FLUID STATICS
HYDROSTATIC FORCES AND BUOYANCY
• Explain the hydrostatic forces applied on
submerged plane surface.
• Hydrostatic forced on submerged curved
surface.
• Define and explain Buoyancy and Stability.
FLUID STATICS
Deals with problems associated with fluid
at rest.
No relative motion between adjacent
layers. Thus, no shear stress (tangential
stress) to deform the fluid.
Liquid – hydrostatic; Gas – aerostatic.
FLUID STATICS
The only stress in fluid statics is normal
stress (perpendicular to surface)
• Normal stress is due to pressure
(Pressure: gravity field-weight of
fluid)
• Variation of pressure is due only to
the weight of the fluid → fluid
statics is only relevant in presence
of gravity fields.
HYDROSTATIC FORCES
F=PA
Only when
the pressure
distribution is
uniform over
the entire
area of
interest.
HYDROSTATIC FORCES
What about
the areas
where the
pressure is
non-uniformly
distributed ??
FIND THE
AVERAGE
PRESSURE AND
WHERE IT ACTS
HYDROSTATIC FORCES ON PLANE
SURFACES
Problem
• To find the resultant force FR (magnitude) and its point of
application (center of pressure) for non-uniformly distributed
pressure.
**Atmospheric pressure Patm can be neglected when it acts on
both sides of the surface.
WHAT IS A RESULTANT FORCE AND CENTRE OF
PRESSURE ?
Engineering Mechanics
Resultant force:
A force that results from the
combination of two or
more forces.
COP:
A point where the entire
forces (Resultant force)
were concentrated at a
single point.
HOW TO DETERMINE THE RESULTANT FORCE
Consider a flat plate
completely submerged
in a liquid.
Plane of the top surface
intersects with
horizontal surface with
an angle Ө, the line of
intersection will be the x
axis.
HOW TO DETERMINE THE RESULTANT FORCE, FR
Absolute pressure at any point of the fluid,
P = Po + ρgh,
h = y sin Ө
= Po + ρgy sin Ө,
FR = ∫PdA =∫ (Po + ρgy sin Ө)dA = PoA + ρg sin Ө ∫ydA
**∫ydA is the first moment of area is related to the y
coordinate of the centroid (or centre) of the surface by ,
yc= 1/A ∫ydA.
FR = (Po + ρgyc sin Ө) A = (Po + ρghc) A
= PcA
The magnitude of the resultant force, FR acting on a plane surface of a completely
submerged plate in a homogeneous (constant density) fluid is = the product of the
pressure Pc at the centroid of the surface and the area A of the surface.
CENTROID AND CENTROIDAL MOMENTS OF
INERTIA FOR SOME COMMON GEOMETRIES
SUBMERGED RECTANGULAR
PLATE : HOW TO DETERMINE
THE RESULTANCE FORCE, FR.
How to determine the location of the COP?
Line of action of
resultant force
FR=PCA does not
pass through the
centroid of the
surface. In general,
it lies below the
centroid because
pressure increase
with depth.
How to determine the location of the resultant
force (Center of Pressure) ?
The vertical location of the line of action is
determined by equating the moment of the
resultant force, FR to the moment of the
distributed pressure force about the x-axis.
Then,
ypFR = ∫yPdA
=∫ y(Po + ρgy sin Ө)dA
= Po ∫ ydA + ρgsin Ө ∫y2dA
= PoycA + ρgsin Ө Ixx,o
Moment of the resultance force = Moment of the
distributed pressure force about the x axis.
y pFR   yPdA
A
  y(Po  ρgysinθ)dA
A
2
 Po  ydA  ρgsinθ y dA
A
A
 Po y c A  ρgsinθIx x , o
Ixx,o = ∫y2dA is actually the second moment of area
about the x axis passing through point O.
Normally, the second moment of area is given
about the axes passing through the centroid
of the area. Therefore, we need parallel axis
theorem to relate the Ixx,o and Ixx,c
I xx ,o  I xx ,c  yc A
2
Therefore,
y p  yc 
I xx ,c
yc  Po /( g sin  )A
If Po = 0
y p  yc 
I xx ,c
yc A
SUBMERGED RECTANGULAR PLATE : HOW TO
DETERMINE THE C.O.P.
I xx ,c
y p  yc 
yc  Po /( g sin  )A
b
ab
yc  s  ; A  ab; I xx ,c 
2
12
3
3
b
ab 12
yp  s  
2  b

 s  2  Po /( g sin  ) ab
If Ө=90
3
b
ab 12
yp  s  
2  b
 s  2  ab
Example 11-1; Calculate FR, C.O.P and discuss whether the driver can open
the door or not.
