Presentation on Fluid Mechanics I : Hydrostatics

1. Welcome To MY Presentation

2. Serial Name ID
01 Morshedul Hasan 14107006
Sec: Day
Sub: MEC 267
Prog: BSME

4. Contents:
 Introduction
 Total pressure
Total Pressure on an Immersed Surface
 Total Pressure on a Horizontally Immersed Surface
 Total Pressure on a Vertically Immersed Surface
 Total Pressure on an Inclined Immersed Surface
 Centre of Pressure
 Centre of Pressure on a Vertically Immersed Surface
 Centre of Pressure on an Inclined Immersed Surface
 Centre of Pressure of a Composite Section
Hydrostatic Force on a Curved Surface
 Application Hydrostatics

5. Hydrostatics is the branch of fluid mechanics that
studies incompressible fluids at rest.
Hydrostatics is fundamental to hydraulics, the engineering of
equipment for storing, transporting and using fluids.
Figure : Hydrostatic Pressure

6. Total pressure : The 'total pressure' is the sum
of the static pressure, the dynamic pressure, and the
gravitational potential energy per unit volume. It is therefore
the sum of the mechanical energy per unit volume in a fluid.
Ptotal = P1+ P2+ P3 …………..
Total Pressure on an Immersed Surface: The total pressure exerted by a
liquid on an immersed surface. The position of an immersed surface may
be:-
Horizontal
Vertical and
Inclined

7. Total Pressure on a Horizontally Immersed Surface:
Consider a plane horizontal surface immersed in a liquid as
shown in figure:
Let,
W= Specific weight of the liquid.
A= Area of the liquid.
X= Depth of the horizontal surface
from the liquid level in meters.
Now, Total pressure on the surface,
P =Weight of the liquid above the immersed surface
= Sp. Wt. of liquid x Volume of liquid
= Sp. Wt . Of liquid x Area of surface x Depth of liquid
= wAx KN
Where w is the specific weight of the liquid in KN/m3

8. Total Pressure on a Vertically Immersed Surface:
Consider a plane vertical surface immersed in a liquid as
shown in figure:
First of all, let us divide the whole
immersed surface into a number of small
parallel strips as shown in figure.
Let, w= specific weight of the liquid.
A= Total area of the immersed surface
x= Depth of center of gravity of the immersed
surface from the liquid surface.
Let us consider a strip of thickness dx, width b and at a depth x from the free
surface of the liquid as shown in figure.
Now, The intensity of pressure on the strip = wx
And area of the strip = b.dx

9. Total Pressure on a Vertically Immersed Surface:
Now, Total pressure on the surface,
P = ʃ wx . bdx
= w ʃ x . bdx
But, ʃ x . bdx = Moment of the surface area about of the liquid level.
= Ax
So, P = wAx
So, Pressure on the strip,
P = Intensity of pressure x Area
= wx.bdx

10. Total Pressure on an Inclined Immersed Surface:
Consider a plane inclined surface immersed in a liquid as
shown in figure:
First of all, let us divide the whole immersed
surface into a number of small parallel strips
as shown in figure.
Let, w= specific weight of the liquid.
A= Total area of the surface
x= Depth of center of gravity of the immersed
surface from the liquid surface.
θ = Angle at which the immersed surface is
inclined with the liquid surface.
Let us consider a strip of thickness dx, width b and at a distaance l from O.
Now, The intensity of pressure on the strip = wl sinθ
And area of the strip = b.dx
So, Pressure on the strip,
P = Intensity of pressure x Area
= wl sinθ.bdx

