Chapter 2 Buoyancy Stability

CHAPTER 2 – BOUYANCY AND STABILITY CC501-HYDRAULIC 2

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CHAPTER 2

BUOYANCY & STABILITY

2.1 Introduction

Whenever a body is immersed wholly partially in a fluid it is subjected to an upward

force which tends to lift (or buoy) it up. This tendency for an immersed body to be lifted up

in the fluid, due to an upward force opposite to action of gravity is known as buoyancy.

The force tending to lift up the body under such conditions is known as buoyant force or

upthrust force. The magnitude of the buoyant force can be determined by Archimedes

Principle.

2.2 Archimedes Principle

When an object is immersed in water, it felt lighter. In a cylinder filled with water, the

action of inserting a mass in the liquid causes it to displace upward. So, this theory has

been discovered by the Greece scientist known as Archimedes of Syracuse (212 B.C). He

simplified this principle: When an object is completely or partially immersed in a fluid,

the fluid exerts an upward force on the object equal to the weight of the fluid displaced

by the object. These phenomena have become as Archimedes Principle.

For better understanding in this principle, refer the Figure 2.1 below. When object of

7kg is gradually immersed in the water, the weight of an object decreased to 3kg.

Subtracting these two scales readings, 7kg and 3kg, results in differences is 4kg. This is

called buoyant force that exerted on the object and it same as the weight of the water of an

object displaced. It is important to note that the buoyant force does not depend on the

weight or shape of the submerged object, only on the weight of the displaced fluid.


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Equation 1

The buoyancy force explains why some object sink and others float. An object will

float if the buoyancy force is greater than its weight. If the buoyancy force of the object is

equal to that of the fluid, the object will neither sink nor float. If the buoyancy force is

lesser than the weight of object, the object will sink. In summarize, an object will sink or

floats depend on how the buoyancy force compares with the weight. In formulas:

Weight of object, W = Weight of water displaced, R

mg = ρgVOBJECT

WeightObject, W

BuoyancyForce, R


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Problem 1

A ship shifts 115m3 of water. Determine:i) Weight of the ship. (Assuming the density of water = 1025kg/m3

ii) The volume of water that has a density of 1000kg/m3 may be displaced by the ship.

Solution:

i) Given V = 115 m3 , = 1025 kg/m3

Weight of object, W = Weight of water displaced, R

W = gV

= 1025 * 9.81 * 115

= 1156.35 kN

ii) Given density of water, ( = 1000 kg/m3 )

Volume displaced, V = W

g

= 1156.35 * 1000

1000 * 9.81

= 118 m3

Problem 2

Determine the volume and relative density of an object that has a value equal to the weightin air 2703N and in water 1909N.

Solution:

i) For stability,

The net value of the buoyancy, (R) = 2703 – 1909

= 794N


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R = gV

Volume, (V) = R/g where; g = 1000 * 9.81= 9810 N/m3

= 794 = 0.081 m3

9810

ii) Relative density (relative.) = substance / water

@ relative. = Mass of objectMass equivalent volume

= 2703/9.81 = 3.404794/9.81

2.3 Stability of Fully Submerged Bodies

For stability of a submerged body, body center of gravity (G) must be located directly

below center of buoyancy (B) fluid slipped. If both point located local, that body is in

equilibrium condition neutral at all position.

For stability cylinder and a float sphere, center of gravity these physical bodies must be lies

beneath center of buoyancy. Other floating object stability is dependent whether righting

moment or overturning moment generated when buoyancy center of gravity and center shift

away from vertical alignment because center position buoyancy move.

Centre of buoyancy will move because if the body or floating object slanting, fluid formslipped will alter past cause his center of gravity also changed.

Consider Figure 2.2(a) and Figure 2.2(b) below.

Figure 2.2(a) : Shift produce

recovery moment.

W W

R R

BG B

G

x


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Figure 2.2(b) : Shift produce reversal moment.

Volume weight body, W = Weight of fluid slipped ,mg ,act through centre of gravity body, G.

Upthrust force, R = ρgV measures through center of buoyancy, B.

If G under B as in Figure 4.1(a), then some shift of equal footing produce recovery momentwhich resulted balance again.

If G be above B as in Figure 4.1(b), then some shift produce reversal moment which resultedunstable body.

2.4 Stability of Floating Bodies

There were 3 types of equilibrium of floating bodies:-

1. Stable equilibrium

2. Unstable equilibrium, and

3. Neutral equilibrium

W

WR

R

BG B

G

x


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1. Stable equilibrium

When a body is given a small angular displacement (i.e. tilted slightly), by some externalforce, and then it returns backs to its original position due to the internal forces (the weight andthe upthrust), such an equilibrium is called stable equilibrium.

