Chapter 3 static forces on surfaces [compatibility mode]

CHAPTER 3
Static Forces on Surfaces
Buoyancy
Dr Yunes Mogheir
CB
A
١

OBJECTIVES
1. Compute the hydrostatic pressures and forces on
submerged surfaces in a static fluid.
2. Use the principle of static equilibrium to solve for
the forces involves in buoyancy problems.
3. Define the condition for stability of submerged and
floating bodies.
٢

INTRODUCTION
§ When a surface is submerged in a Fluid, forces develop on the
surface due to the fluid.
§ The determination of these forces is important in the design of
storage tanks, ships, dams and other hydraulic structures.
From Last Lectures
§ For Fluid at rest, we knew that the force must be perpendicular
to the surface
§ We knew that the pressure will vary linearly with depth
٣

3.1 ACTION OF FLUID ON A SURFACE
¢ Pressure is defined as force per unit area. If a pressure p acts on
a small area δA then the force exerted on that area will be:
¢ Since the fluid is at rest the force will act at right-angles to the
surface.
ApF δ=
٤

GENERAL SUBMERGED PLANE
¢ Consider the plane surface shown in the figure below.
¢ The total area is made up of many elemental areas.
¢ The force on each elemental area is always normal to the surface but,
in general, each force is of different magnitude as the pressure
usually varies.
¢ We can find the total or resultant force, R, on the plane by summing
up all of the forces on the small elements i.e.
∑=+++= ApApApApR nn δδδδ L2211
If the surface is a plane the force can
be represented by one single
resultant force, acting at right-angles
to the plane through the centre of
pressure.
٥

CURVED SUBMERGED SURFACE
¢ If the surface is curved, each elemental force will be a
different magnitude and in different direction but still normal
to the surface of that element.
¢ The resultant force can be found by resolving all forces into
orthogonal co-ordinate directions to obtain its magnitude and
direction.
¢ This will always be less than the sum of the individual forces,
∑ Apδ
٦

3.2 RESULTANT FORCE AND CENTER OF PRESSURE
ON A PLANE SURFACE UNDER UNIFORM PRESSURE
For horizontal plane submerged in
a liquid, the pressure, p, will be
equal at all points of the surface.
Simplest Case: Tank bottom with
a uniform pressure distribution
hghp γρ ==
Now, the resultant Force:
pAR =
A = area of the plane surface (Tank
Bottom)
This force will act vertically
downward and the center of
pressure will be the centroid
of the surface
٧

3.3 RESULTANT FORCE AND CENTER OF PRESSURE
ON A PLANE SURFACE IMMERSED IN A LIQUID
§ Shown a plane surface PQ of an area
A submerged in a liquid of density, ρ,
and inclined at an angle φ to the free
surface.
§ Considering one side only, there will
be a force due to fluid pressure,
acting on each element of area δA,
§ The pressure at a distance y below
the free surface can be written as:
R
D
yy
gyp ρ=
§ Force on elemental area δA: δR = pδA = ρgyδA
§ The resultant force acting on the plane can be found by summing all the
forces on the small element:
R = ΣpδA = Σρgy δA ٨

R
D
yy
§ The quantity Σy δA is the first moment
of area under the surface PQ about the
free surface of the liquid
∑ = yAAyδ
R = ΣpδA = Σρgy δA
Assuming that ρ and g are constant
R = ρg Σ yδA
y
ygAR ρ==ForceResultant
where
§ A = the area of the whole immersed surface and
§ = vertical distance from the free surface to the centroid of
the area, G, of the immersed surface.
Remember: Centroid of the area is defined as the point at which the
area would be balanced if suspended from that point.
٩

R
D
yy
§ It may be noted that the resultant
force, R, is independent of the angle
of inclination φ as long as the depth of
the centroid is unchanged.
§ The point of application of the
resultant force on the submerged area
is called the center of pressure.
y
ygAR ρ=
§ This resultant force will act perpendicular to the immersed
surface at the center of pressure, C at some depth D below
the free surface.
١٠

§ The vertical depth of the center of pressure, D, below the free surface can be
found using the following:
or
where
§ IG = second moment of plane area about an axis through its centroid G
parallel to the free surface
§ A = the area of the whole immersed surface
§ ŷ = vertical distance from the free surface to the centroid of the area A
y
yA
I
D G
+





= φ2
sin
The above equation implies that the center of pressure is always
below the centroid.






