Abj1 4.2.1: Pressure, Pressure Force, and Fluid Motion Without Flow [Q1] 1.Area as A Vector ...

abj 1

4.2.1: Pressure, Pressure Force, and Fluid Motion Without Flow [Q1]

1. Area as A Vector

Component of Area Vector – Projected Area

Net Area Vector for A Two-Dimensional Surface

2. Resultant Due to Pressure

Resultant Force and Moment (on A General Curved Surface)

Questions of Interest

Q1: Given the pressure field/distribution , find the net/resultant pressure force and

moment on a finite surface

-------------------

4.2.2

Q2: Given the pressure field/distribution , find the net pressure force (per unit

volume) on an infinitesimal volume

Q3: Given a motion (fluid motion without flow), find the pressure field/distribution

),( txp

),( txp

),( txp

abj 2

1. Area as a vector

1. Projected area and Component of area vector

2. Net area vector

2. Surface Force: Resultant (Force and Moment) Due to Pressure


Question of Interest

Q1: Given the pressure field/distribution , find the net/resultant pressure

force and moment on a finite surface.

Very Brief Summary of Important Points and Equations [1]

A

),( txp

ApdrFdrMdM

FdrMd

ApdF

ApdFd

Ad

A

later) sum component,by(integrate,

n)integratiovector(direct

iizyx dAAAAA

AdA

abj 3

Special Case: Resultant force due to uniform pressure

Component of the pressure force

Net pressure force

Very Brief Summary of Important Points and Equations [2]

ApF

later) sum component,by (integrate:areaLike,

n)integratioctor (direct ve:areaLike

A

iizyx

A

pdAFFFF

ApdFdF

zyxiAppdAF i

A

ii ,,,area projected theontodueforce

abj 4

Area as A Vector

abj 5

Area as a vector

A

Magnitude of = Magnitude of the area

Direction of = Outward normal (from the system to the surrounding)

A

ne = Outward unit normal vector pointing from the system to the surroundings

neAA ˆ

ne

system

surrounding

neAA ˆ

ne

system

surrounding

abj 6

Component of Area Vector – Projected Area [1]

xxxxx eAeAA ˆcosˆ

yyyyy eAeAA ˆcosˆ

x

y

xAx

yyA

A

l

ne

yyxxn eee ˆcosˆcosˆ

yA

Magnitude of the projected area

yyyy AlwwlA coscos)()cos(

x

y

xA

Magnitude of the projected area

xxxx AlwwlA coscos)()cos(

w

l

ne

neA

A

x

y

z

x

y

abj 7


x

y

z

neAdAd ˆ

zzyyxxn eeee ˆcosˆcosˆcosˆ

xx

xxx

eAd

eeAdAd

ˆcos

ˆˆ

yy

yyy

eAd

eeAdAd

ˆcos

ˆˆ

zz

zzz

eAd

eeAdAd

ˆcos

ˆˆ

In 3D

abj 8


)(,ˆ signedeAddA zz

)(

ˆ

signed

eAddA yy

A

z

x

y

)(,ˆ signedeAddA xx

neAdAd ˆ

abj 9

How to Find The Net Area Vector for A Two-Dimensional Surface

w

l

x

y

z

neAA ˆ

x

y

xAd

yAd

Ad

ne

wdyAd x rd

1

2

wdxAd y

dy

dx

yxyx

y

x

x

x

y

y

A

yyyy

xxxx

yx

yx

xy

yxz

yxz

ewlewlexxweyyw

ewdxewdy

AdA

wdxAdwdxdAewdxAd

wdyAdwdydAewdyAd

AdAd

ewdxewdy

ewdyewdx

ewdyewdxe

edyedxrdrwdeAd

ˆ)cos(ˆ)sin(ˆ)(ˆ)(

ˆ)(ˆ)(

,,ˆ)(:

,,ˆ)(:

ˆ)(ˆ)(

ˆ)(ˆ)(

ˆ)(ˆ)(ˆ

ˆ)(ˆ)(),(ˆ

1212

2

1

2

1

The net area vector can be found simply by summing its projected area components.

yAd

xAd

abj 10

How to Find The Net Area Vector

A

z

x

y

A

zA

yA

A

z

x

y

xA

A

Similar approach can be used in 3-D:

The net area vector can be found simply by summing its projected area components.

zyx AAAA

abj 11

Example: Find The Net Area Vector

Questions:

1. Find the net area vector for the 2-D surfaces (a) and (b). Both have depth w.

2. Are they equal? If so, why?

Surface (b) is plotted here as dotted line in order to compare its size to (a).

