CVS 246:Introduction to Fluid Mechanics
Chapter 4: Hydrostatic Forces on Submerged Surfaces
L. O. Muku
Department of Civil & Structural Engineering
DCSE, MOI UNIVERSITY
3-Feb-25
4.1 Introduction
o Hydrostatics deals with the behaviour of fluids at rest.
o A static fluid element may be subjected to two forces, namely body force (gravity) and
normal surface forces (exerted on the fluid element by the surrounding fluid).
o Force exerted on immersed surfaces by the static fluid is due to pressure distribution on the
surfaces.
o In static fluids, shear stresses are completely absent. Thus, the surface forces are only due to
the action of normal stress, i.e., hydrostatic pressure.
o In many engineering applications, it becomes necessary to determine the pressure forces on
the entire surface of a hydraulic device and structures, such as dams, submarines, ships,
pipes, gates, containers, balloons, tanks, etc.
o This chapter describes the hydrostatic equations and methods required to determine the
magnitude, location and the direction of resultant force acting on a submerged surface
under static fluid conditions.
o The submerged surface may be horizontal plane surface, vertical plane surface, inclined
plane surface or curved surface.
3-Feb-25
4.2 Total Pressure, Centre of Pressure & Centre of Gravity
o In the design of several hydraulic structures and machines, it is often required to calculate
the magnitude of total pressure and to locate its point of application.
1. Total Pressure: A static mass of fluid when it comes in contact with a solid surface (plane
or curved) exerts a force on it. This force always acts normal to the surface, and it is
known as total pressure, denoted by p.
2. Centre of Pressure: The point of application of total pressure on the surface is known as
centre of pressure and it is denoted by C.
3. Centre of Gravity: The centre of gravity (Centroid) is the point where the whole weight
of the body lies, and it is denoted by G.
4.3 Moments of Area and Geometrical Properties
o The determination of first and second moment of areas is necessary in the evaluation of the
resultant force and centre of pressure.
First Moment of Area
o Consider the area A and the moments of area about the
line O–O as shown in Fig.4.1.
o Let hG be the distance of the centre of gravity of area from
the line O–O.
Fig. 4.1: First and second moments of an
1.
area
4.3 Moments of Area and Geometrical Properties….
o The moment of area with respect to the line O–O can be obtained by summing up the
moments of elementary areas (dA) all over the surface with respect to the given axis. The
first moment of area about the line O–O is given in Eq. 4.1.
……………….. (4.1)
o The first moment of area is used to locate the centroid of the area.
2. Second Moment of Area (Area Moment of Inertia)
o It is given by Eq. 4.2.
…………………………. (4.2)
o By parallel axis theorem, we get:
…………..… (4.3)
Where IG is the moment of inertia about an axis G–G passing through the centre of Gravity, G
and parallel to the line O–O.
o Thus, moment of inertia (M.O.I.) of an area about any axis is equal to the sum of the
moment of inertia about a parallel axis through the centroid and the product of the area and
the square of the distance between this axis and the axis passing through centroid.
o The second moment of area is used in the determination of centre of pressure for plane areas
submerged in liquids.
4.3 Moments of Area and Geometrical Properties….
o The moments of inertia and other geometrical properties of some important plane surfaces are
given in Table 4.1 in which GG is the centre of gravity, IG is the moment of inertia about an
axis passing through GG and parallel to base and IO is the moment of inertia about base.
Table 4.1: Moments of inertia and geometrical properties
4.4 Horizontal Submerged Plane Surface
o Consider a horizontal plane surface submerged in a static liquid as shown in Figure 4.2.
Fig. 4.2: Horizontal submerged plane surface
o Let A be the surface area, C be the centre of pressure, G be the centroid, hC be the distance
of centre of pressure from the free surface of liquid, hG be the distance of centre of gravity
from the free surface of liquid, p be the pressure intensity and F be the total pressure force
on the surface.
o For a submerged horizontal plane surface, the points C and G coincides with each other and
thus, hC = hG.
o Since all the points on the horizontal plane surface are at the same depth from the free
surface of liquid, the pressure intensity is constant over the entire surface, and it is as in Eq.
4.4.
…………………………………. (4.4)
o The total pressure force on the surface is given as,
…………………… (4.5)
4.5 Vertically Submerged Plane Surface
1.Total Pressure on a Vertical Submerged Plane Surface
o Consider a plane vertical surface with random shape submerged in a static liquid (Fig. 4.3).
o Let A be the surface area, C = the centre of pressure,
G = the centroid,
hC = distance of centre of pressure from the free surface
of liquid,
hG = the distance of centre of gravity from the free
surface of liquid,
Fig. 4.3: Vertical submerged plane surface
p = pressure intensity and
F = total force on the surface.
 Consider an elementary strip of area dA at a depth h from the free surface of liquid and
parallel to it.The pressure force on the strip is expressed as.
 Total pressure force on the whole surface is given by,
……………………… (4.6)
4.5 Vertically Submerged Plane Surface…
2. Centre of Pressure on aVertical Submerged Plane Surface
 The pressure force on the strip is given by,
 Moment of this pressure force about the free liquid surface is given by,
 Sum of moments of all such pressure forces about the free liquid surface becomes,
……………. (4.7)
 Now moment of total force F acting at point C at a distance hC is given by,
……………………….. (4.8)
 Principle of moments states that the moment of the resultant force about an axis is equal to the sum
of moments of the components about the same axis. Thus, equating Eq.4.7 and 4.8, we get the
following expression.
…….…….. (4.9)
 Now substituting the value of IO from Eq.4.3 in Eq.4.9, we get:
……………….…... (4.10)
 Thus, Eq. 4.10 gives the position of the centre of pressure on a plane surface submerged
vertically in a static mass of liquid.
