stability floating body
TRANSCRIPT
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5.
THE
STABILITY OF A FLOATING BODY
l roduct on
When designing a vessel such as a ship, which is to float on water,
it
is clearly
necessary to be able to establish beforehand that it will float upright in stable
equilibrium.
Fig 5 a shows such a floating body, which is in equilibrium under the action of
two equal and opposite forces, namely, its weight W acting vertically downwards
through its centre
of
gravity G, and the buoyancy force, of equal magnitude W, acting
vertically upwards at the centre
of
buoyancy
B
This centre
of
buoyancy is located at
the centre of gravity
of
the fluid displaced by the vessel. When
in
equilibrium, the
points
G and B lie in the same vertical line. At first sight, it may appear that the
condition for stable equilibrium would be that G should lie below However, this is
not so.
B
w
I
I
G
J
a
b
Stable
Fig 5 J Forces acting
on
jlo ting body
c Unslable
To establish the true condition for stability, consider a small angular displacement
from the equilibrium position, as shown in Figs 5 b and 5 c . As the vessel tilts, the
centre of buoyancy moves sideways, remaining always at the centre of gravity of the
displaced liquid. If,
as
shown on Fig 5 b , the weight
and
the buoyancy forces
together produce a couple which acts to restore the vessel to its init ial position,
the
equilibrium is
stable.
however, the
couple acts
to move the vessel even
further from its initial position, as in Fig S c , then the equilibrium is unstable.
The special case when the resulting couple is zero represents the condition of neutral
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I
ex
L
i _
x
. f UX
cb
x
' , '/
_
x. .J
Fig
5.2
Derivation
o
conditionsor stability
w
a
W=wV
WX
x
g
W
Fig 5.2(a) shows a body
of
total weight W floating on even keel. The centre of
gravity G may be shifted sideways by moving a jockey
of
weight
\Vj
across the width
of the body. When the jockey is moved a distance xi>
as
shown in Fig 5.2(b), the
centre of gravity of the whole assembly moves to G The distance GO', denoted
by
x
g
, is given from elementary statics
as
(5.1)
Experimental Determination
o
Stability
In thc following text, we shall show how the stability may he investigated
experimentally, and then how a theoretical calculation can be used to predict the
results.
stability. It will be seen from Fig
5 1
(b) that it is perfectly possible
to
obtain stable
equilibrium when the centre
of
gravity G is located above the centrc
of
buoyancy B
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The shift of the centre of gravity causes the body to
tilt
to a new equilibrium position.
al a small angl e
e
the vertical, as shown in Fig 5.2 b), with an associated movement
of the centre of buoyancy from B to B
t
The point B
t
must lie vertically below
since the body is in equilibrium in the tilted position. le t the vertical l ine of the
upthrust through B intersect the original line of upthrust S at tbe poi nt M. c al le d the
metacentre We may now regard the
jockey
movement as having caused the floating
body
to
swing about the point
M.
Accordingly, the
equilibrium
is
stable
if the
mctacentre lies
above
G. P ro vi de d that
e is
small, the distance GM is given
by
X
g
~ -
S
where e is in c irc ula r measure. Sub st it ut ing for x
g
from Equation 5.1) gives the
result
W.
GM = .
W S
5.2)
The
dimension GM
is
called the metacentric height.
In
the experiment described
below. it is mea sured direc tly from the slope of a graph of Xj against obmined by
moving a
jockey
across a pontoon.
Analy i al De ermina ion
A quite separate theoretical calculation
of
the position
of
the met acent re ca n be made
as follows.
The movement
of
the centre of buoyancy to B
t
produces a moment
of
the buoyancy
force a bou t the original ce ntre
of
buoyancy
B.
To establish the magnitude of this
moment. first consider the element
of
moment
exened
by a small cl eme nt of change
in
di sp la ce d volume, as indicated on Fig 5.2 c).
An
element of width 8x, lying
at
distance x from
B.
has a n additional depth xdue to the tilt of the body. Its length.
as shown
in
the plan view on Fig 5.3 c), is
L
So the volume
OV cfthe
element
is
8V = S.x.L.ox = SLx8x
and the element of additional buoyancy force 8F is
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8F
w.8V = we Lx8x
where
\\
is the specific weight of water. The element of momenl about B produced by
the element of force is 8M. where
oM = of.x
w8Lx
2
x
The total moment about 8 is obtained
by
integration over the whole of the plan area
of the body. in the plane
of
the water surface:
M = we fLx
1
dX = weI
5.3)
In this,
1
re pre sen ts the s ec on d mome nt , about the a xis of symmetry,
of
the water
plane area of the body.
