HYDROSTATICS
(fluid at rest)
MEC 121 – W.MIA18”
HYDROSTATICS
 Fluid can either be a liquid or a gas.
 Mainly characterized by their negligible resistance to
change in shape and capable of following.
 Liquids and gases are distinguished as follows;


A gas completely fills the space in which it is contained while a
liquid usually has free surface see figures below.
A gas is a fluid which can be compressed relatively easy, while
liquids are considered incompressible.
HYDROSTATICS
Hydrostatics or statics of fluid is a study of force and
pressure in a fluid at rest.
Pressure
Fluid in a closed cylinder (see figure above) can be put
under pressure by applying force P to the piston.
Neglecting the weight of the fluid the pressure is given
by the ratio
P
force P on the piston
area A of piston
or p 
A
HYDROSTATICS
The derived SI unit for pressure is the same as that of
that of stress.
newtons per square meter N m 2
The following multiples shall be used for pressure in
liquids.
k N m 2  103 N m 2
M N m 2  106 N m 2
G N m 2  109 N m 2
A Bar is given as a special multiple of the unit N m2
1bar  105 N m 2
HYDROSTATICS
 The pressure exerted by the atmosphere (atm) is
1.013bar, making the special multiple of the unit
convenient to use among engineers.
Pressure in the fluid has the following
important features
1. The pressure at a point is the same in all directions
2. Pressure exerted at a point on a surface is normal
to that surface
TRANSMISSION OF FLUID PRESSURE
States that pressure intensity at any point of a fluid at
rest is transmitted without loss to all other points in
the fluid.
Demonstrating the principle
Lets consider the hydraulic press machine. It is
required to find the effort in the piston E that will keep
the load W in equilibrium.
Applying the principle of work;
TRANSMISSION OF FLUID PRESSURE
TRANSMISSION OF FLUID PRESSURE
𝑓
𝑎
𝑊
𝐴
is the pressure in cylinder E and is the pressure
in cylinder D hence the pressures are equal.
NOTE
 The principle does not depend on the frictionless
piston. The effect of friction is merely to increase
the effort F, required to hold the load W above the
𝑎
theoretical value F = 𝑤 × .
𝐴
 The principle does not depend on fluid being
incompressible, it is also applicable to gasses as
long as the force is applied gradually.
DENSITY, RELATIVE DENSITY, SPECIFIC
WEIGHT AND SPECIFIC GRAVITY
1. Density p of a substance is mass per unit volume
Unit = 𝒌𝒈 𝒎𝟑 .
Other forms include;
 The density of water is the most important and need
to be remembered. And it is given by;
DENSITY, RELATIVE DENSITY, SPECIFIC
WEIGHT AND SPECIFIC GRAVITY
2. Relative Density of a substance is the ratio of its
density to that of water.
3. Specific Weight of a substance is the weigh per
unit volume. Since weight = mg. the relationship
between specific weight and density is given by;
DENSITY, RELATIVE DENSITY, SPECIFIC
WEIGHT AND SPECIFIC GRAVITY
4. Specific Gravity = Relative gravity
Given by;
PRESSURE IN A LIQUID DUE TO ITS OWN
WEIGHT
Application of hydrostatic pressure
Construction of dams
PRESSURE IN A LIQUID DUE TO ITS OWN
WEIGHT
Suppose that a thin plate with area A m2 is
submerged in a fluid of density ρ kg/m3 at a depth
d meters below the surface of the fluid.
PRESSURE IN A LIQUID DUE TO ITS OWN
WEIGHT
The fluid directly above the plate has
volume
V = Ad
So, its mass is:
m = ρV = ρAd
PRESSURE IN A LIQUID DUE TO ITS OWN
WEIGHT
The pressure P on the plate is defined
to be the force per unit area:
F
P    gd
A
PRESSURE IN A LIQUID DUE TO ITS OWN
WEIGHT
Therefore pressure in a liquid due to its own
weight is directly proportional to its height. As the
depth of a liquid increases the pressure increases.
See figure below;
PRESSURE IN A LIQUID DUE TO ITS OWN
WEIGHT
For an open surface of the liquid, if the atmospheric
pressure is acting on the free surface, the the total pressure
at the bottom is given by;
𝑷 = 𝑷𝒂 + 𝒑𝒈𝒉
the atmospheric pressure is given as 103𝑘𝑁 𝑚2 or 1𝐵𝑎𝑟
PRESSURE IN A LIQUID DUE TO ITS OWN
WEIGHT
 The SI unit for measuring pressure is newtons per
square meter—which is called a pascal (abbreviation: 1
N/m2 = 1 Pa).
 As this is a small unit, the kilopascal (kPa)
is often used.
PRESSURE IN A LIQUID DUE TO ITS OWN
WEIGHT
Example
For instance, since the density of water is ρ = 1000 kg/m3, the
pressure at the bottom of a swimming pool 2 m deep is:
P   gd
 1000 kg/m3  9.8 m/s 2  2 m
 19, 600 Pa
 19.6 kPa
TOTAL THRUST ON A VERTICAL PLANE
SURFACE
 Due to the existence of hydrostatic pressure in a fluid
mass, a normal force is exerted on any part of a solid
surface which is in contact with a fluid.
The individual forces distributed over an area
give rise to a resultant force.
