Statics Part 2

7/29/2019 Statics Part 2

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HYDROSTATIC FORCE ON A PLANE SURFACE

DETERMINATION OF HYDROSTATIC FORCES IS IMPORTANT IN THE

DESIGN OF STORAGE TANKS, SHIPS, DAMS, AND OTHER HYDRAULIC

STRUCTURES

For fluids at rest

The force must be perpendicular to the surface since

there are no shearing stresses present.

The pressure will vary linearly with depth if the fluid is

incompressible.


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HYDROSTATIC FORCE ON A HORIZONTAL PLANE SURFACE

FR = hA

The resultant force of a static fluid on a plane surface is due to the

hydrostatic pressure distribution on the surface


3/57

yyC

yRFR

dF

hhc

Free Surface

Centroid, c

Location of resultant force

(center of pressure, CP)

HYDROSTATIC FORCE ON A INCLINED PLANE SURFACE OF ARBITRARY SHAPE

O - origin

xR

xC

dA

x

y

x


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AA

R dAsinydAhF sinyh

For constant and

sinyAdAysinF cA

R

A

dAy - First moment of the area w.r.tx-axis

AydAy c

A

yc

- y-coordinate of the centroid measured fromx-axis which passes through O

cR hAF

hc - vertical distance from the fluid surface to the centroid of the area


5/57

MAGNITUDE OF FR IS

Independent of angle

depends on specific weight, total area, depth of centroid of the area

below the surface

cc hP

FR IS PERPENDICULAR TO THE SURFACE

Magnitude of the resultant fluid force is equal to the pressure acting

at the centroid of the area Pc multiplied by the total area

RESULTANT FORCE SHOULD PASS THROUGH THE CENTROID OF THE

AREA - ?

cR hAF

APF cR


6/57

yR of the resultant force - summation of the moments aroundx-axis

Moment of resultant force = Moment of the distributed pressure force

AAAA

RR dAysindAsinyydAhydFyyF

2

sinyAF cR

A

2Rc dAysinysinyA

Ay

I

Ay

dAy

yc

x

c

AR

2

A

2dAy- second moment of area (Moment of Inertia), Ix,w.r.t an axis

formed by the intersection of the plane containing the surface and

the free surface (x-axis)


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Parallel axis theorem, 2cxcx AyII

xcI Second moment of the area w.r.t an axis passing through its

centroid and parallel to thex-axis

cc

xcR y

Ay

Iy

Resultant force does not pass through the centroid but is always below

it

Ay

I

Ay

dAy

yc

x

c

AR

2


8/57

xR of the resultant force - summation of the moments around y-axis

AA

RR dAxysindFxxF

Ay

I

Ay

dAxy

xc

xy

c

AR

Ixy

Product of the inertia w.r.t thexand y axes

c

c

xycR x

Ay

Ix

Product of the inertia w.r.t an orthogonal set of axes (x-y coordinate

system) is equal to the product of inertia w.r.t an orthogonal set of

axes parallel to the original set and passing through the centroid plus

the product of the area and thex and ycoordinates of the centroid of

the area

ccxycxy yAxII


9/57


10/57

cRhAF

MAGNITUDE OF THE RESULTANT FORCE

c

c

xycR x

Ay

Ix

LOCATION OF THE RESULTANT FORCE

Direction : FR IS PERPENDICULAR TO THE SURFACE

cc

xc

R yAy

I

y


11/57

The 4-m-diameter circular gate is located in the inclined wall of a large reservoir

containing water The gate is mounted on a shaft along its horizontal diameter. For a

water depth of 10 m above the shaft determine: (a) the magnitude and location of

the resultant force exerted on the gate by the water, and (b) the moment that would

have to be applied to the shaft to open the gate.


12/57

cR hAF 104410008192.FR NFR 1232761

cc

xc

R yAy

I

y

60

10

4

460

10

24

2

4

sin

sin

yR

m.m.yR 551108660

m.yR 611

For the co-ordinate system shown, xR = 0 since the area is symmetrical and the

centre of pressure must lie along the diameter A-A

The magnitude and location of the resultant force exerted on the gate by the

water are 1.23 MN and acts through a point along its diameter AA at a

distance of 0.0866 m (along the gate) below the shaft


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The moment that would have to be applied to the shaft to open the