: \ow this moment represents the movement
of
the upthrust wV from B to B
t
namely,
wV.BB . Equating this the expression for M in Equation 5.3)
wV.BB weI
From the geometry of the figure,
we
see that
BB
=
e.BM
and eliminating BS between these last two equations gives 8M as
BM
I
V
5.4)
For the particular case ofa body with a rectangular planfonn of width 0 and length L,
the second moment I is readily found as:
0 1 0 1 [
]0/1
I;
fLx
dX = L fx
1
dx
= L =
r
Df2 D 2
42
L
12
5.5)
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Now the distance BG may be found from the computed or measured positions of B
and of G, so the metaccntric height GM follows from
Equation
S.4 and
geometrical relationship
GM
M
5.6
This gives an independent check on the result obtained experimentally by traversing a
jockey weight across the floating body.
xperimental rocedure
T he pon toon shown in Fig 5.3 has a re cta ng ul ar p la tf on n, a nd is prov ide d w it h a rigid
sail. jockey weight
t
ma y be traversed in preset steps nd at various heights across
the pontoon, along slots in the sail. Angles
of
tilt are shown by the movement
of
a
plumbline over an angular scale. as indicated in Fig 5.3 a).
The height of the centre
of
gravity
of
the whole floating assembly is first measured.
for one chosen height of the jockey weight.
he
pon to on is s usp en de d from a hole at
one side
of
the sail, as indicated
in
Fig 5.3 b), and the jockey weight
is
placed at such
a position on the line of symmetry as to cause the pontoon to hang with its base
roughly vertical. A p um bl in e is h un g from the su spe nsi on point.
he
height of the
centre of gravity G of the w ho le s usp end ed asse mb ly t he n lies at the point whe re the
plumbline intersects the line of symmetry of the pontoon. This establishes the
position of G for this particular jockey height. he position
of
G for any o th er j oc ke y
height may then be calculated from elementary statics, as will be seen later.
After measuring the external width and length
of
the pontoon. and noting the weights
of
the various components. the pontoon is floated
in
water.
Wilh the j oc ke y weight on the line of symmetry, small magnetic weights are used
trim the a sse mb ly to e ve n keel. i ndi ca te d by a zero reading on the a ngu la r scale.
jockey
is
then moved in steps across the width of the pontoon. the corresponding
t In some equipmenls. two jockey weights
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angle
of
tilt (over a range which
is
typically 8 being recorded
at
each step. This
procedure is then repeated with the jockey traversed at a number
of
different heights.
}
Jockey
weight
Angular
/
_/ scale
a Floating pontoon tilted
by movement
jockey
weight
, Suspension
l
f
l = = -
U
=
- .
-
Plumb
line
b Determination position
o centre
gravity
Fig
5.3
Sketch
a pontoon
Results
and
Calculations
Weight and imensions Pontoon
Weight
of
pontoon (excluding jockey weight) W
p
Weight
of jockey Wj
Total weight
of
floating assembly W
=
W
p
Wj
P
d
I V W 2.821
ootoon ISP acement
=
w
1000
Breadth of pontoon D
Length
of
pontoon L
Area
of
pontoon in plane
of water
surface
A LO
0.3601
x
0.2018
3
0.360 I x 0.20183
Second Moment
of
Area I
=
= ---,.,-- --- -- -
12 12
V 2.821
x
10-
3
Depth
of
immersion OC
= - =
c: ccc:: c
A
7.267 x
1
2
2.430 kgf
0.391
kgf
2.821
kgf
2.821 x IO 3
m
3
201.8
mm
=
0.2018
m
360.1
mm
0.3601
m
7.267 x 10-
2
m
2
3.88
X
10-
2
m
=
38.8 mm
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Height
of
centre
of
buoyancy B above 0
=
BC
= OC
2
Height o Centre Gravity
19.4
mm
When the pontoon was suspended as shown in Fig
5.3 b and with the jockey weight placed in the uppennost slot
of
the sail, the
following measurements were made:
=
G
c
B
o
i 5
I
Y
I
I
Fig 5.4 shows schematically the positions of the
centre of buoyancy B. centre of gravity G. and
metacentre M 0 is a reference point on the
external surface
of
the pontoon, and C is the point
where the axis
of
symmetry intersects the plane
of
the water surface. The thickness of the material
from which the pontoon is made is assumed to be
2 mm. The height of G above rhe reference point
o
is
OG. The height
of
the jockey weight above
o is Yj
Height
of
jockey weight above 0 Yj 345 mm
Corresponding height
of
G above 0 OG 92 mm
The value
of
may now be detcnnined for any other value
of Yj
If
Yj
changes by
6Yj then this will produce a change
in of
Wj ..6y/W. The vertical separation
of
the slots
in
the sail is 60 mm, so
will change
in
steps
[ 391
x 60/2.821
=
8.32
mm Table 5
J
shows the values
of
calculated in this way for the 5 different
heights Yj
of
the jockey weight.
