Good morning, I welcome you all to the session
of fluid mechanics. Now, if you recall, last
time we were discussing about the buoyancy,
phenomena of buoyancy, what is that? Again,
just if you recall it, a body, if a solid
body is either partially emerged or fully
emerged in a fluid, the net horizontal force
in any direction because of the pressure exerted
by the fluid on the surface of the body is
0, but the body experiences net vertical force
in the upward direction. That means, the resultant
of the pressure forces on the surface of the
body, that is, the mass body, either fully
or partially, is in the vertical direction,
upward vertical direction, and its magnitude
is equal to the weight of the displaced volume
of the fluid.
What is the displaced volume? That is the
volume, emerged volume, that is the volume
of this solid emerged in the fluid. In case
of a fully emerged body, that is, submerged
body, then the displaced volume is the volume
of the whole body itself. So, weight of that
volume of the fluid is the magnitude of the
upward hydrostatic pressure force, which is
known as buoyant force and this phenomenon
is known as buoyancy.
Now, then afterwards we recognized that for
an equilibrium condition of the body, first
let us consider this submerged body, but it
is very, it is valid for both, submerged and
partially emerged body, that for equilibrium
the primary condition is, that the weight
of the body should balance the buoyancy force,
upward buoyancy force by magnitude and also
by the line of action. That means, weight
of the body should be equal to that of the
buoyant force and this two forces, which must
be colonial for equilibrium of the body either
in floating condition, that is, partially
an emerged condition or fully emerged or submerged
condition.
Now, the question comes, that probably I have
already read in, earlier in mechanics, that
equilibrium is of three types, one is stable
equilibrium, another is unstable equilibrium,
another is neutral equilibrium. What is meant
by stable equilibrium in general is, that
if a body is in equilibrium under several
forces at a particular instant, now if you
disturb the body to just depart from its initial
equilibrium position, whether the body is
able to come back to its initial position
or not? If the body is able to come back to
its initial position that means force system
acting on the body is such, that it makes
it possible to restore its initial equilibrium
position. Then, this type of equilibrium is
known as stable equilibrium; that means equilibrium
is stable. It is the question of reproduction;
just the body is capable of reproducing its
initial equilibrium position under any small
perturbation to distort, to allow the body
to depart from this initial position. It will
again come back to its initial position.
Now, unstable equilibrium is such, that if
the body is in equilibrium under such several
forces at a particular position, if you disturb
the body to depart from its initial equilibrium
position the force system is such, that it
does not allow the body to come back to its
initial position. It goes on departing from
its initial equilibrium position and in fact,
the entire equilibrium condition of the body
is destroyed. This is known as unstable equilibrium.
So, equilibrium is there at a particular position
and the particular instant, but this is unstable,
unstable equilibrium.
Another type of equilibrium is neutral equilibrium,
which means, that if you give a disturbance
to the body, then the body will neither come
back to its original position nor it will
go on departing further and destroy its equilibrium
condition, neither of these two, but body
will remain again in the equilibrium condition
at that point. That a sleeping person, that
a sleeping person, if you take him from one
position to another position, he will be keeping
the, like that the neutral equilibrium.
I mean, a very simple example, probably we
have read it in school level, that if you
place a marble on a hill or on a table of
convex shape, if u place it, it may be in
equilibrium, that the weight may be balanced
by the reaction force if there is a contact
surface. But if you slightly perturb it, that
means, if you slightly push it, this will
go down because of the convexity of the surface,
that means, this is the unstable equilibrium.
Similarly, if you place a body on a surface,
which is concave inward and you place a body,
it is in equilibrium. The weight of the body
is equal to the reaction force. If you give
a displacement, then it will again come back
to its original position because of the concavity
of the surface. That means, this is a, and
this is an example of stable equilibrium and
neutral equilibrium. Example is very simple,
on a flat table, flat table, if you place
a body for example, this is in neutral equilibrium.
If you place it here it is in equilibrium;
if you place it here, if you just do not consider
the rolling of the, say it is in equilibrium.
So, that means, this a concept neutral equilibrium.