gate

cRR yyFM

08660101230 3 .M

m.N.M 510071


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PRESSURE PRISM

2

hAyAF cR

h3

2

2

h

6

h

2

h

bh.2

h12

hb

y

3

R

ccxcR yAy

Iy

h3

2From the free surface of fluid

h3

1From the base of the fluid


15/57

Rc FAyhbh

hbhVolume

2

2

FR = Volume

Centroid of the volume =3

h


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2F

1212

1F

121R bhhhh2

1hhbhF

2211RR yFyFyF 2

hh

y

12

1 122 hh3

2

y

f i f d i i h f b d l


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Use of pressure prisms for determining the force on submerged plane areas

is convenient if the area is rectangular so the volume and centroid can be

easily determined

For non-rectangular shapes, integration would be needed to find volume and

centroid. Then, previous approach would be better

Effect of atmospheric pressure on a submerged area - Nil

A i d k i il (SG 0 9) d h 0 6 b 0 6 l


18/57

A pressurized tank contains oil (SG = 0.9) and has a square, 0.6-m by 0.6-m plate

bolted to its side, as is illustrated in Fig. When the pressure gage on the top of the

tank reads 50 kPa, what is the magnitude and location of the resultant force on the

attached plate? The outside of the tank is at atmospheric pressure.


19/57

NFFF

NF

..

..Ahh

gF

N

...

AghpF

R

s

25354

954

3602

60819100090

2

24400

360281910009050000

21

2

12

2

11


20/57

m.y

..y

.F.FyF

o

o

oR

2960

20954302440025354

203021

HYDROSTATIC FORCE ON CURVED SURFACE


21/57

Isolate a volume of a fluid that is bounded by

the surface of interest, ie., section BC

2H FF WFF 1V

2V

2HR FFF

Point O summing moments about an appropriate axis

HYDROSTATIC FORCE ON CURVED SURFACE

Figure shows a gate having a quadrant shape of radius of 2m Find the resultant


22/57

Figure shows a gate having a quadrant shape of radius of 2m. Find the resultant

force due to water per metre length of the gate. Find also the angle at which the

total force will act

cx hAF

2

2121000819.hAF cx

cc

xcR y

Ay

Iy

2

3

2

3

2

262

2

12

3

hhhh

bh.h

hb

yR

Horizontal force

liquidtheofsurfacefreethefrommyR3

4

NFx 19620

V ti l f


23/57

Vertical force

0124

10008192 ..VolumeFy

Fy weight of the water (imagined) supported by AB

NFy 30819


OBfrom..R

84803

024

3

4

22223081919620yxR FFF

N.FR 436534

The angle made by the resultant with the horizontal is

1357

19620

30819 o

x

y

F

FTan

Compute the horizontal and vertical forces


24/57

Compute the horizontal and vertical forces

acting on a curved surface AB, which is in the

form of a quadrant of a circle of radius 2 m.

Take the width of the gate as unity.

cx gAhF

512

2128191000 ..Fx

NFx 49050

cc

xcR y

Ay

Iy

633251121511

12

213

....

yR

liquidtheofsurfacefreethefromm.yR 6332

Vertical force


25/57

Vertical force

0125124

10008192

...VolumeFy


NFy 60249


OBfrom..R 84803

024

3

4

Buoyancy Force (Hydrostatic Lift)


26/57

Buoyancy Force (Hydrostatic Lift)

When a body is completely submerged in a fluid, or floating so that it

is only partially submerged, the resultant upward force acting on the

body is called BUOYANCY FORCE

The line of action of the buoyant force passes through the centroid of

the displaced volume. The centroid is called the Center of buoyancy

Buoyancy force on the body = Weight of the fluid displaced by the body

P1


27/57

blPPF 12B

2z

1z

2p

1p

12 dzdpPP

2z

1z

1z

2z

B dzbldzblF

y

P1

x

z

l

P2b

z1

z2

Weight of the fluid displaced by the body

21B zzgblFIf is constant

1z

2z

dzgbl

21

zzblVVgFB



28/57


Archimedes Principle Greek Scientist (287 BC 212 B.C)

Archimedes Principle holds good

for bodies of any general shape

for both gases and liquids

does not require density to be constant

water = 1000 kg/m3

air = 1.225 kg/m3

Buoyancy force in water is thousand times greater than in air

buoyancy force is important

For naval vehicles

lighter than air vehicles hot air balloons

ARCHIMEDES SECOND PRINCIPLE OF BUOYANCY


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ARCHIMEDES SECOND PRINCIPLE OF BUOYANCY

A floating body displaces a volume of fluid equivalent to its own

weight

A body will float if its average density is less than the density of thefluid in which it is placed

sfB VgF

f - Density of fluid

Vs - Submerged volume

Vs

Liquid ( f)


Find the volume of the water displaced and the position of the centre of buoyance


30/57

Find the volume of the water displaced and the position of the centre of buoyance

for wooden block of width 2.5 m and of depth 1.5 m when it floats horizontally in

water. The density of the wood is 650 kg/m3. Length of the wooden block is 6 m.

