Buoyancy and Stability of Floating Bodies.ppt

7/24/2019 Buoyancy and Stability of Floating Bodies.ppt

1/19

Buoyancy and Stability of Floating Bodies

Most problems concerned with totally or partially immersed bodies:

Principle of Archimedes:

The principles of Archimedesstates that the up thrust ( upward vertical

force due to the fluid )on a body immersed in a fluid is equal to the weight of

the fluid displaced, The up thrust will act through the center of gravity of the

displaced fluid, which is called the center of buoyancy.

-A body immersed in a fluid experiences a vertical buoyant force equal to the

weight of the fluid it displaces

A floating body displaces its own weight in the fluid in which it floats


2/19

!hese two laws are easily derived by referring to the figure" the body lies between an

upper curved surface # and a lower curved surface $ %rom &q for vertical force" the

body experiences a net upward force

'p-thrust on immersed body:


3/19

`

- Stability and Metacenter :

Weight W

G

B

Up-thrust ! W

Weight W ! Up-thrust

eight of liquid displaced

Rand W must act in the same vertical straight line.


4/19

# !he body is tilted a small angle " and a new waterline is established for the body to float at this

angle !he new position * of the center of buoyancy is calculated A vertical line drawn upwardfrom * intersects the line of symmetry at a point M" called the metacenter" which is independent of

for small angles

$ +f point M is above ," that is" if the metacentric height M, is positive" a restoring moment is present

and the original position is stable +f M is below , (negative M," the body is unstable and will

overturn if disturbed tability increases with increasing M,

"a#. Tilt the body a small angle "b#Bmoves far out "pointM above G denotes stability#$

"c#Bmoves slightly"pointMbelow G denotes instability


5/19

G

B

%

W

R = W

&. 'f % lies above G a righting moment is produced, G% is regarded as

positive, and equilibrium is stable.

(. 'f % lies below G an overturning moment is produced, G% is regarded as

negative, and equilibrium is unstable.

). 'f % and G coincide the body is in neutral equilibrium.


6/19

G

M

B B1

W

FB1

G

M

B B1

W

FB1

G M

B B

W

FB

GM is +ve positive

Stable

GM is -ve negative

Unstable

GM is 0

Neutral


7/19

Determination of te Meta!entri! eigt

GM " BM - BG

Were#

$%G

M

B B1

W

FB1

BG*

'% = GG

+ &s te 'rea Moment of &nertia of te top vie( about te verti!al a)is

. &s te volume of li*ui$ $ispla!e$ b te bo$,

%* / g .

BB1" BM sin $%

BB1" BM , $%


8/19

)ample

A block of wood 0.2 x 0.5 m cross-section and 0.8 m long as a mass

of !" kg. #an te block float wit 0.5 m side $ertical%

Solution#

0,.

m0,/m

0,

m

0,.m

0,

mG

B0,.

m

0,.

m

0,1

m

Water 2evel

)m+.+&+++

* ==

m+.

+.+.(

+.+h =

=

&ol'me of dis(laced water

)e eigt of s'bmergence


9/19

BG

*

'%G =)e *etacentric eigt

)e center of b'o+anc+ is 0.2 m from te base.

)

m++-(,,.+&(

-.+.+' =

=

m+&,/.++0.++,-.+

++-(,,.+

G% ==)e *etacentric eigt

BG ! +.(0 1 +.( ! +.+0 m

G% ! B% - BG

G% 2 + )e ,lock is stable wit its 0.5 m side $ertical

l 3.4


10/19

)ample 3.4

A vessel has a displacement of $011 111 2g of fresh

water A mass of $1 111 2g moved 3 m across the dec2

causes the lower end of a pendulum 4 m long to move $4cm hori5ontal 6alculate the transverse of metacenteric

height

m

m

m

5,G,

B


11/19

m

m

0.2m

m

WmassW

5,G,

M

%

%

5,G,

/,/.+)

().+3tan == //.+

)