TOTAL THRUST ON A VERTICAL PLANE
SURFACE
Now lets consider a plane surface of area A immersed
vertically in a liquid of density p. the pressure on one
side of the surface is normal to the surface and gives
rise to a resultant force or thrust on that side. See
figure below
TOTAL THRUST ON A VERTICAL PLANE
SURFACE
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝒑 𝑜𝑛 𝑜𝑛𝑒 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑟𝑖𝑝 𝐷𝐸 𝑎𝑡 𝑑𝑒𝑝𝑡ℎ 𝒙
𝑷 = 𝒑𝒈𝒙
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑎 𝑠𝑡𝑟𝑖𝑝 𝑜𝑓 𝑤𝑖𝑑𝑡ℎ 𝒃 𝑎𝑛𝑑 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝒅𝒙 = 𝒃 × 𝒅𝒙
𝑓𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑡𝑟𝑖𝑝 = 𝑷 × 𝒃𝒅𝒙
= 𝒑𝒈𝒙 × 𝒃𝒅𝒙
= 𝒑𝒈𝒃𝒙𝒅𝒙
∴ 𝑡𝑜𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑜𝑛 𝑎𝑟𝑒𝑎 𝐴 =
𝒑𝒈𝒃𝒙𝒅𝒙
= 𝒑𝒈
𝒃𝒙𝒅𝒙
TOTAL THRUST ON A VERTICAL PLANE
SURFACE
But 𝒃𝒙𝒅𝒙 is the total moment of area A about an axis
through O in the water surface is equal to 𝑨 × 𝒙
Where 𝒙 is the depth of the centroid G of the plane
surface below the water line.
Hence
𝑭 = 𝒑𝒈 × 𝑨𝒙 = 𝒑𝒈𝑨𝒙
The total thrust on an immersed vertical plane surface
is proportional to the depth of the centroid of the
wetted area below the free surface.
TOTAL THRUST ON A VERTICAL PLANE
SURFACE
NOTE
However, the total line of action of the total thrust
does not pass through the centroid but through the
point called centre of pressure, which has yet to be
found.
CENTRE OF PRESSURE
The centre of gravity is the point of application of the
resultant force due to liquid pressure on one side of the
immersed surface.
CENTRE OF PRESSURE
To determine the depth 𝒚 of the centre of pressure
employ the principle of moments.
The sum of the moments about the water surface O-O
of the forces on all strips such as DE must equal to;
𝑭×𝒚
𝑓𝑜𝑟𝑐𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑡𝑟𝑖𝑝 𝐷𝐸 = 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 × 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠𝑡𝑟𝑖𝑝
𝑜𝑟 𝒅𝑭 = 𝑷 × 𝒃𝒅𝒙
= 𝒑𝒈𝒙 × 𝒃𝒅𝒙 𝑠𝑖𝑛𝑐𝑒 𝑷 = 𝒑𝒈𝒙
CENTRE OF PRESSURE
Moment of this force about O-O = 𝒅𝑭 × 𝒙
= 𝒑𝒈𝒃𝒙𝒅𝒙 × 𝒙
= 𝒑𝒈𝒃𝒙𝟐 𝒅𝒙
Therefore the total moment about O-O =
𝒑𝒈𝒃𝒙𝟐 𝒅𝒙
This is equal to the moment 𝑭 × 𝒚 of the resultant
force about O-O. Hence
𝑭 × 𝒚 = 𝒑𝒈 𝒃𝒙𝟐 𝒅𝒙 since pg is a constant
CENTRE OF PRESSURE
But since 𝒃𝒙𝟐 𝒅𝒙 is the total second moment of the
area 𝑰𝟎 of the area A about O-O, so that;
𝑭 × 𝒚 = 𝒑𝒈 × 𝑰𝟎
Making 𝒚 subject of the formula
𝒑𝒈𝑰𝑶
𝒚=
𝑭
We know that 𝑭 = 𝒑𝒈𝑨𝒙
𝒑𝒈𝑰𝑶
∴𝒚=
𝒑𝒈𝑨𝒙
CENTRE OF PRESSURE
𝑰𝑶
𝒚=
𝑨𝒙
𝑆𝑒𝑐𝑜𝑛𝑑 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑎𝑟𝑒𝑎 𝑎𝑏𝑜𝑢𝑡 𝑂 − 𝑂
=
𝐹𝑖𝑟𝑠𝑡 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑎𝑏𝑜𝑢𝑡 𝑂 − 𝑂
Now that this refers to the wetted area only, the last
expression is conveniently rewritten as;
Let 𝐼𝐺 = second moment of area A about axis through
G to the water surface O-O
= 𝐴𝑘 2
CENTRE OF PRESSURE
 Where k is the corresponding radius of gyration.
Then the parallel theorem states that
𝐼𝑜 = 𝐼𝐺 + 𝐴𝑥 2
Hence
𝐼𝑜
𝐼𝐺 + 𝐴𝑥 2
𝑦=
=
2
𝐴𝑥
𝐴𝑥
𝐼𝐺
=
+𝑥
𝐴𝑥
𝐴𝑘 2
𝑘2
=
+𝑥 =
+𝑥
𝐴𝑥
𝑥
CENTRE OF PRESSURE
Thus the distance from the center of pressure C and
the centroid G;
𝐺𝐶 = 𝑦 − 𝑥
𝑘2
=
𝑥
This expression tends to be ZERO as 𝑥 becomes very
large. The center of pressure is always below the
centroid of wetted area but tends to coincide with the
centroid at a great depth.
CENTRE OF PRESSURE
The second moment of area 𝐼𝐺 and corresponding 𝑘 2
for rectangular and circular areas are shown below;
TOTAL THRUST ON A VERTICAL PLANE
SURFACE & CENTRE OF PRESSURE
Example
A lock gate has a sea water to a depth 3.6𝑚 on one side
and 1.8𝑚 on the other. Density of sea water =
103 𝑀𝑔 𝑚3
Find;
a. The resultant thrust per meter width on the
gate?
b. The resultant moment per meter width tending to
overturn the gate at its base.
End of topic
Tutorial hand out from Hanna& Hillier Applied
Mechanics