F R PC A
Pc  ρghc  ρg(s  b/2)
2
 84.4kN /m
 FR  Pc A  101.3kN; [ A  1.2m *1m]
3
b
ab 12
yp  s  
2  b
 s  2  ab
 y p  8.61m
The moment acting on the mid point of the submerged
door is 101.3kN x 0.5m = 50.6kNm, which is 50 times
of the moment the driver can possibly generate. CAN’T
OPEN.
Hydrostatic Forces on Curved Surfaces
• Complicated: FR on a curved surface requires
integration of the pressure forces that change
direction along the surface.
• Easiest approach: Determine horizontal and
vertical components FH and FV separately.
Hydrostatic Forces on Curved Surfaces
1. Vertical surface of the liquid block,
BC = projection of the curved surface
on a vertical plane (vertical
projection).
2. Horizontal surface of the liquid
block, AB = projection of the curved
surface on a horizontal plane
(horizontal projection).
3. Newton’s 3rd Law – Action and
reaction. The resultant force acting on
the curve liquid surface = the force on
curved solid surface.
• Assume that the direction to the right and up is
positive,
Horizontal force component on curved surface:
FH – FX = 0.
The horizontal component of the hydrostatic
force (FH) acting on a curved surface is (both
magnitude and line of action) equal to the
hydrostatic force (FX) acting on the vertical
projection of the curved surface.
• Vertical force component on curved surface:
FV - FY – W = 0, where W is the weight of the liquid
in the enclosed block W=gV.
The vertical component of the hydrostatic force
(FV) acting on a curved surface is (both magnitude
and line of action) equal to the hydrostatic force
(FY) acting on the horizontal projection of the
curved surface, plus or minus (depend on
direction) the weight of the fluid block.
Hydrostatic Forces on Curved Surfaces
Magnitude of force FR=(FH2+FV2)1/2
Angle of force is a = tan-1(FV/FH)
A GRAVITY CONTROLLED CYLINDRICAL GATE
Example 11-2 A long solid cylinder, R=0.8, hinged at point A.
When water level at 5m, gate open.
Determine:
1. The hydrostatic force acting on the cylinder and its line of action when the gates open
2. The weight of cylinder per m length of the cylinder
• Friction at hinge is negligible
• The other side of the gate is
exposed to the atmosphere,
therefore, the Patm is
cancelled out.
A) Determine the net horizontal and vertical force, FH and FV respectively.
FH = FX = PCA = ρghcA; hc = 4.2 + 0.8/2
FY = ρghcA; hc = 5m ; W=ρg(R2 – (πR2/4))
FY - FV - W = 0;
Solve for resultant force and angle.
B) Think about the moment acting at point A
due to the cylinder weight and also resultant
force
BUOYANCY
An objects lighter in fluid compare to in an
air.
Fluid exerts an upward force on a body
immersed in it.
BUOYANCY
Buoyant force is caused by the increase of
pressure with depth in fluid.
Figure 10-13
Difference between pressure at the top and
bottom surface is
ρg(s+h)A - ρghA = ρgsA = ρgV,
V = hA, volume of the plate.
Buoyancy
Archimedes principal : The buoyant
force acting on a body immersed in a fluid
is equal to the weight of the fluid displaced
by the body, and it acts upward through
the centroid of the displaced volume.
Buoyancy
The tendency of fluid to
exert a supporting force
on a body placed in the
fluid.
The force = weight of the
fluid displaced by the
body. Its act upward
through centroid of the
displaced volume.
FB=fgV
Density of fluid
Bouyancy of Floating Bodies
FB = W
(The weight of the entire body must be equal to the buoyant force)
ρf gVsub = ρave,bodygVtotal
Vsub / Vtotal = ρave,body / ρf
(The submerged volume fraction is equal to the ratio of
the average density of the body to the density of the
fluid)
Completely submerged when, ρave,body >= ρf
Submerged portion (yellow)
Displaced fluid whose volume =
yellow volume
LIQUID
BOUYANT FORCE
=
WEIGHT OF DISPLACED FLUID
BUOYANCY
Buoyance force by air is so small,
0.1m3, ρair = 1.2 kg/m3 , 1.2N. If mass = 80 kg,
weigth = 788 N. Ignore the buoyancy, error is
0.15%, so small.
Rise of warm air – natural convection currents.
The rise of hot air or Helium balloons, and air
movements in the atmosphere.
Buoyancy Calculation
• Example 11-3 and 4
• Q 11-33 and 11-34