 From Eq.4.10, it is observed that the centre of pressure hC lies below the centroid of the
area and it is independent of the density of the liquid.
4.5 Vertically Submerged Plane Surface…
Example 4.1: A rectangular plate 0.4 m × 1.6 m is immersed in water (Fig. 4.4) . Determine the
hydrostatic force and the centre of pressure when the plate is kept (i) vertical with 0.4 m side
coinciding with water surface, (ii) vertical with 0.4 m side kept 2 m below and parallel to water
surface and (iii) vertical with 1.6 m side kept 2 m below and parallel to water surface.
Solution
Let b = 0.4 m and d =1.6 m.
(i) Refer Figure 4.4(a).
Eq. 4.6
Figure 4.4
(ii) Refer Figure 4.4(b).
4.5 Vertically Submerged Plane Surface…
(iii) Refer Figure 4.4(c).
Example 4.2: A triangular thin plate of base 1 m and height 1.5 m is hinged vertically inside a tank
containing a liquid (specific gravity = 1.2) such that the base coincides with the free surface.
Determine the total pressure acting on the plate and the depth of its centre of pressure.
Solution
Let b =1m, h =1.5. m and S =1.2 .
4.5 Vertically Submerged Plane Surface…
Example 4.3: A circular thin plate of diameter 600 mm is immersed in water vertically such that its top
edge is 2 m below free water surface. Determine the total pressure acting on the plate and the
position of its centre of pressure (Refer to given Figure).
Solution
 Let d = 600 mm = 0.6 m.
Example 4.4: A disc of diameter 2 m which can rotate about a horizontal diameter is used to close a
circular opening of the same size in the vertical side of a tank. If the head of water above the
horizontal diameter of the disc is 3 m, then find (i) force on the disc, (ii) position of centre of pressure
and (iii) torque required to maintain the disc in equilibrium in the vertical position. (Refer to the
given Figure).
Solution
Let d = 2 m, hG = 3 m and T be the torque required.
4.5 Vertically Submerged Plane Surface…
Exercises
1. A pipeline 4 m in diameter containing an oil (specific gravity = 0.9) has a gate valve. The pressure at
the centre of the pipe is 200 kPa. Find (i) the force exerted on the gate and (ii) the position of centre of
pressure. (Refer to given figure). (Ans: F =2513.316 kN, )
, C = 0.044 m below the centre of pipe). All Grps
2. A square aperture in the vertical side of a tank has one diagonal and is completely covered by a plane
plate hinged along one of the upper sides of the aperture. The diagonals of the aperture are 2 m long
and the tank contains glycerine (specific gravity = 1.26). The centre of aperture is 1.4 m below the
free surface. Determine (i) thrust exerted on the plate by the glycerine and (ii) position of its centre of
pressure. (Ans: F = 34609.68 N, hC = 1.519 m)
All Grps
4.5 Vertically Submerged Plane Surface (Exercise)
3. A dry dock is closed by a gate of trapezoidal shape having top and bottom lengths 18 m and 12 m,
respectively and a height of 7.5 m. Determine the total water pressure and the depth of centre of
pressure on the gate if the sea water (specific gravity = 1.02) level is up to the top of the gate on one
side and the other side is empty. (Refer to the given figure)
(Ans: F= 4502.79 kN, hC = 5.156 m) All Grps
4. A circular plate of diameter 1 m with a hole of diameter 0.25 m is immersed vertically in a liquid
(specific gravity = 0.9) with its upper edge 0.5 m below the free surface of the liquid. The centre of
hole is 0.25 m vertically below the centre of the plate. Determine: (i) the pressure force acting on the
plate and, (ii) the centre of pressure. (Ans: F= 6392.23 N, hC = 1.0423 m )
5. A vertical rectangular gate 3.5 m wide and 5 m high contains water on one side to a depth of 2.4 m
and an oil (specific gravity = 0.9) to a depth of 1.5 m on the other side. Determine the resultant
hydrostatic pressure force on the gate and its point of application with respect to the bottom. All Grps
(Ans: F= 64,120.61 N, hC = 0.9626 m from bottom)
4.5 Vertically Submerged Plane Surface (Exercise)
6. A 4 m × 2 m wide rectangular gate is vertical and is hinged at point 0.2 m below the centre of gravity
of the gate. The total depth of water is 6 m. Find out the horizontal force required at the bottom of the
gate to keep it in closed position.
Solution
Let h = 4 m, b = 2 m, x = 0.2 m and h1 = 6 m.
Pressure force acting on the plane surface of the gate is given by,
 Let F1 be the force required to be applied at the bottom of the gate to keep it closed. Taking moments
of all forces about the hinge, we get the following expression.
4.5 Vertically Submerged Plane Surface (Exercise)
7. A sliding gate of height 1.4 m and width 2.8 m lies in vertical plane that weighs 25 kN. Determine
the vertical force required to lift the gate when its upper edge is 6 m below the free water surface
and the coefficient of friction between the gate and guides is 0.15. Determine the position of centre
of pressure acting on the gate. (Refer to the given figure).(Ans: F= 63.6475 kN) All Grps
4.6 Inclined Submerged Plane Surface
4.6.1 Total Pressure on an Inclined Plane Submerged Surface
 Consider a plane inclined vertical surface with random shape submerged in a static liquid (Figure 4.5).
Figure 4.5 Inclined submerged plane surface
 Let A be the surface area, hC be the distance of centre of pressure from the free surface of liquid,
hG be the distance of centre of gravity from the free surface of liquid, p be the pressure intensity, F be the
total force on the surface, C be the centre of pressure, G be the centroid and α be the angle made by the
plane of the surface with free liquid surface.
 When the plane of the surface is produced from point ‘B’, then it meets the free liquid surface at point
‘A’ and it will be perpendicular to the plane of the surface. Let lG and lC be the distances of G and C,
respectively, from the axis AB.
4.6 Inclined Submerged Plane Surface
4.6.1 Total Pressure Force on an Inclined Plane Submerged Surface
 Consider an elementary strip of area dA at a depth h from the free surface of liquid and at a distance l
from the axis AB.
 Pressure force on the strip is given by,
 Total pressure force on the whole surface is given by,
………………... (4.11)
………….. (4.12)
4.6 Inclined Submerged Plane Surface..