In case of buoyancy, this perturbation to
study whether the body is in neutral, stable
or unstable equilibrium depends upon this
perturbation in angular direction. That means,
if you just displace the body or give a small
rotation, then we see its stability. That
is why sometimes this stable equilibrium,
unstable equilibrium, neutral equilibrium
is a couple or is referred to as angular stability
of the body. So, body is linearly stable.
When and buoyancy force are equal in magnitude
and are collinear, but weather in small angular
displacement, allows the body to come back
to its original position or not is referred
to as angular stability or in general, to
stable, unstable and neutral equilibrium.
So, let us now see, that let this be the free
surface of the liquid. Now, let us consider
first submerged body, first consider submerged
body, well, first consider submerged body.
Now, what is condition? Let us see a body
like this, let us see a body like this, let
us consider, now you see with the center of
gravity and center of buoyancy should be in
this line. Now, there may be three options,
dividing point of application of center of
gravity and center of buoyancy. One option
is, that center of gravity may be, let this
is center of buoyancy, center of gravity may
be below the center of buoyancy. Another option
is that center of gravity may be coinciding
with center of buoyancy. Another option is,
there center of gravity will be center of
buoyancy. This depends upon the relative distribution
of mass over the.
When gravity is below the center of buoyancy,
it is bottom heavy. When gravity is above
the center of buoyancy, it is top heavy. And
when the mass is distributed uniformly throughout
this two coincide.
Now, let us consider when the centre of gravity
is below the center of buoyancy what happens?
Through center of buoyancy the buoyancy force
B acts; through center of gravity weight W
acts. They are collinear and equal always,
W is equals to F B, so this is in equilibrium.
Let us give a small tilt or angular of this
body in this direction, towards the right.
Let me draw this figure, now this tilt condition.
Let us see the body, let us see the body.
Now, this is the axis of symmetry of the original
vertical axis, whatever you can tilt. Now,
the centre of gravity with respect to the
body is unchanged. This is, of course, not
true for all cases, I will explain afterwards.
In certain cases if some solid part of the
body moves, for example, when the ship moves,
when a ship moves in a river some of the cargos
within ship moves from one part to other part.
So, therefore, the mass may change, but if
we consider a tight solid body, the center
of gravity remains unchanged and the weight
acts vertically in center of gravity. Now,
this center of buoyancy also remains unchanged
in this case. This is because what the center
of buoyancy is also the center of the volume
and when in submerged bodies the entire volume
is submerged, so here also entire volume is
submerged. In both the conditions the bodies
are submerged, so center of buoyancy also
will not change through which the buoyancy
forces acts.
Now, in this case you see these two parallel
forces create a moment or a couple, which
is now this is the direction of tilt, so a
couple is generated in the opposite direction
whose magnitude is W or F B, whatever you
call, both the forces are equal, times this
distance, times this distance. If you consider
this distance as x, W into x, that means,
simple, we can tell in this circumstances
this F B and W creates a couple, which is
opposite to the direction of the tilt and
this couple is known as restoring couple.
So, this couple is restoring couple that means,
which restores its position. That means this
couple helps the body to come to its equilibrium
position.
Now, if you give a tilt or in the left direction
you will see even in the body buoyancy force
and the gravity force makes a couple in the
opposite direction to the angular of the body
that means creates a restoring couple and
helps the body to come into equilibrium. So,
when G is below B we see the equilibrium is
stable equilibrium. So, this is stable equilibrium,
stable equilibrium, stable equilibrium.
Now, let us consider a case when G is above
B that means, the distribution of the force
is like this. Distribution of the force, sorry,
distribution of the mass over the volume is
such that means, it is top heavy that means
G is, this is G and let B is below G. That
means, through G is the weight of the body
acts and through B the buoyancy force acts,
but it is in equilibrium. That means, W is
F B, this is. Now, if you give a tilt, very
simple, school level thing, so if you give
a tilt in the same direction that means right-wise,
towards the right if you give a tilt, this
is the original vertical axis here, again
in this condition the gravity remains the
same, the center of the gravity and also the
center of buoyancy.