(a) (b)

h

l

y

xzAd Ad

abj 12

Resultant Due to Pressure


Question of Interest

Q1: Given the pressure field/distribution , find the

net/resultant pressure force and moment on a finite surface

),( txp

abj 13

system

surroundingA

Resultant Due to Pressure (on A General Curved Surface): Finding 1) Resultant Force and 2) Resultant Moment

Ad

1 Infinitesimal area element

ApdFd

2 Infinitesimal pressure force on the element

A

ApdF

3 Resultant pressure force on the finite area A

)( Apdr

FdrMd

4 Infinitesimal moment of pressure force about C

A

A

Apdr

FdrM

)(

5 Resultant moment of pressure force

about C

Ad

ne

ApdFd

r

C

FdrMd

Ad

Fd

Minus sign: is always opposite to

is compressively normal to the surface

Fd

abj 14

Some Properties of Pressure Force [1]

ApdFd

Ad

FdFd

1. is always opposite to

is compressively normal to the surface

),( txp

2. If the pressure field is uniform over A

) of vector area(net ) (uniform A p

ApF

AdpApdFA

Aoveruniformisp

A

abj 15

3. Component of the pressure force as a force due to pressure (distribution on the

surface S being projected) onto the projected area.

Some Properties of Pressure Force [2]

area projected theontodueforce

)ˆ(ˆˆ

ppdAF

iAdpiApdiFF

ApdF

A

xx

AA

x

A

xxx

xx

ApdipdAiiFdFd

pdAiApdiFddF

ˆ)(ˆ)ˆ(

ˆˆx

y

z

Ad

xAd p

ApdFd

p

abj 16

Some Properties of Pressure Force [3] How to find the net pressure force via its components

zyx FFFF

z

A

zz

y

A

yy

x

A

xx

AppdAF

AppdAF

AppdAF




A

z

x

y

A

F

zA

yA

A

z

x

y

AxA

xAd

xx ApdFd

abj 17

Example: Find The Net Pressure Force Due to Uniform Pressure

Questions:

1. Find the net pressure force due to uniform pressure in two cases:

a. uniform p1 (on the left side only),

b. both uniform pressures p1 and p2,

on plates (a) and (b).

2. In the corresponding cases, Are they equal? If so, why?

(a) (b)

h

l

y

xzAd Ad

p1

p2

p1 p2

abj 18

Problem

1. Find the resultant pressure force (magnitude, direction, and line of action) due to all

fluid pressures on the curved plate of width w.

2. Find the center of pressure (CP) for the case of a parabolic gate (n = 2), and D = 1

m and a = 1 m-1.

Resultant Due to Pressure (Hydrostatic Force/Moment) on A Curved Submerged Surface

Water,

Air, po

x

D

naxxy )(y

o

abj 19

Resultant Force

jwdxiwdyjdyidxwk

rwdkAd

ˆ)(ˆ)(ˆ)(ˆ)(ˆ

)(ˆ

uuu AdpFd

Force on upper surface

lll AdpFd

Force on lower surface

Adpp

AdAdAdpp

AdAdAdpAdp

FdFdFd

lu

uulu

ullluu

lu

)(

,)(

),()(

Net force on plate due to all fluid pressures

Dygh

Dypp

pp

Dyghp

Dypp

lu

ol

o

ou

0,

,0

0,

,

pressures

Adgh

AdppFd lu

)(

)(

Thus, the net force x

D

naxxy )(y

o

h

rd

1

2’

Ad AdAd u

,

lAd

2

How to write the area vector vectorially for the integration from 1 2(‘)

jdyidxrd ˆ)(ˆ)( To integrate from 1 2, points from 1 2rd

No need for (+/-) signs in dx or dy since dx and dy are components of vectors, they have sign embedded in them.

uuu AdpFd

lll AdpFd

is the result of 90o-rotation of about the axis; hence,Ad

rdw

)( k )(ˆ rwdkAd

abj 20

jwdxiwdyrwdkAd ˆ)(ˆ)()(ˆ

AdghAdppFd lu

)()(

jhdxgwihdygw

jwdxiwdyghFd

ˆ)(ˆ)(

ˆ)(ˆ)(

naxDxhyDyh )()(

You know how to integrate these already.