=
yA
I
D o
φ2
sin
2
AdII Go +=
Remember:
Parallel axis theorem
y
R
D
yy
١١

THE SECOND MOMENT OF AREA ABOUT A LINE THROUGH THE CENTROID (IG) OF SOME
COMMON SHAPES.
١٢

OGaboutelementsallonforcestheofmomentsofSum=×dR
Lateral position of Centre of Pressure
• If the shape is symmetrical the centre of pressure lies on the line of symmetry.
• But if it is not symmetrical its position must be found by taking moments about
the line OG.
( )∑= xAgyδρ AygR ρ=But
yA
Ayx
d
∑=
δ
RDyR
D
yy
C
O
x
١٣

3.4 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:
p = ρgz
Consider the tank shown:
§ We can draw the shown diagram
(graphical representation of the
(gauge) pressure change with depth
on one of the vertical walls.)
§ It increases linearly from zero at the
surface by p = ρgy, to a maximum at
the base of P = ρgH.
§ This is know as a pressure diagram.
١٦

Pressure Diagram: A graphical interpretation of the forces due to a fluid acting on a
plane area. The “volume” of fluid acting on the wall is the pressure diagram and equals
the resultant force acting on the wall.
( )( )( )bhghFR ρ
2
1
=
Volume
( )bghFR
2
2
1
ρ=
Resultant Force:
Location of the Resultant Force, C:
The location is at the centroid of the volume of the pressure
diagram.
Center of Pressure:






3
2
,
2
hb
O
١٧

CB
A
¢ The area of this triangle represents the
resultant force per unit width on the vertical
wall, (N/m). So:
2
2
1
2
1
2
1
gHgHHBCABArea ρρ ==×=
Resultant force per unit width
(N/m)
2
1 2
gHR ρ=
§ This force acts through the centroid of the pressure diagram. For a triangle,
the centroid is located at 2/3 its height, thus the resultant force acts at a depth
of 2/3 H from the surface.
§ The total resultant force can be obtained by multiplying the above
equation with the width of the surface, B.
(N)
2
1 2
BgHR ρ=
١٨

THIS CAN BE CHECKED AGAINST THE PREVIOUS METHOD:
ygAR ρ=
CB
A
2
2
1
2
)1( gH
H
HgR ρρ =×= R
y
yA
I
D
g
+





= φ2
sin





=
yA
I
D o
φ2
sin or
φ = 90o, sin φ =1, y = H/2, Ig =H3/12
3
2
6
4
262)2/()1(
12/3
HHHHH
HH
H
D ==+=+





××
= R
١٩

¢ If the plane surface is inclined and submerged below the surface, the
pressure diagram is drawn perpendicular to the immersed surface
and will be straight line extending from p=0 at the free surface to
p=ρgh2 at depthh2 .
¢ 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 shaded area.
Note
¢ More complex pressure diagrams can be draw for non-rectangular or
non-vertical planes but it is usually far easier to use the moments
method.
٢٠

¢ The same pressure diagram technique can be used when
combinations of liquids are held in tanks (e.g. oil floating on water)
with position of action found by taking moments of the individual
resultant forces for each fluid.
For example:
¢ Find the position and magnitude of the resultant force on this vertical
wall of a tank which has oil floating on water as shown.
٢١

EXAMPLES
A 6-m deep tank contains 4 m of water and 2-m of oil as shown
in the diagram below.
§ Determine the pressure at point A and at the bottom of the tank.
§ Draw the pressure diagram.
A
oil
water
2 m
4 m ρwater = 1000 kg/m3
SG of oil = 0.98
٢٢