11. Total Pressure on an Inclined Immersed Surface:
Now, Total pressure on the surface,
P = ʃ wl sinθ.bdx
= w sinθ ʃ l . bdx
But, ʃ l . bdx = Moment of the surface about O.
=
So, P = w sinθ x
=
sin

xA
sin

xA

xwA

12. Center of pressure:
The center of pressure is the point where the total sum of
a pressure field acts on a body, causing a force to act through that point.
The total force vector acting at the center of pressure is the value of the
integrated vectorial pressure field. The resultant force and center of
pressure location produce equivalent force and moment on the body as
the original pressure field. Pressure fields occur in both static and
dynamic fluid mechanics. Specification of the center of pressure, the
reference point from which the center of pressure is referenced, and the
associated force vector allows the moment generated about any point to
be computed by a translation from the reference point to the desired new
point. It is common for the center of pressure to be located on the body,
but in fluid flows it is possible for the pressure field to exert
a moment on the body of such magnitude that the center of pressure is
located outside the body.

14. Centre of Pressure on a Vertically Immersed Surface:
Consider a plane surface immersed vertically in a liquid as
shown in figure:
First of all, let us divide the whole immersed
surface into a number of small parallel strips
as shown in figure.
Let, w= specific weight of the liquid.
A= Total area of the immersed surface
x= Depth of center of gravity of the immersed
surface from the liquid surface.
Let us consider a strip of thickness dx, width b and at a depth x from the free
surface of the liquid as shown in figure.
Now, The intensity of pressure on the strip = wx
And area of the strip = b.dx
h
G
p

15. Centre of Pressure on a Vertically Immersed Surface:
h
G
p
So, Pressure on the strip,
P = Intensity of pressure x Area
= wx.bdx
Moment of this pressure about the liquid surface,
= (wx . bdx) x = wx2 . bdx
Now, The Some of moments of all such pressures about the
liquid surface ,
M = ʃ wx2 . bdx
= w ʃ x2 . bdx
But, ʃ x2 . bdx= I0 (i.e. moment of inertia of the surface about the liquid level)
So, M= w I0 ..................(i)
We know that, the sum of the moments of the pressure = P x h..............(ii)
Where, p= Total pressure on the surface and
h= Depth of center of pressure from the liquid surface.

16. Centre of Pressure on a Vertically Immersed Surface:
Now, equating equation (i) and (ii)
)...(....................
*
.*
0
0
0
iii
xA
I
h
wIhxwA
IwhP







We know the theorem of parallel axis that, I0 =IG+ Ah2
Where, IG=Moment of inertia of the figure, about horizontal axis through its center of gravity
h= Distance between the liquid surface and the center of gravity of the figure. ( in this case)
Now rearranging the equation (iii),

x









x
xA
I
h
xA
xAI
h
G
G
2

17. Centre of Pressure on an Inclined Immersed Surface:
Consider a plane inclined surface immersed in a liquid as
shown in figure:
First of all, let us divide the whole immersed
surface into a number of small parallel strips
as shown in figure.
Let, w= specific weight of the liquid.
A= Total area of the immersed surface
x= Depth of center of gravity of the immersed
surface from the liquid surface.
θ = Angle at which the immersed surface is
inclined with the liquid surface
Let us consider a strip of thickness dx, width b and at a depth x from the free surface of
the liquid as shown in figure.
Now, The intensity of pressure on the strip = wl sinθ
And area of the strip = b.dx
G p

h

18. Centre of Pressure on an Inclined Immersed Surface:
So, Pressure on the strip,
P = Intensity of pressure × Area
= wl sinθ.bdx
and moment of this pressure about 0,
= (wl sinθ.bdx) l = wl 2sinθ.bdx
So , M= ʃ wl2 sinθ.bdx = wsinθ ʃ l 2.bdx
But, ʃl2 .bdx= I0 (i.e. moment of inertia of the surface about 0)
so, M = wsinθ I0 .....................(i)
We know that the sum of the moments of all such pressure about 0,
)...(....................
sin
ii
hp



Where,
p= Total pressure on the surface and
h= Depth of center of pressure from the liquid surface.

