We continue with our discussions on fluid
statics and what we will first do today is
we will see some other mechanisms by which
you can measure the pressure of the fluid.
In the last class we saw the example of barometer
today we will first see the example of a manometer.
Since this concept is well known to you, we
will briefly recapitulate it through an example.
Let us say that there is a pipe and there
is some fluid which is entering the pipe.
There are 2 different sections say mark by
A and B and we are interested to find out
what is the difference in pressure across
these 2. To do that what we will do we will
be 
considering a U tube connected across these
sections.
This U tube will be the so called manometer
and it will try to measure the pressure difference
between these 2. To see how it actually does
let us say that this device contains some
other fluid, of course there is a fluid which
is passing through the pipe say that fluid
is water. Now the fluid which is a part of
these U tube manometer it is something different
because it is the difference in density of
these 2 fluids which will dictate the principle
of operation of this device.
Let us say that, that fluid is mercury as
an example. Mercury is a very common fluid
for manometers and we just take that as a
specific example. Now when the fluid is flowing
from A to B which side you expect to be a
higher pressure A or B? You expect A to be
a higher pressure, we will see later on that
it is not always true that fluid will flow
from higher pressure to lower pressure it
will definitely flow from higher energy to
lower energy.
And here other components of energy being
unaltered as an example like the kinetic energy
and potential energy. Therefore, the pressure
differential is the sole driving parameter
for the fluid flow to take place. If A has
the pressure greater than B and if we have
some fluid here which is not water but some
other fluids say mercury on which side you
expect it to be more depressed, on A side
or on B side?
So, wherever the pressure is more it will
depress it more. Let us make the sketch accordingly
and let us say this is mercury and let the
difference in level in the 2 sides or the
2 limbs of the manometer will be delta h.
What is our objective? Our objective is to
find out what is the difference in pressure
between A and B. So, we want PA – PB that
to be evaluated. To do that we will just considers
some reference dimensions let us say that
this is a Z1, let us say this is Z2.
The difference between these 2 of course is
delta h. What fundamental principle we will
use here? We will use the fundamental principle
that in the absence of any other body force
when the fluid is at rest within this manometer
the pressure in a connected system is same
along the horizontal line it does not change.
So, if you mark this point say C and D you
must have the pressure at C and pressure at
D as same.
So, how can you write the pressure at C in
terms of pressure at A it is = pressure at
A + C1 * the density of water because this
is filled up with water except for this red
dots the remaining is water. So, this into
rho water * g that is the pressure at C. What
is the pressure at D? It is because of a combination
of some column of water and some column of
mercury. What is the column of water? So,
first pressure at B + Z2 rho water g + delta
h rho of mercury * g.
From here it follows that PA – PB = you
can write this as delta h rho mercury g – 
it will be the rho water g taken as common
* Z1 – Z2. Z1- Z2 is delta h as you can
see from the figure. So, it will be delta
h rho water g. So, it is delta h rho water
g if you take as common it is also written
in this form. So, you can clearly see that
by measuring what is delta h this you can
experimentally measure.
The difference in the height of the 2 limbs
you can clearly find out what is the difference
in pressure between 2 points. What is the
factor that is dictating your result is the
relative density of 1 fluid over the other.
And of course, not only that what is the density
of the original fluid that is also important.
So, this is in a very simple way what is the
working principle of a manometer?
You may have manometers in series connected
from one point to the other point and if you
want to find out the pressure difference between
2 points what you may do is you may start
with the point and traverse along the manometer.
When you were traversing along the manometer
along the same horizontal level if you have
the same connected system you may consider
the pressures to be qual.
So, you may neglect whatever is there below
that and that is how you may go from 1 point
to the other calculate the pressure difference.
Go on calculating the pressure difference,
it is like coming down and going up like a
snake ladder game. So, you come down somewhere
find out what is the increasing pressure again
go up find out what is the decreasing pressure
so on.
Along whatever fluid you are traversing you
have to consider the density of that corresponding
fluid for calculating the pressure differences.