Now, you see, this two forces, W and F B,
they create a moment, which is in the same
direction to that of the angular tilt. That
means, when you give a tilt, in this case
the two forces F B and W, whose colinearity
is destroyed, they create a moment in the
same direction or couple in the same direction
and helps the body to tilt further there.
Both the body departs from original position
further and further and equilibrium is never
attained. This case, it is known as unstable
equilibrium, unstable equilibrium. And this
couple under this condition the value of this
moment that means, W or F B times this perpendicular
distance is known as overturning or destroying
couple; sometimes we call overturning, this
is the terminology in mechanics or destroying
couple, the couple, which destroys the equilibrium.
Therefore, we see, now in this case if you
give a in the left direction you can again
examine, that the couple created by the destruction
of the colinearity between F B and W, the
couple created by this two force again will
be in the same direction of the angular tilt
and creates overturning and destroying couple
and makes the equilibrium an unstable equilibrium.
Now, the last possibility is the possibility,
that G and B coincides, that means, this is
the case when the weight, sorry, mass of the
liquid is distributed, this is G, uniformly
over the entire volume. That means, the weight
acts to the center of gravity and the same
point is the center of buoyancy, that is,
F B. That means, the mass is uniformly distributed
over the entire body. In this case, it is
just like neutral equilibrium, as I have told
earlier.
That means, if you displace the, sorry, if
you displace this under the displaced condition,
that means, if you give angular hill in this
direction, G, B, both will be same, so this
will be the new vertical lines and they will
be again collinear. This is the W and this
is F, so it will be again collinear, that
means, under any angular location and at any
angular displacement and this position it
will be neutral equilibrium. It will neither
go further in this direction or will come
back in this direction. This is known as neutral
equilibrium, this is known as neutral equilibrium.
So, therefore, what we conclude is, that when
a body is totally submerged the condition
for stable equilibrium, that means, angular
stability, that if the body is given a slight
angular hill in either direction, whether
it will come back or not depends upon the
fact, that will G, the center of gravity,
is below the center of buoyancy. The body’s
angular; body has this angular stability means,
stable equilibrium.
If the center of the gravity is above the
center of buoyancy, if the body is in unstable
equilibrium, that means, it does not have
the angular stability. If you give a small
angular hill in either direction or part of
the body, it will go on departing for the
original equilibrium condition and the entering
equilibrium will be destroyed. This is the
case of unstable equilibrium when the center
of gravity is above the center of buoyancy.
And the last case is the center of gravity
and the center of buoyancy coincides, that
is, for a special case when the distribution
of mass is uniform throughout the volume of
the body, in that case the body is in neutral
equilibrium. That means, if you give any position
with any angular position at any inclination,
at any configuration, if you submerge the
body, body will be at that particular configuration.
That is the concept of neutral equilibrium.
Now, we will discuss the same thing under
floating condition, but, there is a difference.
First of all you must know, that why separately
you are discussing this submerged and floating
bodies? What is the key point of difference
between a submerged and the floating body?
First difference is, that as it is very simple,
literally, floating body means a part of the
body submerged, not the entire body submerged
in the fluid. Therefore, the buoyancy force
will be equal to the weight of the displaced
volume of the fluid. That means weight of
that volume of the fluid, which corresponds
to the emerged volume of the solid; that is
number one. Number two is that obviously,
under equilibrium condition of a floating
body, the weight of the body must balance
the buoyancy force and next is that they should
be collinear and buoyancy force is equal to
the weight of the displaced volume.
Therefore, floating body in equilibrium is
that the density of the body has to be lower
than the density of the fluid so that the
weight of the displaced volume of the fluid,
which is the emerged volume, partially emerged,
that means, not the total volume of the fluid
must be equal to the weight of the fluid.
This is very simple, we have rated school
level. Now, what happens, when you give an
angular hill for this floating body what is
the most important is, that here though the
center of gravity may remain unchanged, if
there is no movement of mass within this solid
body, but center of the buoyancy is obviously
changing. This is because though the total
displaced volume remains same to make it equilibrium
and to be equal to weight, but some portion
is getting more emerged and some portion is
getting out of the water level depending upon
the direction of the hill. So, they are both,
the center of buoyancy to distribution of
the emerged volume changes though the total
emerged volume remains same, so that center
of buoyancy moves. It does not become same
as earlier one, which we saw in case of submerged
body, that is the key difference. So, they
come separately for investigation of the angular
stability in case of floating bodies.