19. Centre of Pressure on an Inclined Immersed Surface:
Now equating equation (i) and (ii)








xA
I
h
Iw
hxwA
Iw
hp





2
0
0
0
sin
sin
sin
sin
sin
We know the theorem of parallel axis that, I0 =IG+ Ah2
Where, IG=Moment of inertia of the figure, about horizontal axis through its center of gravity
h= Distance between the liquid surface and the center of gravity of the figure. ( l1 in this case)
Now rearranging the equation (iii),
 









 











x
xA
I
h
lAI
xA
AlI
xA
h
G
xx
GG



2
sin1
2
sin
2
2
1
2
sin
sinsin


20. Hydrostatic Force on a Curved Surface:
On a curved surface the forces pδA on individual elements differ in
direction, so a simple summation of them may not be made. Instead, the
resultant forces in certain directions may be determined, and these forces
may then be combined vectorially. It is simplest to calculate horizontal
and vertical components of the total force.
Horizontal component of hydrostatic force:
Any curved surface may
be projected on to a
vertical plane. Take, for
example, the curved
surface illustrated in
figure.

21. Its projection on to the vertical plane shown is represented by the trace
AC. Let Fx represent the component in this direction of the total force
exerted by the fluid on the curved surface. Fx must act through the center
of pressure of the vertical projection and
is equal in magnitude to the force F on the fluid at the vertical plane. In any
given direction, therefore, the horizontal force on any surface equals the force
on the projection of that surface on a vertical plane perpendicular to the given direction.
The line of action of the horizontal force on the curved surface is the same as that of the
force on the vertical projection.
Vertical component of hydrostatic force:
The vertical component of the force on a curved surface may be determined by
considering the fluid enclosed by the curved surface and vertical projection lines
extending to the free surface. Thus FH= F2 = ρgzs
FV= F1 = ρgV
where V is the volume of the liquid between the free surface liquid and solid curved
surface. The magnitude of the resultant is obtained form the equation
22
VHR FFF 

22. Some Application Hydrostatics :
The Hydrostatics Pressure is either utilized in the working of a
Hydraulic Structure, or a Structure is checked to withstand the
Hydrostatic Pressure exerted on it.
01)Water Pressure on Sluice Gate:-

23. To regulate the flow of Water, in the path of a River, a Sluice Gate is
provided. The Sluice Gate is made to move up and down with the help of
Rollers fixed on Skin Plate, which travel on Vertical Rails called Guide.
These Rails are fixed on Piers or Vertical Walls.
In between these two Skin Plates, a number of I-beams are provided
horizontally to withstand the Water Pressure. The spacing between the I-
beam is lesser at the bottom than that at top of Sluice Gate, as the Water
Pressure varies in the depth.

24. 02)Water Pressure on Lock Gate:-
The Water Level on both sides of the Dam will be different. If it is desired to have Boating in
such river, the Chamber, known as Lock, is constructed between these two different Water
Levels. Two sets of Lock Gates are provided in order to transfer a Boat from Higher Water Level
to Lower Water Level. The Upstream gates are opened, and the Water Level in the Chamber is
rises up to the Upstream Water level. The Boat is then shifted in the chamber, then Upstream
Gates are closed and Downstream Gates are opened and the Water Level in the Chamber is
lowered to the Downstream Water level. The procedure is reversed for the transfer of Boat from
Downstream to Upstream.

25. 03)Water Pressure
on Masonry Walls:-
When Water on one side
of Masonry Walls, the
Water Pressure will act
perpendicular to the Wall.
A little consideration will
show, that the intensity of
Pressure at Water Level
will be Zero and will be
increase by a Straight
Line Law to wH at the
bottom. Thus the Pressure
of the Water on
a Vertical Wall will act
through a point at a
distance H/3 from the
bottom, where H is the
depth of Water.

26. Conditions for Stability of Dam.
01)The Resultant must pass within the base to safeguard Dam against Overturning.
02)The Resultant must pass through the middle third of the base, to avoid tension at the
base.
03)To prevent sliding, the maximum Frictional Force should be more than the
Horizontal Force.
04)The Maximum Stress developed at the bottom of the Dam should be within the
permissible Stress of the site.