2. We have seen the following features of static
fluids in the foregoing lectures:
• Hydrostatic vertical pressure distribution is
linear
• Pressures at any equal depth in a continuous
fluid are equal
• Pressure at a point acts equally in all
directions
• Forces from a fluid on a boundary acts at right
angles to the boundary
3. In this lecture we will look at the effects of Forces
on Submerged Surfaces and Bodies in Static Fluids
The objectives are to:
• Use the facts learned hitherto to analyse and
obtain expressions for forces on submerged
surfaces.
• Reinforce the knowledge that pressure which is
a scalar quantity is equal in alldirections.
• Reinforce the knowledge that force which is a
vector quantity has both magnitude and
direction
• Determine the stability of submerged and
floating bodies.
4. Fluid Pressure on a Surface
Pressure is force per unit area. If pressure 𝒑
acts on a small area 𝜹𝑨, then the force exerted
on the area is:
𝑭 = 𝒑 𝜹𝑨 (1)
Since the fluid is at rest, the force acts at right-
angles to the surface.
5. Horizontally Submerged Plane
𝒑 𝒑 𝒑 𝒑 𝒑
In this case pressure 𝒑 will be the same at all points
of the surface.
The resultant force 𝑹 will be given by:
𝑹 = (pressure) x (area of plane) or 𝑹 = 𝒑 𝑨
6. General Submerged Plane
Now consider a plane surface submerged in a fluid
and inclined at an angle as shown below.
The total area is made up of many elemental areas.
Each force is normal to the elemental area, but may
be of different magnitudes if the pressure varies.
7. The total or resultant force 𝑹 on the plane can be
found by summing up all the forces on the individual
small elements. That is:
𝑹 = 𝒑𝟏𝜹𝑨𝟏 + 𝒑𝟐𝜹𝑨𝟐+. . +𝒑𝒏𝜹𝑨𝒏 = Σ 𝒑𝜹𝑨 (2)
This resultant force 𝑹 will act through the centre of
pressure. That is,
For a plane, the force can be represented by one
single resultant force, acting at right-angles to
the plane through the centre of pressure.
8. Resultant Force and Centre of Pressure on a
Submerged Plane Surface in a liquid
Consider a plane surface totally submerged in a liquid
of density ρ and inclined at an angle φ to the
horizontal. Takepressure at the surface as zero.
9. The quantity ∑yδA is the first moment of area under the
surface PQ about the free surface of the liquid and is equal to
AD, where A = the area of the whole immersed surface PQ and
D = the vertical depth to the centroid G of the immersed
surface. Substituting in equation (1),
Summing the forces on all such elements over the whole
surface, since these forces are all perpendicular to the plane PQ,
(1)
(2)
10. Moment of R about O = Sum of moments of forces on
all elements of area δA about O,
Force on any small element
= ρgyδA = ρgs sin φ × δA (3)
Moment of force on element about O = pgs sin φ × δA × s
= ρg sin φ × δA × s2.
Since ρ, g and φ are the same for all elements,
Sum of the moments of all such forces about O
= ρg sin φ ∑ s2δA.
12. The Parallel Axis Theorem
Suppose a surface of area A is made to rotate
about an axis passing through the centroid.
The surface has 2nd moment of area IGG with
respect to this axis.
The parallel axis theorem states that if the surface
is made to rotate instead about a new axis O
which is parallel to the first (centroidal) axis and
displaced from it by a distance 𝒙, then the
moment of area IO with respect to the new axis is
related to IGG by:
𝑰𝒐 = 𝑰𝑮𝑮 + 𝑨
𝒙 𝟐
13.
14. Lateral Position of the Centre of Pressure
If the shape is symmetrical, the centre of pressure
lies on the line of symmetry.
If the shape is not symmetrical, its position must be
found by taking moments about the line OG in the
same way as we took moments along the line
through O, i.e.