4.6 Inclined Submerged Plane Surface…
Example 4.4: A rectangular plane surface that is 1.5 m wide and 4 m deep is immersed in a liquid (specific
gravity = 0.9) in such a way that its plane makes an angle of 30° with the free surface of liquid.
Determine the total pressure force and position of centre of pressure when the upper edge is 1 m below
the free liquid surface. (Refer to the given figure)
Solution
Example 4.5: Determine the total pressure and position of centre of pressure of a circular plate that has
diameter of 2 m submerged in water whose greatest and least depths below the surface are 1.5 m and 0.5
m, respectively. (Refer to the given figure).
Solution
4.6 Inclined Submerged Plane Surface..
Solution….
Example 4.6: A triangular plate of base 1.5 m and height 2 m is submerged in oil (specific gravity = 0.92).
The plane of the plate is inclined at 30° with free oil surface and the base is parallel and it is at a depth of
1 m from the oil surface. Determine the total pressure and position of centre of pressure on one face of
the plate. (Refer to the given figure).
Solution
4.6 Inclined Submerged Plane Surface..

4.6 Inclined Submerged Plane Surface..
Exercises
1. An annular plate having external and internal diameters of 2 m and 1 m, respectively is submerged in an
oil (specific gravity = 0.92) in such a way that its greatest and least depths below the oil surface are 3 m
and 2 m, respectively. Determine the total pressure and the position of centre of pressure on one face of
the plate. (Refer to the given figure).(F= 53162.941 N, hC = 2.5312 m ).
2. A trapezoidal plate of height 2.2 m and sides of 2.4 m and 3.6 m is immersed in water at an inclination
of 30° to the free surface of the water. The depth of top edge of the plate is at 2 m from the free surface.
Determine the hydrostatic force on the given plate and the centre of pressure. (Refer to the given figure).
(Ans: F= 167368.41 N, hC = 2.6235 m ).
4.6 Inclined Submerged Plane Surface…
Exercises…
3. A 6 m × 2 m rectangular gate is hinged at the base and it is inclined at an angle of 60° with the
horizontal. The upper end of the gate is kept in position by a weight of 55000 N acting perpendicularly
to the gate through a pulley system. If the weight of the gate and the friction at the hinge and pulley is
neglected, then find the level of water when the gate begins to fall. (Refer to the given figure).
(h= 4.232 m ).
4. A 4 m × 2.5 m rectangular sluice gate PQ hinged at point P (Figure) and inclined at an angle of 45°
with the horizontal is kept closed by a weight fixed to the gate. The total weight of the gate and weight
fixed to the gate is 450 kN. The centre of gravity of the weight and gate is at G. Determine the height of
the water h which will cause the gate to open.TG = 0.75 M.
(Ans: h = 4.44 m)
4.7 Curved Submerged Plane Surface