You can clearly see that these devices although
we say these are pressure measuring devices
but that is a very loose way of looking into
these because these actually measure pressure
differences not really absolute pressures.
So, whenever you have a fluid you expect that
they are may be different pressures at different
points and these mechanism is only trying
to find out pressure at one point relative
to the other not in a very absolute sense.
The other important remark is if you see that
the resolution of these device strongly depends
on what is this delta h. Because if this is
very small there can be a lot of reading error.
And that will magnify the error in the determination
of what is PA – PB. So, if this is quite
large or magnified it may be easier for us
to read with better accuracy. To do that sometimes
people use inclined tube manometers. To understand
the working principle we will again take an
example.
In an inclined tube example it did not always
be a U tube. It is basically having a measuring
tube it may be a collection of measuring tubes
may be U tube but the axis of the tube is
inclined with the vertical. It is not vertical.
So, let us say that you have a tank like this
and you have a tube connected to it in this
way.
Initially everything was exposed to atmosphere
and let us say that this red colored dotted
line represents the initial height both in
the inclined limb as well as in the tank.
Let us say that capital D is the diameter
of the tank and small d is the diameter of
the tube, assume both of circular cross section.
Now let us say that you apply a pressure differential.
That is you apply a higher pressure from the
top.
If you do that this level will be depressed,
it will come to a new location let us say
it comes to this location. When it comes to
this location obviously there is a drop in
height and where this extra liquid will go?
It will climb up the incline. So, it will
come say to some height like this and therefore
now have a net difference in the level as
say L. L is the new level difference, earlier
there was no level difference.
But because of the application of pressure
differentials delta P the fluid in the big
tank has gone down the same volume has gone
up along the inclined tube and the level difference
now is L. This is inclined difference and
that is what you can read. If you graduate
with lines on or markers on this inclined
tube you can read the length very easily just
like a scale. Let us say that you have the
angle of inclination as theta with the horizontal.
Now one important principle that may guide
you that what should be the corresponding
rise if you have length up to this much say
L1then you can clearly relate L1 with say
that distance by which the level has gone
down in the big tank. Let us say that that
is delta h. Since the big tank, this may be
small seen this is of smaller cross section
the change will be large. So, we can say from
the conservation of volume that delta h * pi
capital D square by 4 = what?
L1 * pi small d square by 4. So, this is from
the consideration that the volume at state
1 is same as the volume at state 2 that is
what is the principle that is guiding this
very simple expression. From here you can
find out what is L1. So, that is delta h * capital
D square by small d square. Now, what is the
difference in pressure now between the 2 levels?
It is that difference delta P.
Because if one end is exposed to atmosphere
then the other end is subjected to a pressure
of say difference of delta P then that should
give rise to the difference in level. So,
delta P will be what? Delta P will be L sin
theta * rho *g. You can replace L with what
L1 + delta h/sin theta and L1 you can write
in terms of delta h. So, these you can write
as delta h * capital D square/small d square
* sin theta + delta h rho g.
You can take certain terms as common. So,
delta h rho g it will come to. You may of
course replace delta h with L1 again because
at the end it is not delta h that you are
measuring. It is L1 or L that you are measuring.
So, you can write it express in terms of 
say L. So, either way I mean, mathematically
you can express either in terms of delta h
or in terms of L.
But when you come to the final expression
it is more convenient if you express in terms
of L because that is what you can experimentally
read more clearly and that is the whole objective
of keeping it incline. So, when you keep it
incline you can see that the for the same
vertical height you can get more inclined
distance and pressure differential is dictated
not by the inclined distance but the vertical
height.
So, for the same vertical height we will have
the same pressure difference but with more
inclination you can have more inclined length
and that is how you may have a greater length
for greater readability or for better readability
for the pressure difference measurement. If
you have say the same system but with vertical
tube say with water as a fluid then that will
be equivalent to some h water * rho water
* g.
