
00:35:39,790 --> 00:35:44,300
similar situation; fully distributed load;
we want to find out and practically there
300
00:35:44,300 --> 00:35:47,290
is no
physical load there.
301
00:35:47,290 --> 00:35:58,050
Now, this problem, we will just put a force
here P and this force is nothing but will
302
00:35:58,050 --> 00:36:07,060
be
equal to 0. Now, in a similar manner, if we
303
00:36:07,060 --> 00:36:17,450
take x axis from this side, we will have a
moment expression of M of Px omega x square
304
00:36:17,450 --> 00:36:29,320
divided by 2. At any x it is P into x into
omega x square by 2; it will be plus; both
305
00:36:29,320 --> 00:36:34,350
are in the in the same direction. Now, this
P is
306
00:36:34,350 --> 00:36:40,470
0. So, at the beginning, if I put P equal
to 0, this part will not be there. Why it
307
00:36:40,470 --> 00:36:43,920
is
important? That if we carry out that problem,
308
00:36:43,920 --> 00:36:48,800
it will give that field. So, our U is equal
to
309
00:36:48,800 --> 00:37:18,320
half M square EI dx; that part we know. So,
we can write, here it is Px. So, just I have
310
00:37:18,320 --> 00:37:41,330
substituted the value of M Px plus omega x
square by 2 here. Now, next job is, one
311
00:37:41,330 --> 00:37:46,040
option is we can perform the integration finally,
take the derivative. Other option is we
312
00:37:46,040 --> 00:37:49,840
can take the derivative, and then, perform
the integration. I think second option is
313
00:37:49,840 --> 00:38:05,230
better.
So, delta it should be du by dp. Now this
314
00:38:05,230 --> 00:38:14,940
will be half. So, this 2 will come out, and
here
315
00:38:14,940 --> 00:38:23,460
it will be Px omega x square by 2. So, we
are taking derivative of that quantity. So,
316
00:38:23,460 --> 00:38:24,460
it
317
00:38:24,460 --> 00:38:31,280
was square 2 into that quantity, 2 we are
putting it outside, and we have to take the
318
00:38:31,280 --> 00:38:40,340
derivative of this quantity with respect to
P. So, it will be straightaway x divided by
319
00:38:40,340 --> 00:38:56,440
EI
dx. So, this term was with square. Now, we
320
00:38:56,440 --> 00:39:02,119
are taking derivative of that. So, something
square, it will be 2 into that quantity, plus
321
00:39:02,119 --> 00:39:07,310
inside quantity we have to take derivative.
So, if that internal quantity if you take
322
00:39:07,310 --> 00:39:10,710
derivative of P, so it will be simply x, this
part; it
323
00:39:10,710 --> 00:39:15,970
will be constant; w may be 10 kilo Newton
per meter or something. So, it is not a
324
00:39:15,970 --> 00:39:28,740
variable quantity. Now, at this level, now
you can go ahead with the integration or at
325
00:39:28,740 --> 00:39:34,100
this
level we can substitute P is equal to 0, before
326
00:39:34,100 --> 00:39:43,560
that we have to keep P, because this part
we have taken square. So, it is there after
327
00:39:43,560 --> 00:39:46,620
derivative. Now, once the internal part we
have
328
00:39:46,620 --> 00:39:53,680
taken derivative, we were getting x. So, x
it is coming from Px and Px is this one; though
329
00:39:53,680 --> 00:39:58,370
this part is not, there P is equal to 0. So,
at the beginning if we put 0, if we take
330
00:39:58,370 --> 00:40:03,690
derivative, nothing will come. So, artificially
we have to put a P and it will carry. So,
331
00:40:03,690 --> 00:40:09,050
it
will give x here. So, at this level we can
332
00:40:09,050 --> 00:40:11,820
substitute 0 or we can integrate, later on
we can
333
00:40:11,820 --> 00:40:21,150
put 0, anything you can make it.