Solution:
Pressure at oil water interface (PA)
PA = Patm + Poil (due to 2 m of oil)
= 0 + ρoilghoil = 0 + 0.98 x 1000 x 9.81 x 2
= 15696 Pa
PA = 15.7 kPa (gauge)
Pressure at the bottom of the tank;
PB = PA + ρwaterghwater
PB = 15.7x1000 + 1000 x 9.81 x 4
= 54940 Pa
PB = 54.9 kPa (gauge)
Patm = 0
4 m
2 m
PA
PA=15.7 kPa
B
A
oil
water
PB = 54.9 kPA
Pressure Diagram
٢٣

EXAMPLES 3.3 PAGE 70
Air
p=35 kN/m2
Water
h =1.2 m
1.8 m
The length of the wall of tank is 3m
Determine the resultant force on this wall
and the height of the center of pressure
above the base
٢٤

3.5 FORCE ON A CURVED SURFACE DUE TO
HYDROSTATIC PRESSURE
Q Many surfaces in dams, pumps, pipes or tanks are curved
Q General theory of plane surfaces does not apply to curved
surfaces
Q As stated above, if the surface is curved the forces on each
element of the surface will not be parallel and must be
combined using some vectorial method.
Q It is most straightforward to calculate the horizontal and
vertical components.
Q Then combine these forces to obtain the resultant force and its
direction.
٢٥

¢ In the diagram below liquid is
resting on top of a curved
base.
¢ The element of fluid ABC is
equilibrium (as the fluid is at
rest).
Consider the Horizontal forces
¢ The sum of the horizontal
forces is zero.
¢ FAC, must be equal and
opposite to RH
The resultant horizontal force of a fluid above a curved surface
is:
RH = Resultant force on the projection of the curved surface onto
a vertical plane.
٢٦

We know that:
¢ The force on a vertical plane must act
horizontally (as it acts normal to the
plane).
¢ RH must act through the same point.
Note:
RH acts horizontally through the centre of pressure of the
projection of the curved surface onto a vertical plane.
We can use the pressure diagram method to calculate the position and
magnitude of the resultant horizontal force on a curved surface.
So we can say
٢٧

Consider the Vertical forces
¢ The sum of the vertical forces
is zero.
¢ There are no shear force on the
vertical edges, so the vertical
component can only be due to
the weight of the fluid.
The resultant vertical force of a fluid above a curved
surface is:
RV = Weight of fluid directly above the curved surface.
It will act vertically downward through the centre of
gravity of the mass of fluid.
٢٨

Resultant force
¢ The overall resultant force is found by combining the vertical
and horizontal components vectorialy:
¢ And acts through O at an angle of υ.
¢ The angle the resultant force makes to the horizontal is:
¢ The position of O is the point of intersection of the horizontal
line of action of RH and the vertical line of action of RV .
22
VH RRR +=






= −
H
V
R
R1
tanθ
٢٩

Consider the Horizontal forces
¢ There are two horizontal forces on
the element which is in equilibrium:
¢ The horizontal reaction force RH
¢ The pressure force on the vertical
plane FB.
The resultant horizontal force of a fluid below a curved surface is:
RH = Resultant force on the projection of the curved surface onto a
vertical plane.
What are the forces if the fluid
is below the curved surface?
٣٠

Consider the Vertical forces
¢ If the curved surface were removed and
the area it were replaced by the fluid, the
whole system would be in equilibrium.
¢ Thus the force required by the curved
surface to maintain equilibrium is equal
to that force which the fluid above the
surface would exert - i.e. the weight of
the fluid.
The resultant vertical force of a fluid below a curved
surface is:
RV = Weight of the imaginary volume of fluid
vertically above the curved surface.
What vertical force
would keep this in
equilibrium?
٣١

EXAMPLES
Find the magnitude and
direction of the resultant
force of water on a quadrant
gate as shown below.
٣٢

EXAMPLES
A 2m long cylinder lies as shown in the figure
(diameter = 2m also) , holding back oil of relative
density 0.8. If the cylinder has a mass of 3250 kg.
A
B
Find:
• the reaction at A
• the reaction at B
٣٣