So, the L/h water is an indicator of the sensitivity
of this device. That h water for a vertical
tube may be very small but for an inclined
tube the corresponding L which shows the same
pressure difference may be quite large. So,
it adds to the sensitivity of the device.
So, L/ that equivalent h water is also an
indicator of the sensitivity of the device.
Now it is not just the manometer that is commonly
used for measuring pressure differences.
If the pressure differences are not very large
sometimes there are inexpensive means of measuring
pressure differences and one such example
is a pressure gauge. So, pressure gauge it
is an indicator of pressure at a point relative
to some other reference pressure. Very often
that reference pressure is the atmospheric
pressure and therefore it implicitly many
times reads the gauge pressure.
We have earlier defined that the gauge pressure
is the pressure relative to atmospheric pressure
at that location. So, we see an example with
a demonstration to see pressure gauge.The
example that we will be seeing here it is
known as a bourdon gauge. So, if you look
at this one, we will see the example once
more. But if you see that it has a tube and
the tube is connected to a mechanical arrangement
and one end of the tube is fixed and the tube
is of say elliptical cross section. It is
a deformable tube. When fluid enters the tube
then what happens?
When fluid enters the tube the tube gets deformed
and the section of the tube tends to get more
circular as compared to the elliptical one.
So, what is happening is there is a fluid
that is entering the tube one end of the tube
is fixed the other end of the tube is moving
or movable. As the fluid enters the tube there
is a deformation in the tube and that deformation
is being read by a dial indicator.
So, the deflection of the indicator in the
dial gauge it gives an indicator of the deformation
of the tube and the deformation of the tube
has taken place because of application of
a pressure differential. Earlier it was 0
deformation or a base state deformation now
with application of a pressure difference
there is a change in deformation from that.
So, if this is calibrated then one may calibrate
the deformation as a function of the applied
pressure difference.
It is just like calibrating a spring because
of application of a force. So, it is just
like a spring mass type of arrangement. So
this, name of this device is bourdon gauge.
So, that is an interesting on a simple way
of measuring pressure. The more advance ways
of measuring pressure are not these ones.
The more advance ways are by utilizing principles
of certain things known as transducers.
So, what the transducers do they convert may
be one form of signals and mechanical form
of signal into an electric form of signal.
So, there are certain piezoelectric pressure
sensors.
So what do these do, say piezoelectric pressure
sensors. So, these are made up of materials
which when subjected to a pressure difference
converts that into an electrical voltage signal.
So, that electrical voltage signal is read
by a convenient mechanism and that is how
you may have a digital output which does not
directly show the pressure difference but
it shows the corresponding electrical voltage
that is developed.
And if the pressure difference is calibrated
with that voltage then it is possible to very
accurately describe that what is the pressure
difference that actuated that voltage, one
obviously needs to calibrate these type of
device. But once it is calibrated it may be
very very accurate. So, it is based on the
transformation of signal in one form say mechanical
form into a signally another form like the
electric form.
In a summary what we may say that we have
discussed very brief about different pressure
measuring devices starting from the simple
barometer to manometer, pressure gauges and
pressure sensors in piezoelectric form. With
these pressure sensitive devices what we may
experimentally find out is what is the value
of pressure at a point may be relative to
some reference. What we will do with that
pressure?
When we are discussing about fluid statics
one of our objectives will be that to find
out because of that pressure what is the force
that is acting on a solid surface? The solid
surface may be a plain one or may be a curved
one.
We start with an example of force on a plane
surface. What is the special characteristic
of the plane surface that is such a surface
that we are considering that is emerged in
fluid at rest. We will try to make a sketch
of the arrangement let us say that this is
the free surface of the fluid that means on
the top there may be atmosphere in a bottom
there may be water as an example. There is
a surface the h view of the surface is like
this. It is a plane surface.
So, how does it look? Let us try to have a
visualization of this assume that the surface
is like this, okay. So, this is a plane surface
and when you are seeing its projection in
the plane of the board it looks like just
the edge view that is the line. So, whenever
we draw a line you keep in mind that it is
some kind of arbitrarily distributed flat
surface. The edge of which is represented
by this line.