j

a

DD

n

ni

DgwF

n

ˆ1

ˆ2

)(/12

jhdxgwihdygwFx

x

y

y

ˆ)()(ˆ)(2

1

2

1

x

D

naxxy )(y

o

h

rd

1

2’

Ad 2

AdppFd lu

)(

abj 21

NOTE on Direction of Integration, , and rd

Ad

x

D

naxxy )(y

o

h

rd

1

2’

Ad 2

jwdxiwdy

jdyidxwk

rwdkAd

ˆ)(ˆ)(

ˆ)(ˆ)(ˆ

)(ˆ

jdyidxrd ˆ)(ˆ)( 1 2

Ad

rdw

)( k: Rotate about

x

D

naxxy )(y

o

h

rd

2

1’

Ad 1

jwdxiwdy

jdyidxwk

rwdkAd

ˆ)(ˆ)(

ˆ)(ˆ)(ˆ

)(ˆ

jdyidxrd ˆ)(ˆ)( 1 2

Ad

rdw

)( k: Rotate about

abj 22

CP

F

Resultant Moment and Line of Action

x

D

naxxy )(y

o

h

rd

1

2’

Ad

2

AdppFd lu

)(

p

R

r

Fd

Infinitesimal force

FdrMd o

Infinitesimal moment due to Fd

is the position vector to any point p

on the line of action of

jYiXR ˆˆ

F

Net/Resultant moment due to Fd

FdrM o

FdrFR

Principle of Moment

kYFXF

jFiFjYiXFR

xy

yx

ˆ)(

)ˆˆ()ˆˆ(

kydFxdF

jdFidFjyixFdr

xy

yx

ˆ)(

)ˆˆ()ˆˆ(

jhdxgwihdygw

AdghAdppFd lu

ˆ)(ˆ)(

)()(

)( xyxy ydFxdFYFXFThus,

)(:

,)()/1()/()(

XYequationLinear

ydFxdFFXFFXY

numberconstanta

xyx

slope

xy

This equation is in fact a linear equation Y(X),

representing the locus of points on the line of

action of F

abj 23

)( xyxy ydFxdFYFXF

jhdxgwihdygw

AdghAdppFd

yx dFdF

lu

ˆ)(ˆ)(

)()(

n

w

aDn

w

aD

w

aD

wy

a

DDgw

n

n

xdxaxDgwxdxyDgw

dxhxgwxdF

nn

n

/2

)/(

0

)/(

0

)/(

0

)()2(2

)()()()(

)()(

/1/1

/1

3

0

0

)(6

1

)()(

)()(

Dgw

ydyyDgw

dyhygwydF

w

D

w

D

wx

n

y

x

a

DD

n

ngwF

DgwF

/1

2

1)(

,2

)(

D

a

D

Dn

nX

a

D

Dn

nXY

nn

3

11

)2(

1

1

2)(

/2/1

Thus, the line of action of the resultant force is given by

(1)

naXY (2)CP is the intersection of this line with the surface:

Thus, solve 2 equations in 2 unknowns X and Y, we have for the parabolic gate

myx )215.0,464.0(),(

abj 24

Special Case: Resultant Pressure Force/Moment on A Flat Surface

abj 25

x

y

O

Problem: Find the resultant force (magnitude, direction, and line of action) on a plate

submerged in fluid with pressure at the free surface of po.

Resultant Force Due to Pressure (Hydrostatic Force) on A Plane Submerged Surface

y

O gpo

zhF ApdFd

Interest in force on this side of the plate.

C = Centroid of the plate area

CP = Center of pressure

x

CP

y

r

• Consider an area element located at r

Ad

dA

r

• This area, located at depth h, has infinitesimal

force

acting on it.

ApdFd

C

hC

y x

xC

yC

If it is not too confusing, recall that we also have another characteristic point that has a role to play, the centroid of the volume of pressure distribution.

Ad

Draw first to identify the system of interest.Ad

• Let the resultant force, which is in the +z

direction, be denoted by .F

r • It acts through the point on the surface called the

center of pressure (CP), which is located at . • As we shall later see, another characteristic point is the centroid C of the plate area.

• Let this centroid be located at depth hC and point (xC, yC)

yxC ˆˆ

• As we shall also see, the parallel coordinate

system whose origin is at the centroid C

will also have a role to play.

abj 26

x

y

O

y

Ogpo

z

dA

CP

C

y x

RF ApdFd

Interest in force on this side of the plate.

r

r

CCP

Ad

abj 27

Assumptions:

• Static fluid.

• Gravity is the only body force.