Weight of the block = g Volume = 650 9.81 2.5 1.5 6 = 143471.25 N

Buoyancy force on the body = Weight of the fluid displaced by the body =

weight of the wooden block

1000 9.81 Volume of the water displaced = 143471.25

Volume of the water displaced = 14.625 m3

Volume of the wooden block in water = volume of the water displaced

14.625 = 2.5 h 6 h = 0.975 m

Position of the centre of buoyancy = h/2 = 0.975/2 = 0.4875 from the base

A stone weighs 392 4 N in air and 196 2 N in water Compute the volume of stone


31/57

A stone weighs 392.4 N in air and 196.2 N in water. Compute the volume of stone

and its specific gravity

Weight of stone in air Weight of stone in water = weight of water displaced

392.4 196.2 = 1000 9.81 Volume of the water displaced

Volume of the water displaced = 0.02 m3

Mass of the stone = weight in air/g = 392.4/9.81 = 40 kg

Density of the stone = Mass of the stone/volume = 40/0.02 = 2000 kg/m3

Specific gravity of the stone = 2000/1000 = 2

Hydrometer to measure the specific gravity of a liquid


32/57

Hydrometer to measure the specific gravity of a liquid

Hydrometer in water Hydrometer in other liquid

oOH VW 2 AhVSGW oOH2

AhVSGVW oOHoOH22

AhV

VSG

o

o

SG of buffalo milk = 1.0323

SG of cow milk = 1.0317

Buoyancy force per unit of

submerged volume is greater, so less

volume of submergence is necessary

to balance the hydrometers weight

Mercury or lead to keep it

upright

STABILITY


33/57

STABILITY

Stable equilibrium a body returns to its equilibrium position, when

displaced

Unstable equilibrium a body moves to new equilibrium position,

when displaced (even slightly)

Neutral Unstable Stable

Stability is important for submerged or floating bodies since the centreof buoyancy and gravity do not necessarily coincide

A small rotation can result in either a restoring or overturning couple

STABLE centre of gravity is BELOW the centroid of displaced volume


34/57

STABLE centre of gravity is BELOW the centroid of displaced volume

Disturbing couple

W W

W

SUBMERGED BODY

UNSTABLE centre of gravity is ABOVE the centroid of displaced volume


35/57

UNSTABLE centre of gravity is ABOVE the centroid of displaced volume

Disturbing couple

W W

W

Floating Bodies


36/57

Disturbing

couple

Floating Bodies

Floating body stability problem is more complicated, since the body rotates

the location of the centre of buoyancy may change (which passes through

the centroid of the displaced volume)

BODY SHORT AND WIDE STABLE EQUILIBRIUM

W

W


37/57

Disturbing couple

W W

TALL AND SLENDER BODY UNSTABLE CONFIGURATION

DETERMINATION OF STABILITY OF SUBMERGED OR FLOATING BODIES


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DETERMINATION OF STABILITY OF SUBMERGED OR FLOATING BODIES

IS DIFFICULT

For complicated geometry and weight distribution of the body

Inclusion of other external forces like wind currents

STABILITY IS IMPORTANT

Design Of Ships, Submarines Naval Architects

METACENTRE


39/57

G

C

G

C C1

M

W

FB

METACENTRE

Metacentre is defined as the point about which a body startsoscillating when the body is tilted by a small angle

Metacentre may also be defined as the point at which the line of

action of the force of buoyancy will meet the normal axis of the body

when the body is given a small angular displacement

To determine distance GM


40/57

x-co-ordinate of the centroid of the displaced volumex

2211VxVxVx

1V Area DOE times the length

2V Area AOB times the length

The cross section is assumed to be uniform so that the length lis constant

DOE AOB


41/57

2211VxVxVx

21VV

VxdVxdVx

dATanxVd Volume 1 - DOE

dATanxVd Volume 2 - AOB

21

22

AA

dAxTandAxTanVx

A

dAxTanVx 2 TanIVx o

Io second moment (moment of inertia) of the water line area about an axis passingthrough the origin O. Waterline area = length of the body (l) length AE