With that let us see, that what happens in
case of a floating body. Good, now, let us
consider this is the free surface and let
us consider the floating body, typical cross-section
of a ship, let us consider a floating body
like this. This is the axis of symmetry, a
symmetrical body like this. Now, let us consider
a floating body, this is the part, which is
emerged in the fluid and this is outside the
fluid. Let us consider this is the initial
position and here we will see one interesting
thing.
Now, first of all, if gravity is below this
center of buoyancy, center of gravity, it
has always been in stable equilibrium, that
you can see by yourself, I am not going to
that. Let us consider the gravity is above
its center of buoyancy. Now, we have seen
in case of submerged bodies, we immediately,
obviously, whenever the gravity is above the
center of buoyancy, then obviously it is unstable
equilibrium, but in this case it may not be.
Then, angular stability may not always be
destroyed if the center of gravity is always
above B, you cannot tilt them. Now, center
of buoyancy is the center of this volume and
we consider this center of gravity is above
B. One thing is true, that the weight and
the buoyancy force, that is always true, they
balance each other. Now, in this case what
happens?
Now, what I have told just now, you see, that
if you give a small tilt in this direction
and let me draw the figure like this. Now,
let in displaced position this is the old
line of symmetry or the line, which contains
the center of gravity and the center of old
vertical line containing center of gravity
and old center of buoyancy. Now, it is obvious,
that if I retract or retrace this line, let
this line be, this line, so therefore, we
see, that this portion of the body portion
has gone out of the free surface, which was
earlier there underneath the free surface.
And this portion of the body, which was above
the free surface, has gone down below the
free surface. That means, that if we give
a tilt here more of the volume in this portion
will go under the free surface and some volume
in this portion will automatically come up.
So, therefore, we see the distribution of
the emerged volume change. It obviously, if
this be the old center of buoyancy a new center
of buoyancy will be shifted somewhere here
in this direction. B dash is the new center
of buoyancy, so what it does?
Now, if this be the G, now the force acting
on this body is W, vertically downward through
G and then we see, that this F B. Now, it
is not very easy to say, that always it will
be overturning moment or a destroying moment.
In this case what do you see, that if I make
projection of the vertical through G like
this under this particular condition this
is which moment? Restoring moment, very good,
so under this case it is the restoring moment,
that means, it is angularly stable, but can
we tell always if B shifts in this direction,
just commonsense, you tell always there will
be restoring moment?
No, if this B dash comes here, that means,
if I draw the vertical projection or a vertical
line from the centre of gravity and if B cannot
cross this vertical line towards this direction,
direction of hill, then it cannot create a
restoring moment so long it is here. So, this
force will always create a overturning moment,
that whether there will be a restoring moment
or an overturning moment is not conclusively
decided by the position of G above B, but
this shifting of B with respect to G. If the
shifting of B takes place like this, that
it crosses the vertical line in that direction,
then it creates a restoring moment, otherwise
it creates an overturning moment. So, this
is the condition.
This is better stated analytically by this
fact, that if we make a vertical projection
from the new center of buoyancy, make a vertical
line, extended vertical and let it intersect
with the old vertical line containing the
center of gravity at the old center of buoyancy
at the point A and this point A means metacentre.
This is very important concept, metacentre
and usually, for a small angular hill, if
we do not consider a very big angular hill,
that means, small perturbation with small
angular hill in either direction. So, you
can see, if you give a tilt in that direction
the same condition will, that means, B either
create an overturning or restoring couple
depending upon its relative displacement with
respect to G. Now, this M metacentre is usually
for a small angular hill, is a geometrical
criterion depending upon the dimension of
the body and its geometrical shape, so it
becomes a geometrical criterion.