𝑹 𝒙 𝒅 = sum of the moments of the force on all
elements of δA about OG
= Σ𝝆gy𝜹Ax
15. Example: A trapezoidal opening in the vertical wall of a tank
is closed by a flat plate which is hinged at its upper edge. The
plate is symmetrical about its centreline and is 1.5 m deep. Its
upper edge is 2.7 m long and its lower edge is 1.2 m long. The
free surface of the water in the tank stands 1.1 m above the
upper edge of the plate. Calculate the moment about the hinge
line required to keep the plate closed.
16. Solution
To find the position of the centroid G, take moments of area
about BB′, putting the vertical distance GB = y:
17.
18. Submerged Vertical Surface – Pressure Diagrams
For vertical walls of constant width, it is possible
to find the resultant force and centre of pressure
graphically using a pressure diagram.
We know the relationship between pressure and
depth is:
𝒑 = 𝝆𝐠𝒛
19. Gauge pressure increases linearly from zero at the surface
by 𝒑 = 𝝆gz, to a maximum at the base of 𝒑 = 𝝆gH. So we
can draw the diagram below:
The area of this triangle represents the resultant force per unit
width on the vertical wall, R. Using SI units, this would have units
of Newtons per metre.
20. 𝑨𝒓𝒆𝒂 =
𝟏
𝟐
𝒙 𝑯𝒆𝒊𝒈𝒉𝒕 𝒙 𝑩𝒂𝒔𝒆
𝟐
= 𝟏
𝑯𝝆g𝑯
So, resultant force per unit width,
𝟐
𝐑 = 𝟏
𝝆g𝑯𝟐
The force acts through the centroid of the pressure
diagram. For a triangle, the centroid is at 2/3 down
its height.
Hence, for a vertical plane the depth to the
centre of pressure is given by:
𝟑
𝑫 = 𝟐
𝑯
21. This result could also have been obtained from
equations (2) and (5), since, for unit width,
and, in equation (5), φ = 90°, sin φ = 1, 𝑦 = H/2, 𝑘𝐺
2
=H2/12
22. If the plane surface is inclined and submerged below
the surface, the pressure diagram is drawn
perpendicular to the immersed surface and will be a
straight line extending from
p = 0 at the free surface to p = ρgH at depth H.
As the immersed
surface does not extend
to the free surface, the
resultant force R is
represented by the
shaded area, instead of
the whole triangle, and
acts through the
centroid P of this area.
23. Example: A closed tank, rectangular in plan with vertical
sides, is 1.8m deep and contains water to a depth of 1.2 m. Air
is pumped into the space above the water until the air pressure
is 35kN/m2. If the length of one wall of the tank is 3 m,
determine the resultant force on this wall and the height of the
centre of pressure above the base.
24. Solution
RH2O will act at h = 0.4 from the base. If x is the height above the
base of the centre of pressure through which R acts,
25. Resultant Force on Curved Submerged Surface
For a curved surface, each elemental force will be a
different magnitude and in different direction but
still each acting normal to the surface of that
particular element.
It is easier to calculate the horizontal and vertical
components and combine these to obtain the
resultant force and its direction.
This can be done for all three dimensions, but here
we will only look at one vertical plane.
The diagram on the next slide shows a forces acting
on a curved surface (a) Fluid on top of the surface
and (b) fluid at below the surface.
27. The element of fluid ABC is in equilibrium since
the fluid is at rest.
Horizontal forces
Considering the
horizontal forces,
none can act on CB
as there are no
shear forces in a
static fluid so the
forces would act on
the faces AC and AB
as shown here.
28. We can see that the horizontal force on AC,
FAC, must equal and be in the opposite
direction to the resultant force RH on the
curved surface.
As AC is the projection of the curved surface AB
onto a vertical plane, we can generalise this to
say:
𝑹𝑯 = Resultant force on the projection of
the curved surface onto a vertical plane.
29. We know that the force on a vertical plane FAC
must act horizontally (as it acts normal to the
plane) and that RH must act through the same
point. So we can also say:
RH acts horizontally through the centre of
pressure of the projection of the curved
surface onto a vertical plane.