4.7 Curved Submerged Plane Surface..
 The total resultant force on the curved surface is given,
……………………….... (4.19)
 The resultant force F passes through the intersection of its two components and its inclination with
horizontal is given by Eq. 4.20.
………………………..…….... (4.20)
 Note: When the underside of a curved surface is subjected to hydrostatic pressure as shown in Figure 4.6
(c), the force FV will be equal to the weight of the imaginary fluid supported by PQ upto the free surface
of liquid and its direction will be taken in upward direction.
4.7 Curved Submerged Plane Surface..

4.7 Curved Submerged Plane Surface..

4.7 Curved Submerged Plane Surface..

4.7 Curved Submerged Plane Surface..

4.7 Curved Submerged Plane Surface

Buoyancy and Stability of Floating Bodies

4.8 Buoyancy and Stability of Floating Bodies
Proof of Archimedes’ Principle..
 Let dA be the cross-sectional area of a small vertical element, dv be the volume of the small
element, p1 and p2 be the intensity of pressures at depths h1 and h2, respectively.
 The force acting on the top face of the element is equal to:
 The force on the bottom face of the element is equal to:
 The net force on the element is equal to the buoyant force dFB which acts upwards (h2 > h1)
and it is given as.
 The total buoyant force is given by,
…………….. (4.21)
v is the volume of the submerged body which is equal to the volume of fluid
displaced by the body and Wd is the weight of the fluid displaced by the body.
 Thus, buoyant force is equal to the weight of the fluid displaced by the body. It acts through
centre of buoyancy which coincides with the centroid of the displaced volume.
 For a fully submerged body, the centre of buoyancy (B) coincides with the CG (G) of the body.
 Further, the lines of action of both the buoyant force and the weight of the body must be along
the same vertical line, so that their moment about, any axis is zero.
Buoyancy and Stability of Floating Bodies