Now, say let us keep it. So, this will cancel.
334
00:40:21,150 --> 00:40:27,230
So, it will be one part will be, we can say,
let
335
00:40:27,230 --> 00:40:37,000
us put the integral it will be Px square.
Another part is omega x cube divided by 2,
336
00:40:37,000 --> 00:40:46,230
whole
quantity divide by EI into dx; or we can write
337
00:40:46,230 --> 00:40:57,550
it will be Pl cube divided by 3 EI plus. So,
this x square x square will be x cube divided
338
00:40:57,550 --> 00:41:00,690
by 3. If we put the limit, it will be l cube
by
339
00:41:00,690 --> 00:41:13,000
3 divided by EI, and here, this omega x cube
will be x 4 divided by 4. So, it will be
340
00:41:13,000 --> 00:41:55,619
omega l 4 divided by 8 EI.
Now, at this level, we can make P is equal
341
00:41:55,619 --> 00:41:57,770
to 0. So, this part will be the deflection.
So,
342
00:41:57,770 --> 00:42:18,619
what will be the deflection? How much we have
derived earlier? We can check it out; no
343
00:42:18,619 --> 00:42:41,071
problem. It is ok. It is 8 here. So it is
ok. So, I was suspecting there might be some…
344
00:42:41,071 --> 00:42:48,220
because we are calculating, we drop some 2,
so the whole thing will be a different value.
345
00:42:48,220 --> 00:42:55,880
Now, the actual problem was omega and the
deflection under the load it should be wl
346
00:42:55,880 --> 00:43:01,240
to
the power 4 divided by 8 EI. Now, this will
347
00:43:01,240 --> 00:43:11,369
be equal to… because your P is equal to
0.
348
00:43:11,369 --> 00:43:25,300
Now, I kept it intentionally, because in previous
problem we have determined that
349
00:43:25,300 --> 00:43:29,740
deflection is Plq by 3 EI. So, this Plq by
3 EI part is also there.
350
00:43:29,740 --> 00:43:37,300
Say a beam is subjected to this load plus
this load omega plus P. So, this part is due
351
00:43:37,300 --> 00:43:40,450
to
the load - point load; this part is due to
352
00:43:40,450 --> 00:43:43,270
uniformly distributed load. Noe, if that part
is not
353
00:43:43,270 --> 00:43:51,170
there, it will be dropped out; if that part
is there… So, this is the value. So, I could
354
00:43:51,170 --> 00:43:52,170
put it
355
00:43:52,170 --> 00:43:56,380
0 earlier, I think. So, we need not carry
out one of the component. So, unnecessarily
356
00:43:56,380 --> 00:43:59,900
I
have kept it, just get the feeling that on
357
00:43:59,900 --> 00:44:12,280
this part is also there; it will come after.
So, the objective of taking this problem is,
358
00:44:12,280 --> 00:44:18,530
say, there is no… we want to get the
deflection and we found there is no force
359
00:44:18,530 --> 00:44:21,040
there. So, you need not bother; you take a
force
360
00:44:21,040 --> 00:44:29,201
P; go ahead, you know it will be 0. So, you
write the expression in the form of P; you
361
00:44:29,201 --> 00:44:32,430
put
it there; take the derivative; after taking
362
00:44:32,430 --> 00:44:37,231
derivative, at this level you could put it
is 0; but
363
00:44:37,231 --> 00:44:45,270
if you put 0 at the beginning, you are not
supposed to get this one. So, for getting
364
00:44:45,270 --> 00:44:48,040
this
one, at least at this level you have to keep,
365
00:44:48,040 --> 00:44:51,130
after taking the derivative, then only you
can
366
00:44:51,130 --> 00:44:54,920
make P is equal to 0; before that you cannot
make it.