Thus we can use the pressure diagram method
to calculate the position and magnitude of the
resultant horizontal force on a two dimensional
curved surface.
30. Vertical forces
In Fig. (a), the vertical component Rv will be entirely due to
the weight of the fluid in the area ABDE lying vertically
above AB. There are no other vertical forces, since there
can be no shear forces on AE and BD because the fluid is at
rest. Thus:
Vertical component, Rv = Weight of fluid
vertically above AB,
and will act vertically downwards through the centre of
gravity G of ABDE.
31. 𝑹𝑽 will act vertically downward through the
centre of gravity of the mass of fluid.
Resultant force
The overall resultant force is found by combining
the vertical and horizontal components vectorialy.
Resultant force 𝑹 = 𝟐 𝟐
RH + RV
This resultant R acts through O at an angle of θ.
The angle the resultant force makes to the
horizontal is 𝜽 = 𝐭𝐚𝐧−𝟏 𝑹𝑽
𝑹𝑯
32. The position of O is the point of interaction of the
and the vertical line of
horizontal line of action of 𝑹𝑯
action of 𝑹𝑽.
A typical example application of this is the
determination of the forces on dam walls or
curved sluice gates.
So ….What are the forces if the fluid is below
the curved surface?
This situation may occur on a curved sluice gate
for example.
33. In the special case of a
cylindrical surface, all the
forces on each small element
of area acting normal to the
surface will be radial and will
pass through the centre of
curvature O. The resultant
force R must, therefore, also
pass through the centre of
curvature O.
FIGURE: Resultant
force on a cylindrical
surface
34. Example: A sluice gate is in the form of a circular arc
of radius 6 m as shown in Figure below. Calculate the
magnitude and direction of the resultant force on the
gate, and the location with respect to O of a point on
its line of action.
36. Since the surface of the gate is cylindrical,
the resultant force R must pass through O.
37. Buoyancy
We have see that in a liquid, pressure increase
with depth.
Therefore, a body submerged in a fluid (e.g.
submerged in water) will experience an
upthrust (upward force) as a result of the
pressure difference.
38. Consider a cylindrical object immersed in a
liquid as shown below.
p1
h1 ρgh1
ɭ
h2
ρgh2
p2
Horizontal forces on the sides balance and cancel
each other, so horizontally the resultant force is
zero.
39. Vertically, pressure p2 > p1 , hence the object
experiences an upward force (an upthrust), or
buoyancy.
Upward force is: FB = p2A – p1A
= ρ gh2A – ρgh1A
= ρ g(h2- h1)A
Hence: 𝑭𝑩 = 𝝆𝐠𝑽
Note that 𝝆𝐠𝑽 = weight of the fluid displaced.
Hence, equation (21) does not depend on the
geometric shape of the immersed object but only
on the volume of fluid displaced.
40. The above fact was established over 2000 years
ago by the Greek philosopher Archimedes.
The two statements of the above facts
attributed to Archimedes are as follows:
1. The upthrust force experienced by a body
immersed in a fluid equals the weight of the
displaced fluid
2. A floating body displaces its own weight in
the fluid in which it floats
41. Stability of Immersed and Floating Bodies
Immersed bodies
Stability of a submerged body depends on:
Centre of gravity of body.
Centroid of the displaced volume of fluid,
called Centre of Buoyancy.
B B G G
G B
(a) (b) (c)
42. Consider immersed bodies as shown in (a), (b)
and (c) above where B is the centre of buoyancy
and G is the centre of gravity.
(a) B above G: Stable – any tipping moment
produces righting couple.
(b) B coincident with G: Neutrally stable.
(c) B below G: Unstable – any tipping moment
produces increasing overturning moment.
Note that only if a fully submerged vessel has its
centre of mass above its centre of buoyancy will it
be unstable.
43. Assuming that the fluid is of constant density, the
upwards buoyancy force on any submerged object
must act at the centroid of the fluid displaced.
This point is called the Centre of Buoyancy.