Buoyancy and Stability of Floating Bodies
Example 4.14: A metallic body weighs 500 kN in air and 250 kN in water. Determine the volume of
body and its specific gravity.
Solution
 Let W = 500 kN and W1 = 250 kN. Let v be the volume of body which is equal to the volume of
water displaced by it and S be its specific gravity. The reduction in weight of the metallic body when
immersed in water is due to the buoyant force (FB).
But
 The specific weight of metallic body is given by,
Example 4.15: An iceberg of relative density 0.92 floats in sea water (specific gravity = 1.03). Find the
weight of the iceberg if the volume of ice above the water surface is 10 m³.
Solution
Buoyancy and Stability of Floating Bodies
 Weight of iceberg = Weight of water displaced by iceberg
 The weight of iceberg is given by,
Example 4.16: A wooden body of height 73 mm floats in a water tank of height 25 mm projecting above
the water surface. The same wooden body when placed in glycerine tank is projected 37.5 mm above the
surface of glycerine. Find (i) the relative density of the wooden body and (ii) the relative density of
glycerine.
Solution
Buoyancy and Stability of Floating Bodies
 Let d = 73 mm, y1 = 25 mm and y2 = 37.5 mm. Let Swood be the relative density of wooden body and
Sg be the relative density of the glycerine.
(i)
Weight of wooden body = Weight of water displaced
(ii)
Weight of wooden body = Weight of glycerine displaced
Buoyancy and Stability of Floating Bodies
Exercises
1. A metallic body floats at the interface of mercury (Hg) and water (H2O) in a tank such that 35% of its
volume is submerged in mercury and 65% in water. Find the density of the metallic body. Take density of
mercury as 13600 kg/m³ and density for water as 1000 kg/m³. (Refer to given figure).
(Ans: 5410 kg/m³)
2. A football of diameter 40 cm fell into a water tank, 20% of its volume is found under water. Determine
the density of the football.
3. A wooden block (specific gravity = 0.65) that is 2.5 m long, 1 m wide and 0.5 m high floats in a water
tank. Determine the volume of concrete of specific weight 24.5 kN/m³, that may be kept on the block
and immerse the (i) block completely in water and (ii) block and the concrete completely in water. Take
weight density of water as 9.81 kN/m³. (Refer to given figure).
Buoyancy and Stability of Floating Bodies
Exercises
4. A metallic cube has side 0.25m and it weighs 250 N when lowered into a tank containing a two-fluid
layer of water and mercury. Determine the position of block at mercury-water interface when it has
reached equilibrium. (Refer to given figure).
5. A wooden block (specific gravity = 0.64) that is 0.12 m square in cross-section and 2.6 m long floats in
a water tank. Determine how much lead (specific gravity = 12.5) is to be attached at the lower end of the
block so that it floats vertically in water with 0.6 m length out of the water. (Refer to given figure).
Metacentre

Metacentric Height and Methods of Its Determination
 Metacentric height is the distance between the centre of gravity G and the metacentre M of a
floating body.
 In Figure 4.8(b), GM is the metacentric height.
 2 methods applied in determination of GM: Analytical method and (ii) Experimental method.
1. Analytical Method (Assignment to be submitted in 1 week's time)

Derive (Analytically) the expression for Metacentric height which is given by:
Valid when G lies above B

If G lies below B, then the metacentric height is given as.
2. Experimental Method
 The Figure 4.9(a) illustrates a floating body in equilibrium at the water surface in which points
G and B lie on the normal vertical axis and the top surface of the body is horizontal.
 Let w1 be a movable weight placed centrally on the floating body and W be the total weight of
the body including the movable weight w1.
Fig. 4.9 Experimental method for metacentric height of
a floating body
Metacentric Height and Methods of Its Determination

Metacentric Height and Methods of Its Determination
(i) Weight of the wooden block = Weight of water displaced
(ii) Moment of inertia of the top view at water surface about y -y is given by,
Metacentric Height and Methods of Its Determination
Exercises
1. A rectangular barge of dimensions 10 m × 3 m weighs 75 tons and its centre of gravity lies 1.3 m above
the bottom. Determine the metacentric height when it floats in fresh water. (Refer to figure in Example
4.17).
2. A rectangular pontoon of length 20 m and weight 2750 kN floats in fresh water of specific weight
10 kN/m³. Its centre of gravity lies 25 cm above the centre of cross section and for 10° angle of heel its
metacentric height is 1 m. If 0.6 m height portion of the pontoon is lying outside water, then determine
its breadth and height.
3. A cube of side 2 m floats in a liquid with half of its volume immersed and the bottom face being
Horizontal. The weight 360 N is moved on to the middle point of one of the top edges of the cube. Find
the angle through which the cube tilts under the action of weight, if the centre of gravity of the cube is
0.65 m below the geometric centre in a vertical line through it. (Refer to given figure).
End of the Lecture
3-Feb-25