• = g is constant wrt depth.

Resultant Force (Magnitude and Direction):

where pc is the hydrostatic pressure at the depth, hc, of

the centroid of the plate area; A is the total area of

the plate.

CoCC

CCCoCo

A

CCo

A

o

A

o

o

AA

o

A

o

A

ghppApkF

yhAghpkAgypk

ydAAyAygApk

yhydAgApkhdAgApk

dAkdAAdghppghdAdApkkdAghpApdF

);(ˆ

sin;)(ˆ)sin(ˆ

:Centroid;sinˆ

sin;sinˆˆ

0,ˆ;;ˆˆ)(

x

y

y

z

CP

dA

O

O

C

y x

F

g

ApdFd

po

r

r C

CP

Ad

abj 28

Resultant Moment (Line of action of the resultant force):

Moment about point O:

iydFjxdFiFyjFx

AApFkdFjyixkFjyix

dAkdAAdpdAdFkdFkpdAApdFdFdrFr

C

ˆˆˆˆ

0,;)ˆ()ˆˆ()ˆ()ˆˆ(

0,ˆ;0,ˆˆ)(;

Application 4.2: Hydrostatic Force on A Plane Submerged Surface

xdFFx ydFFy

x

y

y

z

CP

dA

O

O

C

y x

F

g

ApdFd

po

r

r C

CP

Ad

abj 29

Application 4.2: Hydrostatic Force on A Plane Submerged SurfaceResultant Moment (Line of action of the resultant force):

Moment about the x-axis (that passes through point O):

Consequence: Since the second term on the RHS is always >= 0, the equation indicates that the y-location of

the center of pressure (y’, the point on the plate at which the resultant force acts) always lies at

equal or lower depth than that of the plate centroid, yC.

)(; sin

sin

sin)(

sin)sin(

sin sin

axis.ˆcentroid/ about area of ,); (sin

axis.about area of

intertia, of momentor moment, Second,; sin

area of centroid theof Coordinate,;sin

sin;sin

)(

,

:

ˆˆ

ˆˆ

ˆˆ

ˆˆ

ˆˆ2

ˆˆ2

ˆˆ2

ˆˆ

2

2

2

ApFyyAp

Igyy

IgAyp

IgghpAy

IggypAy

IgAygAyp

xAIIAyIIAyIgAyp

xA

IdAyIIgAyp

AyyAyydAdAygAyp

yhdAgyydAp

ghdAydAypdAghpyypdAydFFy

pdAdFydFFy

MM

CRCC

xxC

xxCC

xxCoC

xxCoC

xxCCo

xxCxxxxCxxCo

xxxxxxCo

CCCo

o

ooR

AR

FDoxRDox

NOTE: If the other side of the plate is exposed to free

surface pressure po and the net force due to pressure

distributions on both sides of the plate is

desired, we have

C

xxC

C

xxC

Cg

xxC Ay

Iy

Agy

Igy

Ap

Igyy ˆˆˆˆˆˆ

sin

sinsin

abj 30

Resultant Moment (Line of action of the resultant force):

Moment about the y-axis (that passes through point O):

NOTE: If the other side of the plate is exposed to free

surface pressure po and the net force due to pressure

distributions on both sides of the plate is

desired, we have

C

yxC

C

yxC

Cg

yxC Ay

Ix

Agy

Igx

Ap

Igxx

ˆˆˆˆˆˆ

sin

sinsin

ApFAp

Igxx

IgAxp

IgghpAx

IggypAx

IgyAxgAxp

yAxIIyAxIgAxp

A

xyxyxydAIIgAxp

xdAAxxydAgAxpdAgxyxdAp

ghdAxdAxpdAghpxxpdAxdFFx

MM

CRC

yxC

yxCC

yxCoC

yxCoC

yxCCCo

CCyxxyCCyxCo

xyxyCo

CCoo

ooR

FDoyRDoy

; sin

sin

sin)(

sin)sin(

sin sin

); (sin

area of inertia, of

moment cross-or moment, second cross; sin

;sinsin

)(

:

ˆˆ

ˆˆ

ˆˆ

ˆˆ

ˆˆ

ˆˆˆˆ

Abj1 4.2.1: Pressure, Pressure Force, and Fluid Motion Without Flow [Q1] 1.Area as A Vector ...

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Transcript of Abj1 4.2.1: Pressure, Pressure Force, and Fluid Motion Without Flow [Q1] 1.Area as A Vector ...