3.6 BUOYANCY
¢ When a body is submerged or floating in a static fluid, the resultant force
exerted on it by the fluid is called the buoyancy force.
¢ Buoyant force can be defined as the resultant fluid force which acts on a
fully submerged or floating body.
A
B
D
E
C
F H
G
F
V
V
F
• Consider the vertical plane VV
• The projected area of each of the two
sides on this plane will be equal.
• The horizontal forces F will be equal
and opposite.
No resultant horizontal force on the
body due to the pressure of the
surrounding fluid ٣٦

In the vertical direction:
¢ A force is exerted by the fluid on the immersed body called the buoyancy or
upthrust force.
A
B
D
E
C
F H
G
• This force will act vertically upward through the
centroid of the volume of fluid displaced, known
as the centre of buoyancy
Upthrust on the body = Weight of fluid displaced
by the body
• Upthrust on the body = Upward force on lower surface ADEC
- Downward force on the upper surface ABCD
= Weight of volume of fluid AECDGFH
- Weight of volume of fluid ABCDGFH
= Weight of volume of fluid ABCDE
Archimedes’ principle ٣٧

Buoyancy: Archimedes’ Principle
Archimedes (287-212 BC)
Archimedes’ Principle states that the
buoyant force has a magnitude equal to
the weight of the fluid displaced by the
body and is directed vertically upward.
BodybyDisplacedFluidb gVFR ρ==
٣٨

If a body is immersed in two different fluid as shown
111 gVR ρ=
Fluid of
density ρ1
Fluid of
density ρ2
G1
V2
V1
G2
Acting through G1, the centroid of V1
222 gVR ρ=
• Upthrust on lower part:
• Upthrust on upper part:
Acting through G2, the centroid of V2
221121 gVgVRRR ρρ +=+=
• Total Upthrust on the body:
٣٩

EXAMPLE 3.5 PAGE 75 (BOOK)
A rectangular pontoon has a width B of 6m, a length L of 12m,
and a draught D of 1.5m in fresh water (ρ =1000 kg/m3).
Calculate:
a) The weight of the pontoon.
b) Its draught in sea water (ρ =1025 kg/m3)
c) The load that can be supported by the pontoon in fresh water if
the maximum draught permissible is 2 m
٤٠

3.7 EQUILIBRIUM OF FLOATING BODIES
1. The upthrust force R acting through the
center of buoyancy B.
2. The weight of the body W acting through
its center of gravity G.
For equilibrium:
1. R and W must be equal.
2. R and W must act in the same straight line.
RW =
W = mg
R
G
B
V = volume
displaced
When the body (Ship) floats in vertical equilibrium in a liquid,
the forces presented are:
Vgmg ρ= ρ
m
V =
٤١

Four solid objects are floating in corn syrup.
Rank the objects according to their density, greatest first.
٤٢

¢ Stable
¢ Unstable
¢ Neutral
¢ Depending upon whether, when given a small displacement, it
tends to return to the equilibrium position, move further from it
or remain in the displaced position.
The equilibrium of a body may be:
For floatingfloating body (such as ship) stability is of major
importance.
٤٣

3.8 STABILITY OF A SUBMERGED BODY
Stable Equilibrium: if when displaced returns to equilibrium position.
Unstable Equilibrium: if when displaced it returns to a new equilibrium position.
Remember
• The upthrust force R acting through the center of buoyancy B.
• The weight of the body W acting through its center of gravity of the body G.
Note: as the body is totally immersed, the shape of the
displaced fluid is not altered when the body is tilted and so
the centre of buoyancy remains unchanged relative to the
body.
Whatever the orientation of the body, these two points will
remain in the same positions relative to the body.
٤٤

R
W
G
B
R
W
G
B
υ
It can be seen that: A small angular displacement υ from the
equilibrium position will generate a moment W.BG.υ.
If G is below B, this will generate righting moment and
the body will tend to return to its equilibrium position
Stable
٤٥

R
W
G
B
R
W
G
B
υ
If G is above B, this will be a overturning moment
and the body is unstable
Unstable
٤٦