367
00:44:54,920 --> 00:45:03,760
Now, it may be some value; positive, negative,
and 0; up to there you have to maintain;
368
00:45:03,760 --> 00:45:10,800
after that you have to really substitute the
value. If there is a problem of say 10 kilo
369
00:45:10,800 --> 00:45:15,760
Newton, so you have to write P, because you
have to take the derivative; you cannot take
370
00:45:15,760 --> 00:45:21,430
10, then take the derivative above 10. So,
you cannot make del U by del 10. You have
371
00:45:21,430 --> 00:45:24,550
to
give some variable name. So, it will be P.
372
00:45:24,550 --> 00:45:28,790
So, at this level you can substitute the value.
So, here integration will be only in terms
373
00:45:28,790 --> 00:45:34,610
of x. Here w, P, everything you can put
numerical value; 0 or any non 0 value; then
374
00:45:34,610 --> 00:45:40,230
integrate; you will get the value of
deflection.
375
00:45:40,230 --> 00:45:50,190
Now, if we take the problem of rotation, slope,
we are talking about only deflection.
376
00:45:50,190 --> 00:45:54,250
Now, there might be a feeling that this method
will be valid for finding out the deflection
377
00:45:54,250 --> 00:46:00,020
only; it cannot be used for slope; it is also
valid for finding out the slope. And we will
378
00:46:00,020 --> 00:46:10,420
take a case like this, where there is no direct
load under this. So, if we want to get slope,
379
00:46:10,420 --> 00:46:15,330
so what should be the corresponding force?
There should be a moment and if we take a
380
00:46:15,330 --> 00:46:29,310
case where there is no applied moment, say,
same problem we can take.
381
00:46:29,310 --> 00:46:44,280
So, it has l, E, I; these are all. Now what
will be the rotation at this point - we are
382
00:46:44,280 --> 00:46:56,250
interested. Now, here just we will put something,
say, MB. Say if it is A, if it is B, I
383
00:46:56,250 --> 00:47:05,920
could put it is M, but we will write expression
of M as M. In order to have a different
384
00:47:05,920 --> 00:47:13,790
notation, say, this end is A this end is B.
So, there we put MB, and we know these values
385
00:47:13,790 --> 00:47:18,680
will be equal to 0.
But we have to put that and we have to get
386
00:47:18,680 --> 00:47:23,940
that expression for the bending movement.
So, expression of M, at any station, it will
387
00:47:23,940 --> 00:48:07,220
be MB omega. So, your U half. So, just write
MB. Now, if we calculate delta, it is basically
388
00:48:07,220 --> 00:48:18,640
du by here it will be MB. So, we are
interested for finding out the slope at B.
389
00:48:18,640 --> 00:48:23,920
Corresponding to slope, we are suppose to
give
390
00:48:23,920 --> 00:48:32,690
some movement M. So, it is at B MB. So, this
delta here is nothing but your, we can
391
00:48:32,690 --> 00:48:50,380
write, these are there, write theta B. So,
theta B will be dU by d MB. Now, this half
392
00:48:50,380 --> 00:48:59,200
0 to
l. So, this part will be MB omega x square
393
00:48:59,200 --> 00:49:10,750
by 2 into 1 into EI into dx. And there will
be a
394
00:49:10,750 --> 00:49:22,330
a 2, that we can put outside. So, this 2 2,
we can cancel. So, square 2 into this quantity
395
00:49:22,330 --> 00:49:26,130
and if you take derivative about MB; MB is
MB; so, it will be 1.
396
00:49:26,130 --> 00:49:32,570
Now, MB we have to carry at least for getting
1; other wise, if that is not there, whole
397
00:49:32,570 --> 00:49:38,710
things should be equal to 0. Now, you can
keep MB or we can make MB is equal to 0 at
398
00:49:38,710 --> 00:49:44,859
this level and perform the integration. So,
if MB is 0, this part will not be there. So,
399
00:49:44,859 --> 00:49:57,420
remaining part we can take. So, this will
be equal to integral 0 to l omega x square
400
00:49:57,420 --> 00:50:09,000
2 EI
dx. So, it will be x square, means x cube
401
00:50:09,000 --> 00:50:15,780
divided by 3. And if you put the limit omega
l
402
00:50:15,780 --> 00:50:29,790
cube 3 into 2 6 EI; I think this value also
we can check. So, distributed load slope omega
403
00:50:29,790 --> 00:50:37,490
l cube by 6 EI.