Now, we see that the way we appreciate, that
if B crosses this vertical line in the direction
of the hill, then it will give a restoring
moment otherwise overturning moment can be
told with respect to this. Very important
point, metacentre, that if metacentre is above
the centre of gravity, in this case, it is
that then the body is under stable equilibrium
and if the metacentre is below, that means,
in this case if this is the shape, then metacentre,
then the vertical line will always intersect
at a point, that the same point known as M,
which is below G. That means, when metacentre
is below G, then we can tell, that the body
is not in stable equilibrium and the height
GM, that means, the height from the centre
of gravity to the metacentre, if it is below,
this is the height, this is known as the,
along this old vertical line or the line of
symmetry you can tell, is known as the metacentrentric.
That means, GM, this distance is known as
the metacentric height, which can be also
written as GM is equal to BM minus BG.
If you write like, that when metacentre is
below the centre of gravity, then GM is negative.
That means, we can tell GM is greater than
0, GM less than 0 and GM equal to 0, three
cases. When GM is greater than 0, metacentre
is above the centre of gravity. By this formula
we define as analytically, that it is greater
than 0, M is above G and the body is in stable
equilibrium. When GM less than 0, means, M
is below G, so we define it as BM, so that
GM is less than 0, then the body is in unstable
equilibrium, that the couple created because
of the tilt by the weight and the buoyancy
force is an overturning one, that is, the
unstable equilibrium. And when metacentric
height coincides with G, that means, B is
displaced in such way, that the vertical line
through B just intersects G that means, again
in that displaced position F B and W become
collinear, if it satisfies that, that means,
GM is 0. That means GM is equal to BG, which
means, the vertical projection from new center
of buoyancy intersects this point, then this
is the neutral equilibrium.
So, therefore, it is not conclusive only from
G above B, which was therefore, submerged
body, it will be under unstable equilibrium
for a floating body. The key point is, that
in this case, since the center of buoyancy
moves towards the direction of the hill it
gives an advantageous condition, that means,
the center of buoyancy moves towards the direction
of the hill, it gives an advantageous condition.
If even if it is top heavy, that means, center
of gravity is above the center of buoyancy
till it is in equilibrium, stable equilibrium,
provided the metacentre is above G, that means,
the buoyancy shifted in such a way, that it
crosses the vertical line from the center
of gravity, so that it creates the restoring
couple to allow the body to come back to original
position. Well, this concept is now clear.
Now, with this concept in metacentric height
I will derive now a very important, what exactly
is required for designing a floating body
impact. This for example, a ship, that the
metacentric height should remain positive,
so that for any angular floatation or angular
disturbance of the ship about any horizontal
axis, ship is stable. Now, this metacentric
height, I have told of the metacentre position
of this, metacentric is the function of the
geometrical shape of the ship or the geometrical
shape of the floating bodies. So, if we know
beforehand by, calculate this metacentric
height from the, perform a functional relationship
of the metacentric height with the geometrical
dimension of the body. Accordingly, we can
design the ship, that make the geometrical
dimension, show that the metacentric height
is always positive and it should be as big
as, that means, MG should be high, high positive
value, so that stability is ensured even under
a greater perturbation. So, for that we tried
to seek or we seek a functional relationship
between the metacentric height or the position
of the metacentric in terms of the geometry
of the body.
Let us do that. Well, again we, again we make
the same thing, that this is the body, we
have floating bodies body like this, the floating
body is like this, that is the cross section
of the ship, rather taken a cross section
of ship, this is the plane area. Now, after
the hill let us, it could have been better,
different, so this is after hill is in the
displaced position. Now, let us see, first
we consider a simple case always, that G is
above B. That is not the simple case; simple
case is G remains unchanged.
Let us consider this is G, but the old B,
obviously, when you give a tilt in this direction
what will happen? Old B will be shifted here,
B dash. So, the weight act in through this
and this is acting F B, F B and this is W.
Now, if we trace this old line of, line for
the plane floatation we can write or we can,
sorry, draw line like this, let this is the
point O, point of intersection of the original
plane of floatation, well this is the original
plane of floatation and the new plane of floatation
and let they intersect along there centroidal
axis.
