Lecture 4 - Fluid 1 - Hydrostatic Forces on Submerged Plane Surfaces.pdf

ef

1. Fluid Mechanics 1 –
Arab Academy for Science, Technology
and Maritime Transportation
Dr. Aly Hassan Elbatran
Associate Professor
[email protected]
[email protected]

Course Assistant Lecturer:
Eng. Islam mohamed

2. Lecture 4:
Hydrostatic Forces on Submerged
PlaneSurfaces
Fluid Mechanics 1

3. Hydrostatic forces on submerged plane surfaces
Graphical Solution
(Prism Solution)
Analytical
Solution

4. Hydrostatic forces on submerged plane surfaces
Analytical
Solution

5. Hydrostatic forces on submerged plane surfaces
Aplateissubjectedtofluidpressure
distributedoveritssurfacewhen
exposedtoaliquid;suchasagatevalve
inadam,thewallofaliquidstorage
tank,orthehullofashipatrest.
Onaplanesurface,thehydrostatic
forcesformasystemofparallelforces,
andweoftenneedtodeterminethe
magnitudeoftheResultantforceand
itspointofapplication,whichiscalled
thecenterofpressure.
Whenanalyzinghydrostaticforceson
submergedsurfaces,theatmospheric
pressurecanbesubtractedfor
simplicitywhenitactsonbothsidesof
thestructure.

6. Hydrostatic forces on submerged plane surfaces
Whenanalyzinghydrostaticforcesonsubmergedsurfaces,theatmospheric
pressurecanbesubtractedforsimplicitywhenitactsonbothsidesofthe
structure.
Effect of atmospheric pressure on the resultant force acting on a plane vertical wall.

7. General submerged plane

8. General submerged plane
Thetotalareaismadeupofmanyelementalareas.
Theforceoneachelementalareaisalwaysnormaltothe
surfacebut,ingeneral,eachforceisofdifferentmagnitude
asthepressureusuallyvaries.
WecanfindthetotalorResultantForce,F
R,ontheplane
bysummingupalloftheforcesonthesmallelements.ApApApApF
nn
 ....
2211
R 

A
PdAFR

9. Hydrostatic forces on SubmergedHorizontal Plane
Surface
•Forahorizontalplanesubmergedinaliquid,thepressure,p,
willbeequalatallpointsofthesurface.
•Thepressureatthebottomofthecontainerisuniformacrossthe
entireareapAFR

10. Hydrostatic forces on Submerged Horizontal Plane
Surface
•Forahorizontalplanesubmergedinaliquid,thepressure,p,
willbeequalatallpointsofthesurface.
•Thepressureatthebottomofthetankisuniformacrossthe
entireareapAFR h AF R

11. sinyzh Where:
Given a plane surface AB entirely submerged in the liquid. The surface is
inclined an angle θto the liquid surface. The centroid of area is located at C.
The vertical distance of C below the liquid surface is:
Hydrostaticforces on SubmergedInclinedor Vertical
Plane Surfaces

12. Pressure at a point at z
below the liquid surface
is:
p(z) =*z
(Gage pressure)
Orin terms of y
(distance along the plate)
pressure at point z is:
p(z) = *y*sin
The differential force
on the differential area
dAis dF(z)
Hydrostatic forces on Submerged Inclined or Vertical
Plane Surfaces
Pressure forces on an elemental area dA:

13. The differential force on the differential area dAis dF(z)

 dAyzdF
dAzzdF
dAzpzdF
yzzp




)(sin
)sin()(







Hydrostaticforces on SubmergedInclinedor Vertical
Plane Surfaces
Where:
Pressure forces on an elemental area dA:

14. Hydrostatic Pressure Distribution
Hydrostatic forces on Submerged Inclined or Vertical
Plane Surfaces

sin

yp
zp



15. Thetotalforce(F
R)ontheareawillbeobtainedbyintegratingthe
differentialforceovertheentirearea: dAydFF
R
sin 
 
and(sin )are constants 

 AyAhF
AhAzF
AyF
AyF
dAyF
ccR
R
R
R
R
sin

sin
sin
sin









  c
c
h
y








h
y
ydAyA
dAA
Remark
The total hydrostatic force on a planar inclined surface
Hydrostatic forces on Submerged Inclined or
Vertical Plane Surfaces

16. The total hydrostatic force on a planar inclined surface
Hydrostatic forces on Submerged Inclined or
Vertical Plane Surfaces

17. y
R
oftheresultantforcecanbedeterminedbysummation
ofmomentsaroundthexaxis.Thatis,themomentofthe
resultantforcemustequalthemomentofthedistributed
pressureforce,or dAydFF
R
sin 
 