So, such a surface is there in a fluid and
we are interested to find out what is the
force on this surface because of the pressure
distribution in a fluid. With that objective
in mind let us try to maybe draw the other
view of the surface. It just may be very very
arbitrary. So, it is a plane surface of some
arbitrary geometry. So, let us say that this
is the section of the surface when look parallel
to it.
We will set up certain coordinate axis let
us say we extend this and it meets the free
surface at this point. So, we will call this
as y axis and maybe an axis perpendicular
to that as x axis. Our problem is actually
a very simple problem. It is a problem of
finding out the resultant of a distributed
force because pressure distribution gives
rise to a distributed force. Why it is a distributed
force because pressure varies linearly with
the height.
Different elements of the plate are located
at different heights from the free surface.
So, it is a distributed force. The advantage
of handling with the plane surface is that
this distributed force is the system of parallel
forces. So if you have, for example if you
considered a small element at a distance y
from the axis, let us say the thickness of
the element is dy. So, what is the force due
to pressure that acts on this element?
Let us say that dF is the force that acts
on this element due to the pressure distribution.
This is the fluid at rest. So, what is dF?
dF is the local pressure on the element times
the elemental area. Let us say that the elemental
area of which we are talking that can be represented
in the other view completely. Let this be
dA. It just corresponds to this dy. So, dF
is the local P at the location y * dA. What
is the local P?
Yes, rho g * y sin theta, if you call it,
call these angle as theta. Because y sin theta
is nothing but the vertical depth from the
free surface to the location under conserve.
That multiplied by dA. So, what is the total
force that acts on the plate? It is integral
of dF, so it is integral of, so rho g sin
theta those are like invariance with respect
to the integration. So integrating with respect
to dA integral of ydA and integral is over
the entire area.
Say capital A is the shaded red colored area.
Pressure at the free surface is the reference
pressure. Say, if the water was not there
at the bottom. Just consider this example.
Let us say that this is a surface. Now there
is no water say it is surrounded by air from
all sides. If it is in equilibrium that means
air pressure is cancelling the effect from
all the sides together. The net effect is
that the sum total force is 0.
So, whatever force is acting on the surface
is because of the difference from the atmospheric
pressure. So, whenever you are calculating
the resultant force on a surface remember
what that you are implicitly dealing with
the gauge pressure not the absolute pressure.
Because the atmospheric pressure if atmosphere
was existing otherwise it would have kept
it in equilibrium.
Now, you are having some pressure over and
above that why atmospheric pressure? Any other
pressure, if there is a uniform pressure acting
on a close surface we will see that later
on that if you have uniform pressure acting
on a close surface then the resultant force
of that is 0. That may be proved by a very
simple mathematical consideration that you
have a distributed force which is always normal
to the boundary then integral of that over
a close boundary is 0, if the intensity of
that pressure is uniform throughout.
So, if you have, if you want to find out what
is the resultant force you may eliminate that
common part and consider only that part which
is over and above that. That is why we are
considering only the water effect. Now, you
can clearly see that what does integral ydA
represents, moment about x axis moment of
what? The moment of area so to say. So, the
first moment of area and first moment of area
gives what give the centroid of the area.
So, if you recall the formula for centroid
say y coordinate of the centroid of the area
that is integral of ydA/A. We will use the
formula here and we can straight away write
this as rho g A yc sin theta. Let us say that
the centroid is somewhere here, so what we
are talking about we have some distance yc
and this height which we may give just a name
say hc this is yc sin theta which is the vertical
depth of the centroid of the plane surface
from the free surface.
So, we can just write this as rho g Ahc. Now,
this gives the resultant force but since it
is a distributed force we also need to find
out what is the point through who is the resultant
of the distributed force passes. So, the point
of application of the distributed force. To
find out that let us say that there is point
P we give it an mp and say that P is the point
over through which the resultant of this distributed
force passes.