Now, one thing is true, so long W is equal
to F B, well, W is equal to F B, that means,
this is the W, this always we consider equilibrium,
only we are considering the stability, that
is angular stability, then the buoyancy force
W is unique; so, buoyancy force is unique.
So, therefore, by the principle of Archimedes
the emerged volume should remain same whether
it is given a tilt or not, which means, the
volume, which has come up, that means, this
hatched portion mass be equal to the volume
at this hatched portion, which has gone inside
the liquid. This is the key point, nothing
else, so this volume is equal to this. So
this is alright.
Now, if we simply define the center of buoyancy
with respect to any frame of reference, now
we will have to do. Let us consider this as
O in this original figure, we consider the
axis like X and this is as Z, vertically downward
Z and the horizontal axis is on the plane
of floatation and take OY, the perpendicular
to the plane. That means, if we see this cross-section
like this, a cross-section like this, cross-section,
just an arbitrary cross-section like this,
so Y will be along this direction. That means,
if we define O somewhere here, this is the
OY that means, this is in this perpendicular
XY and vertically downward Z, I think it is
alright.
Now, if we define this center of buoyancy
X coordinate as x B, as x B, which is 0 under
the condition because this is the Z-axis.
That means, this is from Z-axis, so x B. Then,
we can write, that V, emerged volume, into
x B is nothing but if we take here a prismatic
body of the fluid or the volume emerged whose
cross-sectional area is d A and Z is the coordinate,
that means, this is the height of this prismatic
element, that means, a Z coordinate of this
point, whatever you can tell, that this becomes
equal to z into d A into x, sorry, Z into
d A into x, rather we can write Z x dA, alright.
Now, you see, that here what happens, that
is the dA is the volume and moment of the
volume, that is, the center of buoyancy. Now,
this case, it is 0 if we define center of
buoyancy on this line, this case it is 0.
Now, this case what we can do? We can, now
the axis are defined like this, therefore
Z and x will be like that, but here if we
take, now this is emerged here, so here if
we take a prismatic volume like this what
is its height? So, this dA we can show it
here, this is the dA, this is the typically
elemental area in this plane that means, this
view. Now, where here if we take the same
prismatic element dA here, the height is equal
to Z, that means, A is this one with respect
to this coordinate axis x and this is Z plus
this extra amount, which has become emerged
in the fluid.
So, if this is the x, that means, the x coordinate
of this prismatic element from the coordinate
axis origin, then this will be x tan theta
where theta is the angle of hill. That means,
this is theta, if we make this theta or the
vertical through this O with the old vertical
theta because angle between two lines is equal
to angle between their perpendicular, so theta
is the angular hill, that means, theta is
the angle between the old plane of floatation
and the new plane of floatation or between
the old vertical line and the new vertical
line, so this is the theta.
So, therefore, if this is x and this is theta,
so this extra part is what? x tan theta, that
means, now we can write V x B dash. If x b
dash is this, the B dash, that is, this x
coordinate of the new center of buoyancy,
what we can write? We can write that is equal
to Z plus x tan theta into dA. Simply, we
subtract this upper one from this one, we
get V x b dash minus x b is equal to tan theta
is constant because angular hill is constant
throughout the body. So, therefore, we can
take tan theta into what? x square dA, sorry,
x, one x is missing for the moment, so Z x
plus tan theta, dA is the volume, elementary
volume and moment for, moment, simply x square
tan theta, alright. So, this is equal to,
now you see 
we can write this. Is it ok?
Now, we can write, well, now what we can write?
We can write that V into x B dash minus x
B is equal to what? Tan theta, tan theta into
x square. Now, x B dash minus x B, for small
angle of theta if you make a projection from
this to this axis, this is the axis z for
small angle of theta, this can be written
as B, this is B and if this vertical line
intersects at a point, let this is M, so this
BB dash can be written as BM tan theta, BM
tan theta, for small angle we can write this.
If we take, if we make a perpendicular from
B dash on this old vertical line, which is
the z axis rather, then we can tell for small
angle this perpendicular does not make any
change from B to this point of projection.