Now, here after this level, MB we have made
404
00:50:37,490 --> 00:50:46,750
0. So, only one term we have kept and we
are getting this value. So, this problem,
405
00:50:46,750 --> 00:50:50,071
we are interested not deflection, it is slope,
and
406
00:50:50,071 --> 00:50:55,020
there should be a moment. First of all, we
should know there should be a moment there.
407
00:50:55,020 --> 00:51:02,560
So, force is corresponding to displacement
and moment is corresponding to slope. So,
408
00:51:02,560 --> 00:51:08,410
there should be some corresponding moment
and there is no moment. So, we have put
409
00:51:08,410 --> 00:51:13,869
arbitrarily one moment MB, though the value
is equal to 0. We have processed that and
410
00:51:13,869 --> 00:51:16,260
finally, we got the value omega l cube by
6 EI.
411
00:51:16,260 --> 00:51:18,849
Sir, before integrating MB you already considered
it as 0.
412
00:51:18,849 --> 00:51:21,660
Yes.
Can we assume?
413
00:51:21,660 --> 00:51:29,780
Not assume; it is actually 0. So, at this
level, we can afford for putting the value
414
00:51:29,780 --> 00:51:36,140
MB is
equal to 0.
415
00:51:36,140 --> 00:51:40,360
Sir, for finding slope we should take a moment
and then we can find out at that point?
416
00:51:40,360 --> 00:51:43,140
So, there are two aspects here: number 1,
for getting deflection we have to find…we
417
00:51:43,140 --> 00:51:48,270
have to think for a force; there should be
a force corresponding to that. So, in that
418
00:51:48,270 --> 00:51:53,140
direction, we want a displacement; there should
be a force in that direction. If we want to
419
00:51:53,140 --> 00:51:59,320
get a slope along this; so, there should be
a moment along this. If there is a force or
420
00:51:59,320 --> 00:52:02,280
there
is a moment they are fine; we can go ahead,
421
00:52:02,280 --> 00:52:07,140
but that should be in a variable form. And
finally, we will actually put the real value
422
00:52:07,140 --> 00:52:15,350
of that; it may be 10 Newton or 100 kilo
Newton meter. But if it is not there, we have
423
00:52:15,350 --> 00:52:18,250
to take the thing in the variable form, but
at
424
00:52:18,250 --> 00:52:27,670
the end, we have to put that is equal to 0.
So, this level or up to this level, it will
425
00:52:27,670 --> 00:52:31,270
be there. In the next level, we can make it
0; this
426
00:52:31,270 --> 00:52:37,000
level or at the end also we can make it 0.
So, if you do not take that this is 0 at this
427
00:52:37,000 --> 00:52:40,260
level,
it should be MB into x integration; after
428
00:52:40,260 --> 00:52:43,640
putting the limit it should be MB into l;
it should
429
00:52:43,640 --> 00:52:53,300
be Mb into l. Now you could put MB equal to
0. So, finally, it should be 0. Or if you
430
00:52:53,300 --> 00:53:02,680
retain, there should be a case combination
of there is a moment, there is a distributed
431
00:53:02,680 --> 00:53:05,650
load.
432
00:53:05,650 --> 00:53:27,010
Now, so, with that, at least we came to that
level that we have to write the expression
433
00:53:27,010 --> 00:53:29,700
of
bending moment in terms of force or moment;
434
00:53:29,700 --> 00:53:37,080
then, put in that energy expression; then,
take the derivative; at that level, we can
435
00:53:37,080 --> 00:53:39,053
really put the value of the load and perform
the
436
00:53:39,053 --> 00:53:48,900
integration; finally, we will get the deflections.