Recallthat: dAyydFyyF
RR


 sin
Sothemomentis:AyF
cR
sin
Recallthat:

18. Theintegralinthenumeratoristhesecondmomentofthearea(momentof
inertia),“Ix”.Thus,wecanwrite:
Usingtheparallelaxistheorem:
where“Ixc”.isthesecondmomentoftheareawithrespecttoanaxis
throughitscentroidandparalleltothexaxis.Thus:
So,fromtheaboveEquation,itisclearthattheresultantforcedoesnotpass
throughthecentroid.butfornonhorizontalsurfacesisalwaysbelowit,since
(I
xc/ y
cA) > Zeroc
c
xc
c
c
xc
h
h
I
y
y
I

A
hOr
A
y
RR

19. The centroid and the centroidal moments of inertia for
some common geometries

20. Aheavycarplungesintoalakeduringanaccidentandlandsatthebottomofthelakeonits
wheelsasshownintheFigure.Thedooris1.2mheightand1mwide,andthetopedgeof
thedooris8mbelowthefreesurfaceofthewater.Usingtheanalyticalandprismmethod,
determinethehydrostaticforceonthedoorandthelocationofthepressurecenter.
Examplem 8.61 )]
2
1.2
( [8
1.2) *1( )]
2
1.2
( [8
0.144

144.0)121.2 *1( )12(I
h
h
I

kN 101.24
kN )1*2.1(*)]
2
1.2
( [8 * 9.81
kN/m 9.81 and
433
xc
c
c
xc
3








R
R
R
R
wcwR
h
mba
A
h
F
F
AhF 
Using the Analytical Method

21. Hydrostatic forces on submerged plane surfaces
Graphical Solution
(Prism Solution)

22. Pressureactsnormaltothesurface,andthe
hydrostaticforcesactingonaflatplateofany
shapeformavolumewhosebaseistheplate
areaandwhoselengthisthelinearlyvarying
pressure.
Thisvirtualpressureprismhasaninteresting
physicalinterpretation:itsvolumeisequalto
themagnitudeoftheresultanthydrostatic
forceactingontheplatesinceF
R=PdA,and
thelineofactionofthisforcepassesthrough
thecentroidofthishomogeneousprism.
Theprojectionofthecentroidontheplateis
thepressurecenter.
Therefore,withtheconceptofpressureprism,
theproblemofdescribingtheresultant
hydrostaticforceonaplanesurfacereducesto
findingthevolumeandthetwocoordinatesof
thecentroidofthispressureprism.
Thehydrostaticforcesactingona
planesurfaceformapressure
prismwhosebase(leftface)isthe
surfaceandwhoselengthisthe
pressure.
Pressure Prism

23. Analternateapproachofdeterminingthehydrostaticforceisby
meansofapressureprism.Consideraverticalplanesubmergedin
astaticfluid,asshowninthefigure.Thepressureincreases
linearlywiththedepth.Onecantheneasilyconstructa
correspondingthree-dimensionaldiagramofthepressure
distribution,andsuchavolumeiscalledapressureprism.The
resultantforceisthetotalvolumeofthepressureprism,thatis:
F
R= Volume = 1/2 (ρgh) (bh)
Pressure Prism

24. F
R= Volume = 1/2 (ρgh) (bh)
Theresultantforcepassesthroughthecentroidofthepressureprism.Forthis
particularexample,thecentroidofatriangularelementislocatedata
distanceofh/3fromitsbaseandliesintheverticalsymmetryaxis.
Asillustrated,thismethodisparticularlyconvenientwhentheshapeofthe
pressureprismisacommongeometry,inwhichthevolumeandcentroidcan
bereadilyobtained.
Pressure Prism

25. Aheavycarplungesintoalakeduringanaccidentandlandsatthebottomofthelakeonits
wheelsasshownintheFigure.Thedooris1.2mheightand1mwide,andthetopedgeof
thedooris8mbelowthefreesurfaceofthewater.Usingtheanalyticalandprismmethod,
determinethehydrostaticforceonthedoorandthelocationofthepressurecenter.
Using the Prism Method  
 
m 61.8)59.02.18( that So
; 59.0
* 24.011 )32.1(*063.7)22.1(*176.94
*)32.1(*)22.1(* 0
kN 24.011
kN 063.7)1*2.1(*2)82.18( * 9.81
kN 176.94kN )1*2.1(*8 * 9.81
2 and ,
21
21
2
1
12211








R
Rstst
ststR
st
st
wstwst
h
mh
h
hFFFM
FFF
F
F
AhhFAhF 
Example