Survey in immersion of a the afloat body is dependent to weight (or his volumedisplacement) and to his body shape. Stability on the other hand set by force which acted whenbody harassed of static balanced position.

Figure 2.3(a) show a cross section vessel in static equilibrium. Centre of gravity (G), bevertical above center of buoyancy (B) - only in case when vessel lowest under waterline, whereB be above G.

Figure 2.3(b) show vessel that slant in angle θ. With fixed load, centre of gravity relativeposition will not change, but due to redistribution in body body, centre of buoyancy moving fromB to B'. Displacement sure will not change and consequently what happens was shaped watervolume wedge which represented by DOD section' has migrating across centre axis to EOE'.

D EG

O

B

W

Figure 2.3(a): Vesselin static equilibrium.

dA x

D’D’

E

E’

M

W

R = W

G

OB B’

Figureh 2.3(b): Vesselwith small slant angle.


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M was point of intersection for line thrust on just been through B' with original axissymmetry. M point known as meta and position center its relative to G would control balancevessel that. If M be above G, then same and contrary force with G and B', will produce couplewhich tends to return balance. On the other hand, if M be under G, vessel that will slant againuntil it overturned. GM distance called metacenter height, to cant at one small angle, it may bespecified by analysis according to following way.

M was point of intersection for line thrust on just been through B' with original axissymmetry. Volume Point each ‘wedge' can be gained with integrated waterline entire surface towide horizontal element dA (Refer Rajah 4.2(b)) times with height x tan θ. As such, moment oncenter axis was W tan θ x2 dA. But x2 dA @ I is second moment wide for surface waterline inlongitudinal axis, and further by taking moment on center axis for force buoyancy, found that;

WV x MB sin θ = W tan θ I ……………………… Equation 2

with V was volume of water jostled by vessel that, MB was center radius meta and I is secondmoment wide. Because θ was very small, sin θ tan θ θ, then, metacenter radius was;

MB=I ………………………………………………… Equation 3V

And metacenter height was;

GM = I – BG …………………………………… Equation 4V

When G be under B, BG blended, in height case metacenter must be positive and hisbalance is stable. Increased stability degree with height metacenter, but period roll off alsodepend especially to GM and too big value would be inclined to generate fast overthrow that isundesired.

Like those expected, most important metacenter height in architecture ship. When cargomove or ballast transported, center of gravity vessel will be shifted in direct equal to center ofbuoyancy, fugginess reduce stability. Tanks or room wall on the other hand give contrary effect.In work civil engineering marine, pontoon commonly used by way of something transport andsurvey in immersion controlled by a ballast tank system water designs closely.


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A vessel which possess wall pay heart and bring same liquid as in which it float,indicated in figure 2.3(c) . Moment reduction back original because inside wedge prism positionchange every room was W tan θ1 IB and with IB was second moment wide for free surface incolumn vessel in longitudinal axis of rotation.

Then Equation 2 (Moment Equation) rectified be;

WV1 x M1B1 sin θ1 = W tan θ1 (I – 2IB) …………… Equation 5

where; I = second moment wide for surface waterline in axislengthways.

IB = second moment wide for free surface in column vessel inlongitudinal axis of rotation.

V1 = volume of displacement vessel increase ballast.

Through approximation such as just now,

M1B1 = I – 2IB …………………………………………………… Equation 6V1

and

G1M1 = I – 2IB – B1G1 ………………………………………………… Equation 7V1

or

W1

M1

G1

O1

B1 B1’

W1

1

1

Figure 2.3(c): Vessel with ballast.


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Recovery Moment

In static equilibrium condition, some shift of equal footing produce recovery moment convergeto again balance

Recovery Moment = Wx = W.GM.sin ……………… Equation 8

2. Unstable equilibrium

If the body does not return to its original position from the slightly displaced angularposition and heels farther away, when given a small displacement, such an equilibrium is calledan unstabled equilibrium. For unstable equilibrium , the position of metacenter M is remainslower than G.

G

B

W

R = W

x

WR

M

GB B’

Figure 2.4(a) : Body inbalance static. Figure 2.4(b) : Body in

shift with angle andproduce recovery moment.

Note!

When M was sitting on G, then GM valued positive ( +ve).


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Reversal Moment = Wx = W.GM.cos(900-) …………… Equation 9

3. Neutral Equilibrium

If a body, when given a small angular displacement, occupies a new position and remainsat rest in this new position, it is said possess a neutral equilibrium. For Neutral equilibrium,the position of metacenter M coincides with G.