So, what we can do to find out what is the
location of P, location by location of P we
mean the y coordinate of the point P. So,
our objective is to find out what is the y
coordinate of the point P through which the
resultant of the distributed force due to
pressure passes.
For that we will just use the very simple
principle which we have learned in basic statics
that if you consider an axis with respect
to which you take the moment of forces then
the moment of the resultant force with respect
to that axis is nothing but the summation
of the moments of the individual components
of that forces with respect to the same axis.
This is known as Varignon's theorem and we
will try to use that for finding out the location
of yP.
So, if F is the resultant force and the moment
that we are trying to take this moment with
respect to the x axis then F * yP gives the
magnitude of the moment of the force F. This
is same as the sum of the moments of the individual
components. So, individual component is like
you have a dF that is an individual component.
So, what is the moment of dF with respect
to x, so that is just dF * y local y. So,
the total moment is integral of ydF.
So, rho g sin theta integral of y square dA.
So these are second moment of area. Sometimes
also loosely called as moment of area moment
of inertia just by virtue of its similarity
with mass moment of inertia. This is not really
a fundamentally moment of inertia it is better
to call us second moment of area. So, we can
write this as the second moment of area with
respect to the x axis. So, Ixx. So, we can
write what is yP rho g sin theta Ixx/F.
What is F? F is rho g * A into Yc sin theta.
So, we cancel the common terms Ixx represent
the second moment of area with respect to
some arbitrary axis. It is more convenient
to translate that to an axis which is parallel
to x but passing to the centroid and because
centroid is a reference point with respect
to a particular surface and to do that we
may use the parallel axis theorem to translate
it to C.
So, if we consider an axis which passes through
C and parallel to x with respect to that axis
we can write that Ixx is nothing but Ic that
by c we mean this axis which is passing through
c and parallel to x. That axis is totally
visible from the view parallel to the surface.
So, this is Ic + A * Yc square that /AYc.
From these we get a very important expression
that Yp is Yc + Ic / AYc.
This point P is given a special name in consideration
of fluid statics and that name is center of
pressure. So, center of pressure is the point
through which the resultant of distributed
force due to pressure passes that is known
as center of pressure. We can clearly see
from this expression that Yp is > Yc because
Yp = Yc + the positive term that means the
center of pressure in terms of depth lies
below the centroid, right and that is a very
important observation.
So, the 2 things that we learned from this
simple exercise one is to find out what is
the resultant force on a plane surface which
is emerged in a fluid due to the pressure
distribution and where is the point through
which this resultant force acts. We will consider
a simple example to begin with to demonstrate
that how we may calculate this.
Let us say that you have a surface which in
its sectional view is like this and this is
a vertical surface emerged in a fluid. So,
this is the fluid and this subject is a, or
like this object is a rectangular section
with the dimensions b and h. What is the resultant
force due to pressure acting on this? Let
us give a dimension of this say a. This is
example 1. The question is what is the resultant
force on this shaded surface because of pressure?
So, take an example like this. So, this is
like a surface what are our fluid some other
fluid is acting on it from one surface and
you are interested to find out what is the
resultant force because of pressure distribution
on this. So, this is a special case of the
inclined situation. So, the inclined situation
was like this. Now we had made it vertical,
okay. So, it is also an inclination with theta
= 90 degree.
So, for such a surface now if you want to
find out what is the resultant force that
acts on the surface. What is that? You look
at the formula, this one rho g Ahc. So, what
is A is b * h. What is hc? So, c is the centroid
of this area. So, if it is a homogeneous area
it is a + h/2. What is the location of the
center of pressure? Yp = Yc + first letters
write the formula and then we will substitute
the value.
What is Yc? Here y and h are the same because
it is just a vertical one. So, a + h/2 is
Yc + Ic. So, you have to now figure out first
that is it second movement of area with respect
to this vertical axis or this horizontal axis.