().+3 ==or

B


12/19

m

m

0.2m

m

WmassW

5,G,

M

%

%

5,G,//.+)

().+3 ==

mmo$ing= 20 x /0 kg

mtotal= 25 x /05 kg

B

B6

m4.+G%hieghtc%etacentri

G%)

().+

&.4+++(0++4&.4(++++

momentrightingmomentgoverturnin

==

=


13/19

Liquid in Relative Equilibrium

Horizontal acceleration:

- f te $essel containing a li1'id is at rest or mo$ing wit constant

linear $elocit+ te li1'id is not affected b+ te motion

- 3l'id (ress're is e$er+ were normal to te s'rface on wic it acts.

- ,'t if te container is gi$en contin'o's acceleration tis will be

im(arted to te li1'id wic will take '( a new (osition.

1,m1,m

A * 6moving (it !onstant linear velo!it

at rest or


14/19

1,m1,m

1,m

a

7

A*

6

%

and is constant for all (oints on te free s'rface.

f te linear acceleration is gi$en b+ a = m4sec2 te tank is m

long and te de(t of water wen te tank is at rest is /.5 m.

#alc'late a- )e angle of te water s'rface to te ori6ontal.

b- )e maxim'm (ress're intensit+ on te bottom.

c- )e minim'm (ress're intensit+ on te bottom.

3or e1'ilibri'm 3 = W . tan

)ample

== tan.Wa.m5

g

atan

tan.g.ma.m

=

=


15/19

Solution#7a

o

&/3 =

A = /.5 9 /.5 tan = /.! m

7b

7c

)e (ress're intensit+ at A = 6 = .8/ x /0 x /.! = /.2 :;4m2

,= /.5 < /. 5 tan = /.0"

)e (ress're intensit+ at , = B = .8/ x /0 x /.0"= /0.2 :;4m2

*axim'm (ress're intensit+ occ'rs at (oint A

)e minim'm intensit+ of (ress're occ'rs at (oint ,.

&.4

)

g

atan ==


16/19

- ertical acceleration:

# ,

7rism !ross-se!tion

area is " a

aAccelerating force at = 3

force d'e to (ress're weigt of (rism 3 = > . A - .g. .A

,+ ;ewton?s 2nd law

3 = mass of (rism x acceleration =

7

a.h.g.

#g

a&"h.g.7 +=


17/19

- Forced vorte!:

r

7

A

6

*

8

Axisof

rotation

8

%

rd

dy3tan

r..g

W5 (

=

=

g

r9

g

a

rd

dy3tan

(

===

3or a constant $al'e of 3 will $ar+

wit r since te centrif'gal acceleration is

82. r and

ntegrating te

ttanconsg(

rdr

g

ry

((r

+

(

+

=

=


18/19

f + is meas'red from te A, i.e + = 0 wen rx = 0 .

#onstant = 0

A c+lindrical tank is s('n at 00 r.(.m. wit its $ertical axis. )e tank

is 0.! m ig and "5 #m diameter and filled of water before s(inning.

@ow tat te water s'rface will take te form of (arabolic wen tecontainer is s('n and calc'late

7a )e s(eed at wic te water s'rface will 'st to'c te to(

rim and te center bottom of te tank.

7b )e le$el to wic te water will ret'rn wen te tank sto(s

s(inning and te amo'nt of water lost.

)ample

g(

ry

((=


19/19

Solution#

7a Wen te s'rface 'st to'ced te rim of te base + = 0.! m and

r = 0.225 m

.:sec.rad0(&0.;

yg(9

( ==

; = /"5.! r.(.m

7b n fig're te $ol'me of te (arabolic will be alf tat of te containing

c+linder A,#B . Wen te water s'rface to'ces te rim and te center

of te bottom

$ol'me of water left in tank = alf of te original $ol'me

Be(t wen tank ceases not rotate = 0. m

= CD 70.2252x 0.! = 0.0"E5 m

Amo'nt of water trown o't = 0.0"E5 m

r(

Buoyancy and Stability of Floating Bodies.ppt

Documents