G

B

W

R = W

Figure 2.5(a): Body inbalance static.

G B

W

Figure 2.5(b): Shiftproduce reversal moment.

M

R

B’

x

Note!

When M lies beneath G, then GM valued negative (-ve).


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where;

GM = 0, (no rotation moment)

object position would be neutral.

2.5 Determine the Location of Metacenter of Floating Bodies

The position of the so-called metacenter, metacentric height, is of crucial significance to stabilityof a floating body. The metacentric height is an essential factor w assessing the stability of aship in waves.

There are 2 method on Determining Metacenter Height (GM):-

Method 1;

Consider a body being overwhelmed liquid. This body prismatic as in figure 2.7 below. Equationwould appear legal also to body non prismatic. When small shift happened, B buoyancy's centermoved to B'. Distance horizontal this movement is r. Shift in axis through O, produce prismwhich arise in right and submerged left.

G

B

W

Figure 2.6: Body in balanceneutral.

R

B’

M

Note!

When M overlaps with G, then GM = 0.


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Balanced effort;

F = 0

FB = W

where;

FB = bouyant force

W = body weight

Take moment from axis through B;

Figure 2.7

M

O

G

B

B’

F

F

W

FB

I

r

s


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F I = FB r

= W r ………………………… (1)

Moment from F coupling namely FI could be evaluated by analyzing moment to one smallelement in the prism. Say this element has vast dA section, and his distance from O were x.

Element volume

= (dA)(h)

= (dA)(x tan )

= x dA(for small)

Bouyant force on element

= (specific weight)( element volume)

= ()( x dA)

This force moment from O

= (force)( element distance)

= ()( x dA)(x)

= ()( x2 dA)

FI = ()( x2 dA)

x

h

dA

Figure 2.8


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= ( ) ( x2 dA)

where;

( x2 dA) = second moment of area from axis through O

= Io

FI = Io ………………………… (2)

Include equation (2) in (1)

Io = Wr

Io = ( V)r

Io = Vr ………………………… (3)

From Figure 4.5,

r = MB sin

= MB ………………………… (4)

Equation solution (4) and (3);

Io = V MB

MBIV

o

Metacenter height;

GM = MB GB


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GMIV

GBo

Negative signal applies if G be above B, and positive sign if G be under B. Recovery moment

(restoring couple) can be counted based on association following;

M = W (GM) sin ………………………… Equation 10

Problem 3

Determine metacenter height of a pontoon vehicles carrier to across a straits which possesshis seawater density 1150 kg.s/m3. Pontoon that measuring 27m long, 19m broad and ashigh as 9m while heavy pontoon on the other hand is 500 metric tones.

Solutions:

Weight of pontoon = 500 x 103(9.81)

= 4905000 N

Water weight slipped (faulted), W = gV

= 1150(9.81)(27 * 19 * d)

Because; water weight slipped = weight of pontoon

1150(9.81)(27 * 19 * d) = 4905000

11281.5(513d) = 4905000

5787409.5d = 4905000

d = 4905000

5787409.5

= 0.85m


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OB = d/2 = 0.425m

BG = OG – OB

= 9/2 – 0.425

= 4.5 – 0.425

= 4.075 m

MB = I/V where; I = bd3

12

= 27 (193)

12

= 185193

12

= 15432.75 m4

MB = 15432.75

436.05

= 35.392m

Metacentre height, GM

GM = MB – BG

= 35.392 – 4.075

= 31.317 m


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Method 2;

GM can also be determined if tip due to P burden 's movement to distance x above promenade

measured.

Reversal moment due to P burden = Px

If W = mg was ship and burden total weight P,

Recovery moment = W. GM.θ

For balance;

Reversal moment = recovery moment

W. GM.θ = P

GM = P

Wθ…………………………………………… Equation 11

Centre reversal meta was when value θ 0

M

GG’

W

x

P P’

Figure 2.9 : High determination metacenter using method 2.


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Problem 4

A pontoon is sailing inside a lake shift as heavy as water 2 x 105 kg and as heavy as buffer20 tones shift 8.5m across the deck. Because of that pendulum 5.4m move 0.2m horizontal,determine metacenter height for pontoon to verify information that was stated above.

Solutions:

W. GM. sinθ = Px (tan = ; for angle small)

GM = PxWsin

= 20(1000)(8.5)2 x 105(0.2/5.4)

= 1700007407.41

= 22.95 m

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Chapter 2 Buoyancy Stability