So, we can take simply from this right angled
triangle, this vertical line meets at M, so
this is BM tan theta. So, we can write BM
tan theta for small angle; tan theta, because
this projection may not coincide with B, that
is why small angle concept. We can write,
this is, this is almost equal to BM tan theta,
so tan theta tan theta cancels, well, so we
get what? BM, sorry, this is tan, BM tan theta.
So, V is there, so BM is equal to x square
dA divided by V.
Now, what is this x square dA? What is this
x square dA? If we just see here x square
dA, now we see to this figure, that means,
x is the distance of any elemental area here
from this y x, that means, this x square dA,
which is there in the numerator of this expression
of BM represents what? The moment of, second
moment of area of the plane of floatation,
this is the plane of floatation. That means,
I see here as a section from, here if I take
a section from here and look from top, that
is sectional plane we see, that is plane of
the floatation. So, x square dA is the second
moment of area of the plane of floatation
about the centeroidal axis about the centeroidal
axis, which is perpendicular to the plane
of floatation. This is the plane of floatation,
that means, this is the second moment of area,
be very careful, this one, second moment of
area of the plane of the floatation about
this OY, which means physically about the
centeroidial axis. That means, this axis here,
which is perpendicular to the plane of floatation,
plane of floatation is the exit plane, so
y axis perpendicular to that or you can tell
the centroidial axis of rotation about which
there rotation is taking.
So, if you can understand this nomenclature,
x square dA, as we defined, as I simply, I,
we make I y, again y makes a confusion, that
you will have to take this as the y-axis.
So, simply I tell I, so that I can write BM
is equal to, this is a very important formula,
I by V, so nomenclature is like this. BM means,
what is BM? It is the distance from the center
of buoyancy to the metacentre along the old
line vertical line where center of buoyancy
and center of gravity was acting; so, along
this B end, where the GM is the metacentric,
height BM. Therefore, this is the definition.
And usually it is told as metacentric radius,
not very common, but we will see in my book
I have written, that not all the books they
tell because it is not a very common terminology
as GM metacentric height, but BM you can,
line. So, this is BM, which is equal to I
by V.
So, we will have to recognize the nomenclature
like this, V is the emerged volume, that is,
the volume of the solid, which is emerged
in the fluid displaced volume. And I is the
moment of inertia of the plane of floatation,
moment of the inertia of the plane of the
floatation about a, about the centroidial
axis, which is perpendicular to the plane
of rotation or the axis of the rotation. That
means, if you simply, plane of floatation
is the horizontal projection of the horizontal,
that they, this plane and this perpendicular
axis is the centroidial axis perpendicular
to the plane of floatation. So, the nomenclature
of I should be very important.
Now, you see, in case of a ship if you see
the structure of a ship, now the ship, if
you see a ship, now this is the, now if you
see the ship, this is the elevation if you
see its plane. Now, the structure of the ship
is such, that like this there may be two axis,
that means, about the longitudinal axis, that
means, in this direction is known as pitch.
about the transverse axis, that means, moves
like this and rolling is about is longitudinal.
When rolling, angular stability with rolling
is considered, then the I corresponds to the
second movement of area about OY. Similarly,
when the angular stability with respect to
pitching is considered, that means, it is,
its angular movement about x axis, then the
moment of inertia about the axis x, OX is
considered. Let this is. Now, you see from
the typical geometry of the shape, the angular,
the second movement of area about y axis is
much lower than the second movement of area
about axis because the longitudinal dimension
is more. So, therefore, I is lower in case
of rolling as compared to that in case of
pitching, clear.
Therefore, so BM is I by V. So, therefore,
in case of rolling BM is small because I is
small. Rolling is about OI axis, about OI,
the inertia moment of inertia is small. Therefore,
stability with respect to rolling is more
important than this stability with respect
to pitching. See, if it is almost stable with
respect to the rotational motion or with respect
to pitching about x axis, the moment of inertia
of any element about the x axis is much higher.
So, rolling, for rolling the stability is
more important. That means, we will have to
see the angular stability point of view, that
it is the moment of the inertia of the plane
of floatation about an axis makes this more
vulnerable for its angular stability in a
particular direction.
Well, thank you.