So, try to recall that when this was the inclined
plate the second moment of area was taken
with respect to this axis, right. So, now
when it is vertical you have a second moment
of area with respect to this axis and that
is translated to the centroidal axis.
So, that means the correct axis should be
this one. Just try to visualize it will be,
it is trivial but it is very very important.
Because based on the orientation you have
to figure out that with respect to which x
axis you are having to calculate the second
moment of area. Because centroidal axis is
not something which is unique you have different
axis passing through the centroid.
So, Ic will be based on this axis what will
be that bh cube/12. This is the expression.
If you want to find out or if you just make
a sketch of how this force is distributed
let us try to make a sketch of the pressure
distribution. To make a sketch of the pressure
distribution what we note the pressure varies
with the depth. So, here the pressure is 0
which is the reference pressure not 0 in an
absolute sense but relative to atmosphere.
And then it will linearly increase with the
depth. So, at this height this will be the
pressure at the bottom height this will be
the pressure and it is a distributed force
like this which varies linearly with the height.
And you can clearly see that the area under
this loading diagram will eventually give
you what is the force. These kinds of examples
you have already gone through in basic engineering
mechanics and you can verify it for the case
of fluid at rest very very similar.
Now, what is the state of stress in which
fluid at different depths they are the fluid
elements at different depths they are subjected
to. To consider that we will refer back to
the stress tensor which we introduced earlier
when we were discussing with the traction
vector. We are trying to relate the traction
vector with a stress tensor.
So, whenever you have a fluid element at rest
let us say that we are interested to write
the 6 independent components of the stress
tensor. So, now you tell, so we have the diagonal
elements and off-diagonal elements. If you
recall the diagonal elements represent normal
components of stress and off-diagonal elements
represent the sheer components of stress.
So, what will be the off-diagonal components
they will be 0 because it is fluid at rest.
So it is not subjected to sheer because with
sheer fluid will deform. So, these are all
0’s. When you come to the diagonal element
you have only the state of stress dictated
by the normal component which is just pressure
and that acts equally from all directions.
So, all the 3 components will be – P – P
– P. Tau 11, tau 22 tau 33 and the reason
of putting – is obvious the positive sign
convention of normal stress is tensile in
nature, whereas, pressure by nature is always
compressive.
Now let us say that we have the task of drawing
a more circle of distribution of state of
stress. If you recall what is the more circle,
so if you consider that there is an elemental
area which has an inclination say theta with
respect to some reference that has a resultant
force and that resultant force is given by
the traction vector components area. You can
decompose it let us say that the traction
vector component is like this.
You can decompose it into two parts, one is
a tangential component another is a normal
component. Let us say we call it sigma n and
let us say we call it tau. So, depending on
how you orient the area you will get different
combinations of sigma n and tau. If you draw
the locus of that then that is what constitutes
the more circle, right.
So, more circles gives you the visual appeal
or a visual feel of what is the state of stress
of different locations or at different locations
based on the choice of different choice of
orientation of the area or maybe at the same
location with different choice of orientation
of the area. So, when we are drawing one particular
more circle we are concentrating on one particular
point but changing the orientation of the
area to get the feel of the normal and tangential
components of the forces on that.
So, we are having two axis, one is sigma n
and there is tau. Our locus is or our objective
is to find the locus of sigma n versus tau
or tau versus sigma n. So, how will the more
circle look like for such a state of stress.
This state of stress is a very unique one
and this is known as hydro static state of
stress. The reason is obvious it represents
a hydrostatic physical situation. So, how
will the more circle look like?
See at a particular point you have the normal
component of stress that is because of pressure
and it acts equally from all directions. That
means if you change the orientation of the
plane sigma n will not change. Sigma n will
be unique and what will be tau? Tau will be
0. That means no matter whatever plane you
choose sigma n = - P and tau = 0. So, the
locus of all states of stress converts to
a single point with coordinate – P, 0.
So, the more circle becomes a point say – P.