Table
5 1
Heights
o G
above base
Do pontoon
Yi
OG
mm
mm
105
58.7
165
67.1
225
75.4
285
83.7
345
92.0
45
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BM BG+GM OG-OB+GM OG+GM-19.4mm
Table 5.2 ngles
of
lilt caused byjockey displacement
45.7
.391
x
330.0
2 821
5.76 mm/deg 5.76 x 57.3 3 0.0 mm/rad
GM
dX
l The preset sleps in
Xj
shown in the table are 15 rnm To provide accuracy, this has been reduced
to
7.5 mm in later versions of the equipment.
Inserting this into Equation 5.2 ,
This value, and corresponding values for other jockey heights, are entered
in
the
fourth column
of
Table 5.3. Values
of 8M
are also shown, derived as follows refer
to
Fig 5.4 for notation :
xperimental determination ofmetaeentric height
Jockey Jockey Displacement from Centre, Xj
Height
mm
mm
-45 -30
-15
0 15 30
45
105
-7.8 -5.2 -2.7
0.0
2.6 5.2 7.8
165
-6.2
-3.1 0.0 3.2 6.2
225.
-7.7
-3.8
0.0 3.9 7.8
285
-5.2
0.0 5.2
345 -7.5 -0.1
7.4
Table 5.2 shows the re.sults obtained when the pontoon was tilted by traversing the
jockey weight across its width
l
.
These results are shown graphically on Fig 5.5. For each of the jockey heights, the
angle of tilt is proportional to the jockey displacement. The metacentric height may
now
be found from Equation 5.2 , using the gradients of the lines
in
Fig 5.5. For
example, when 105 mm, the gradient is
330.0 mm/rad
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40
E
2
-
-
E
e
O l ~ ~ : : : . .. = ~ I
~
is.
.
~
20
u
o
..,
-40
-8 -6 -4
-2 0 2
Angle
of
tilt
4 6
8
ig Variation
0
angle
of
tilt
with jockey displacement
~ M
-
-
-
0
-
-
-
-
-
~
E 60
c >
~ .
~
40
+-
6
2 0 - - - - - - - - - - - - - - - - - - - - - - - - -
o
radient ofstability IiDe dx de mml )
ig 6 Variation st bilitywith me acentric height
47
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Jockey
OG
xj 9
Metacentric
BM
height
height GM
mm)
mm)
mm/ )
mm)
mm)
105 58.7 5.76 45.7 85.0
165
67.1 4.82 38.3 86.0
225 75.4 3.88 30.8 86.8
285 83.7 2.88 22.9 87.2
345 92.0
2 01
16.0
88
Table
5.3
Me/acentric height derived experimentally
As 8M depends only on the mensuration and total weight
of
the pontoon its value
should be independent
of
the jockey height and this is seen
to
be reasonably verified
by the experimental results. The value computed from theory is
8M
1
V
2.466 X 10
4
2.821 X 10
8.74 X 10
2
m
87.4
mm
which is in satisfactory agreement with the values obtained experimentally.
Another way of expressing the experimental results is presented in Fig 5.6 where the
height BG
of
the centre
of
gravity above the centre
of
buoyancy is shown as a
function of the slope
x e
The experimental points lie on a straight line which
intersects the BG axis at the value 90 mm. As BG approaches this value x S .
Namely the pontoon may be then tilted by an infinitesimal movement of the jockey
weight;
it
is in the condition of neutral stability. Under this condition the centre of
gravity coincides with the metacentre viz. BM
=
BG. So from Fig 5.6 we see that
BM =
90 mm. This experimentally detennined value again is in satisfactory
agreement with the theoretical value
of
87.4 mm.
Di cussiOIl sults
The experiment demonstrates how the stability
of
a floating body is affected by
changing the height of its centre of gravity and how the metacentric height may be
established experimentally by moving the centre of gravity sideways across the body.
The value established in this way agrees satisfactorily with that given by the
analytical result BM =
JlV