Now, here, this energy theorem, what we are
437
00:53:48,900 --> 00:53:54,790
doing, you try to observe here. U equal to
M square; we are putting it here. So, after
438
00:53:54,790 --> 00:53:57,790
taking derivative, this part will be there
into
439
00:53:57,790 --> 00:54:11,100
this one. Now, what is this quantity? This
quantity is basically M. This one. So, this
440
00:54:11,100 --> 00:54:13,859
was
M square. So, after taking derivative you
441
00:54:13,859 --> 00:54:21,890
are basically getting M and this part is
derivative of M with respect to MB, MB or
442
00:54:21,890 --> 00:54:30,500
P anything. Except this one, earlier problem,
so, it was M square; this is M; and this part
443
00:54:30,500 --> 00:54:37,650
is derivative of this with respect to P.
Artificial or real?
444
00:54:37,650 --> 00:54:45,589
Artificial or real, it should be the force
or moment where we want to get the deformation
445
00:54:45,589 --> 00:54:53,160
or slope. So, whole thing can be redefined
in a little different manner and from there
446
00:54:53,160 --> 00:54:58,330
the
concept of unit load will come. So, that part,
447
00:54:58,330 --> 00:55:22,210
we can take care in the next class.
Preview of the Next Lecture
448
00:55:22,210 --> 00:55:29,339
And I was talking there is a method called
unit load method. It is basically derivative
449
00:55:29,339 --> 00:55:33,690
of
the, your unit energy method. So, energy method,
450
00:55:33,690 --> 00:55:42,119
if we look in a separate angle, we will
get this method, which we are trying to define
451
00:55:42,119 --> 00:55:51,079
as unit load method.
452
00:55:51,079 --> 00:55:59,201
So, it is not a separate method. It is basically
energy method, but looking the whole thing
453
00:55:59,201 --> 00:56:09,270
is looked in a different angle. Idea is to
make the treatment much more systematic or
454
00:56:09,270 --> 00:56:16,000
much more mechanical. So, we need not start
from the very basic thing. So, energy
455
00:56:16,000 --> 00:56:24,339
principle is basically its background, but
in a different form we want to apply. Rather
456
00:56:24,339 --> 00:56:30,230
I
was trying to explain you here.
457
00:56:30,230 --> 00:56:37,820
This problem we have solved. M square part
we have put there, after taking derivative
458
00:56:37,820 --> 00:56:43,859
we are getting this plus another component.
So, this part was basically the moment and
459
00:56:43,859 --> 00:56:53,200
this part is the derivative of that moment
about that P. Now, if we write here, in general,
460
00:56:53,200 --> 00:57:15,470
U equal to half integral M square EI dx. So,
delta, we are trying to tell this is du, dU
461
00:57:15,470 --> 00:57:30,820
by
dP; it will be this half 2 M dM dP divided
462
00:57:30,820 --> 00:57:40,210
by EI dx. So, U is equal to half M square
EI
463
00:57:40,210 --> 00:57:51,620
dx, where this M was say function of P . So,
P I am writing; it is not the P what we have
464
00:57:51,620 --> 00:58:00,849
considered earlier; P may be P MB or any other
quantity. So, it is a variable. So, if P is
465
00:58:00,849 --> 00:58:10,420
some function of, if M is function of P, so
M square by EI dx; this part if we take
466
00:58:10,420 --> 00:58:17,710
derivative, so it will be dU by dP. So, here,
this M square will be 2 M, 2 it will cancel,
467
00:58:17,710 --> 00:58:21,820
it
will be M, and this part will be dM by dx
468
00:58:21,820 --> 00:58:26,560
EI into dx. Now, this part, we can write as
this
469
00:58:26,560 --> 00:59:48,999
is capital M, and this derivative part, if
we write small m divided by E EI into dx.