So, this is just a point not a circle I have
encircled it but it is just like to show that
it is a point, okay. So, this is a very important
interesting limiting case when the more circles
shrinks to a point signifying that there is
no change in state of stress we change in
orientation. Next, we will consider another
example on force on a plane surface.
Example 2. We will make this example a bit
more involved than the previous one. Let us
say that there is again a free surface there
is some fixed structure. There is a base and
there is a gate like this. So, it is something
like say a sluice gate. So, on one side there
is water and it is saw that this water is
being confined in a particular place and this
gate has a tendency to move because there
is a resultant force of water from the left
side.
And there must be some mechanism, some holding
mechanism by which this gate is kept at rest
at equilibrium. So, there is some force which
is acting on the gate say the force is this
F by some support or whatever there is some
force F which acts on these to keep it in
equilibrium. This gate may be of different
shapes and let us take an example which the
gate shape as a semicircular one like this.
So, this is the section of this gate.
This is like it is fixed with something. So
it is, there is a separate structure that
goes within. So, we are not going to concentrate
so much on the upper one. We are, I mean it
may be extending even beyond the free water
surface and so on. But we are more interested
with the gate. So, on this gate there is a
force due to water and that force you may
calculate by considering some dimensions.
One of the dimensions say this is capital
H let us say that the radius of this semicircular
gate is capital R and the density of the fluid
of course is given and G. This gate is hinged
at this point say O. So, this is a situation
where if you want to find out what force F
should keep it in equilibrium you must calculate
what is the resultant force due to pressure
acting on the gate. This is a plane surface.
Shape of the plane is a semicircle but it
is still a plane surface and you may use the
formula for force on a plane surface. So,
let us try to draw the free body diagram of
this gate. So, our objective is that what
should be this force to keep the gate in equilibrium.
So, we draw the free body diagram of the gate.
You have a force F, what other forces you
have? You have the hinge reactions say let
us say Ox and Oy, the 2 components this has
its own weight.
So, some mg and the force due to pressure
distribution in water, let us say Fp which
passes through the center of pressure. So,
for equilibrium the resultant moment of all
forces with respect to O should be 0. The
reason of choice of O as moment center is
obvious it eliminates all unknowns except
the F that we are interested to find out.
So, what it will give? It will give capital
F * R – Fp * say these distance Y1 = 0 and
how can you calculate Y1?
Y1 is capital H – Yp, where Yp is the location
of the center of pressure from the top surface,
right. Fundamentally, you can calculate Yp/the
formula that is Yp = Yc + IC/AYc. So, what
is Yc? Yc is the y coordinate of the centroid
of these semicircular area. So, this part
the location of the centroid from the bottom
is 4R/3 pi, okay. So, Yc should be H – 4R/3
pi. To that extend find A.
What is A? Pi R square/2 that is also quite
clear Ic something which we always forget,
right if you ask me to recall I will be in
a great tension. I have really forgotten what
is Ic, I mean second moment of area with respect
to an axis of a semicircular thing which passes
through its centroid. Of course you may derive
it but we will see that this derivation is
not necessary.
Because we will try to avoid this route of
this formula base determination of this and
we will just do it from the fundamental method
of integration by which you find out the resultant
of a distributed force, the resultant moment
of distributed force and so on. And the entire
reason is that there are certain simple areas
for which we may remember the expressions
for the second moment of area with respect
to the centroidal axis quite easily.
But it is not so convenient for many complicated
areas. This is not complicated as such but
even for that we should not tax our brain
by remembering that. I mean that is not a
very special information that we should remember.
So, what we will do is in the next class we
will try to see we will keep this problem
in mind. We will see the alternative way by
which we will be solving this problem.
You can of course solve this problem by substituting
the value of Ic here expression and this expression
is given in the appendix of the text books
in statics. So, you can find out the expression
or you may even derive it if you want and
just substitute it to get what is Yp and from
that you can get F. But we will see that whenever
possible and whenever convenient it may also
be alright if we just find it out by simple
integration of the distributed forces.
So, that we will do in the next class let
us stop here. Thank you.