TIV


42/57

TanIVx o

TanCMxCM

xTan

V

ICM o

CGV

I

GMo

CGCMGM

A floating object is in Stable Equilibrium if M is above G - MG is positive

Unstable Equilibrium ifM is below G - MG is negative

Metacentric height

MG Metacentric height Distance between metacentre and centre


43/57

g

of gravity

yG

B

x

z

z

z

Plan ofwater

surface

BG

V

IMG

s

z

zI

Moment of inertia about the roll axis zz of the

plan view area of the ship at the water line

sV Submerged volume

A floating object is in

Stable Equilibrium if M is above G

MG is positive

Unstable Equilibrium ifM is below G

MG is negative


44/57

NEUTRAL

EQUILIBRIUM

STABLE

EQUILIBRIUM

M is above GDisturbing couple

anticlockwise

Restoring couple

clockwise

GM positive

UNSTABLE

EQUILIBRIUMM is below G

Disturbing couple

anticlockwise

Restoring couple

Anticlockwise

GM negative

A solid cylinder of diameter 4 m has a height of 4 m. Find the metacentric height of


45/57

the cylinder if the specific gravity of the material of the cylinder = 0.6 and it is

floating in water with its axis vertical. State whether the equilibrium is stable or

unstable.

VICM

444

6464dI

42444

22 .hdV

m.

.

V

IBM 41670

4244

464

2

4

CGV

IGM m...GM 3833021241670

Negative sign implies metacentre is below centre of gravity. Hence, the cylinder is in

unstable equilibrium

PRESSURE VARIATION IN A FLUID WITH RIGID BODY MOTION


46/57

a

k

pEven though a fluid may be in motion, if it moves as a rigid body there

will be no shearing stresses present

xax

Pya

y

Pza

z

P

Example :Container of a fluid accelerates along a straight path

Fluid is contained in a tank that rotates about fixed axis

xy

z

LINEAR MOTION


47/57

LINEAR MOTION

az

ay

0ax existaanda yy

;0xp ;a

yp y zaz

p

P =f(y, z)

dp

dp

dp p


48/57

dzz

pdy

y

pdp ;a

y

py za

z

p

dzzagdyyadpAlong a line of constant pressure, dp = 0

dzzagdyya0

z

y

ag

a

dy

dz

Special case 0ay 0az 0dydz

zag

zd

pd

The tank shown in Fig. a is accelerated to the right. Calculate the acceleration ay


49/57

needed to cause the free surface shown in Fig. b to touch point A. Also, find PB

The angle of the free surface is found by equating the air volume (actually, areas

since the width is constant) before and after since no water spills out.

m.xx.. 6670212

1202

ayay

adz


50/57

z

y

ag

a

dy

dz

0za 816670

21

66700

021

0..

.

.

.

g

a

g

a

dy

dz yy

266178181981 s/m....gay

26617 s/m.ay

dzzagdyyadp

dyadp y

AByAB yyapp

0266171000 .pB PapB 35300

RIGID BODY ROTATION


51/57

zr i

z

Pi

P

r

1i

r

Pp

iP

iP1

iP


52/57

r

2

r

i ra 0a 0az

zr iz

ir

ir

p

akp

2rrP 0

P

z

P

P =f(r, z)

dzz

pdrr

pdp

dzdrrdp 2


53/57

dzdrrdp

Along a line of constant pressure,dp = 0

gr

gr

drdz

22

ttanconsg2

rz

22

Constant Pressure lines are parabolic

dzdrrdp 2

ttanconsz2

rp

22

Constant

pressure at some point ro and zo

r22


54/57

ttanconsz2

rp

22

12

2

1

2

2

2

12

2zzgrrpp

It two points are on the constant pressure surface, such that that the free surface

locating point 1 on the z-axis so that r1

= 0

12

2

2

2

2zzg

r


55/57

PRESSURE VARIES WITH DISTANCE FROM AXIS OF ROTATION

AT A FIXED RADIUS, PRESSURE VARIES HYDROSTATICALLY IN THE

VERTICAL DIRECTION

The cylinder shown in Fig is rotated about its centerline. Calculate the rotational

d th t i f th t t j t t h th i i O d th t


56/57

speed that is necessary for the water to just touch the origin O and the pressures at

A and B.

No water spills from the container, the air volume remains constant

122

1210

22 R cm.R 775

12

2

2

2

2

zzgr

0

100

12819

2

1077522

..

s/rad.626

Volume of the paraboloid

of revolution is one half of

a circular cylinder with

the same height and

radius

22

2

zzgrrpp


57/57

121212

2zzgrrpp

22

2

2 oAA rrp

Pa..

rrp oAA 35400102

626

2

22

2

22

2

1208191000 ..zzgpp ABAB

Pa..pB 236012081910003540

PapB 2360

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Statics Part 2