The weight of the object acts downwards at its Centre
of Mass - and that does not necessarily coincide with
the Centre of Buoyancy.
The rotational stability of any body floating or
submerged in a fluid actually depends on whether the
rotational "couple" set up by the weight and buoyancy
acting at these two points when the body is tilted tends
to return it to its original position (rotationally stable)
or tends to overturn it (rotationally unstable).
44. Floating bodies
In floating bodies, only a portion of the body is
submerged.
Weight of the fluid displaced equals weight of
the floating body.
For equilibrium, the buoyant force and the
weight of the body are collinear.
If tilted, the centre of buoyancy may change but
the centre of gravity for the body will remain
the same.
45. When a body floats in a static fluid, the vertical forces
acting on it are in equilibrium.
If it is displaced upwards a small distance, the
hydrostatic force supporting it (its buoyancy) is
decreased so that the upward force is now less than
the body's weight. There is thus a net downward
force which tends to return the body to its original
position.
Similarly, if the body is initially displaced downwards
a small distance, the forces go out of equilibrium and
set up a resultant upward force which tends to
restore the original position again.
Such a body has vertical stability.
46. Floating vessels frequently have their centre of mass
above the centre of buoyancy but most of them
remain stable.
This is because the centre of buoyancy moves
horizontally as the floating vessel tilts, thereby again
producing a couple which tends to restore the vessel to
equilibrium.
47. Stability
An unstable body will overturn at the first
opportunity to seek equilibrium.
Engineers must design to avoid floating instability
and make sure that “small” disturbances will
result in restoring moments to return the body to
its stable condition.
The basic principle for the determination of static
stability is as follows:
48. 1. The basic floating position is calculated from the
equation:
𝑭𝑩 = 𝝆 𝐠 𝑽 = Weight of floating body
The body’s centre of mass G and the centre of
buoyancy B are calculated.
Remember that for equilibrium, the buoyant force
and the weight of the body have to be collinear.
2. The body is tilted a small angle θ and a new
waterline established for the body to float at this
angle. The new position B’ for the centre of
buoyancy is calculated. A vertical line drawn from B’
intersects the line of symmetry at a point M, which
is independent of θ for small angles.
50. The point M where the line of action of the buoyancy force
of the tilted vessel meets the line through the centre of mass
G which was originally vertical is known as the Metacentre,
and the distance between the centre of mass and the
metacentre, GM, known as the Metacentric Height. This is
of particular importance to engineers interested in the
stability of floating bodies, as the restoring moment is equal
to 𝒎𝒈 𝑮𝑴 𝒔𝒊𝒏𝜽.
51. The metacentric height GM is taken as positive if
M is above G and negative if below.
• If M is above G (positive GM), a restoring
moment is present and the original position
is stable.
• If M is below G (negative GM), the body is
unstable and will overturn if disturbed.
Thus, the metacentric height is a property of the
cross section for the given weight, and its value
gives an indication of the stability of a body.
52. The metacentric height of a vessel can be
determined if the angle of tilt θ caused by moving
a load 𝑷 a known distance 𝒅 across the deck is
measured.
𝒅
𝑷
Turning moment M due to movement of 𝑷 is
𝑴 = 𝑷𝒅
53. If GM is the metacentric height and 𝑾 = 𝒎𝐠 is
the total weight of the vessel including 𝑷, then
Righting moment = 𝑾 𝑮𝑴 𝜽
For equilibrium in the tilted position, the righting
moment must equal the overturning moment.
Therefore: 𝑾 × 𝑮𝑴 × 𝜽 = 𝑷 𝒅
Hence Metacentric height 𝑮𝑴 =
𝑷𝒅
𝑾𝜽
The true metacentric height is the value of GM as θ → 0.
54. It can be shown mathematically that for a vessel of known
shape and displacement, the position of the metacentre
relative to the centre of buoyancy is given by:
𝑩𝑴 =
𝑰
𝑽
where:
I = 2nd moment of area of waterline plane OO
V = volume of fluid displaced
BM is known as the Metacentric Radius.