3.9 STABILITY OF FLOATING BODIES
¢ It is slightly more complicated as the location of the center of
buoyancy can change:
¢ The volume of liquid remains unchanged since R=W, but the shape of
this volume changes and therefore its centre of buoyancy.
¢ When the is displaced through an angle υ the center of buoyancy move
from B to B1 and a turning moment is produced.
¢ Let M (metacentre) is the point at which the line of action of the
upthrust R cuts the original vertical line through G, then:
¢ GM known as the metacentric height
θ××= GMWgeneratedMoment
٤٨

¢ If M lies above G, a righting moment is produced, equilibrium is stable and
GM is regarded as positive.
¢ If M lies below G, an overturning moment is produced, equilibrium is
unstable and GM is regarded as negative.
¢ If M coincides with G, the body is in neutral equilibrium
It can be seen that:
W = mg
R = W
W = mg
B
G
R = W
W = mg
B1
G
M
υ
M
x
Stable
R = W
W = mg
B
G
UnstableUnstable
R = W
W = mg
B1
G
Mυ
x ٤٩

3.10 DETERMINATION OF METACENTRIC HEIGHT
¢ The metacentric height of a vessel can be measured if the angle of tilt
υ caused by moving a load P a known distance x across the deck is
measured.
PxGMW =×× θ x
P
θW
Px
GM =
Where:
W: is the weight of the vessel including P
The overturning moment due to movement of load P = Righting moment
٥١

3.11 Determination of the Position of the Metacentre
Relative to the Centre of Buoyancy
When a vessel is tilted trough a small angle υ then:
• The centre of buoyancy B will move to B’.
• The total weight displaced remains unchanged.
The distanced BB’ can be calculated as follows:
V
I
BB
θ
='
V
IBB
BM ==
θ
'
Where:
υ : tilted angle of the vessel
I : second moment of area of the waterline plane
about OO
V : Volume of liquid displaced
The metacentric radius BM :
٥٢

EXAMPLE 3.6 PAGE 80 (BOOK)
A cylindrical buoy 1.8-m in
diameter, 1.2-m high and
weighting 10-kN floats in salt
water of density 1025-kg/m3.
Its center of gravity is 0.45-m
from the bottom. If a load of
2-kN is placed on the top.
Find the maximum height of the center of gravity of this load
above the bottom if the buoy is to remain in stable equilibrium
٥٣

EXAMPLE
(a) Magnitude of the hydrostatic
force acting on the door and
its centre of pressure,
(b) The moment required by the
shaft to open the door.
§ Use ρ water = 998 kg/m3 and
g = 9.81 m/s2.
§ A circular door having a diameter of 4 m is positioned at the inclined wall
as shown in Fig. which forms part of a large water tank. The door is
mounted on a shaft which acts to close the door by rotating it and the door
is restrained by a stopper. If the depth of the water is 10 m at the level of
the shaft, Calculate:
٥٤

§(a) The magnitude of the hydrostatic force FR is
FR = ρgyA
= (998)(9.81)(10) [ ¼ π(4)2]
= 1.230 x 106 N
= 1.23 MN
§For the coordinate system shown in Figure E2.3(b), since circle is a symmetrical
shape, Ixy = 0, then xR = 0. For y coordinate,
yR = 1xx + yC = ¼ πR4 + yC
yC A π yCR2
= ¼ π (2)4 + 10
π (10/sin 60°)(2)2 sin 60°
= 11.6 m
§Use moment equilibrium ΣM - 0 about the shaft axis. With reference to Figure
E2.3(b), the moment M required to open the door is:
M = FR ( yR - yC )
= (1.230 x 105) (0.0866)
= 1.065 x 105 N • m
= 107 kM • m
٥٥

EXAMPLE
¢ The 2-m diameter cylinder shown is 5m long
into the paper and rests in static equilibrium
against smooth wall at point B.
¢ Compute the weight of the cylinder assume zero
wall friction at point B
٥٦

EXAMPLE
¢ A gate 5m wide is hinged at
point B and rests against a
smooth wall at point A,
Compute:
¢ The force on the gate due to
water
¢ The horizontal force P
exerted by the wall at point
A
15 m
8 m
8 m
٥٧

EXAMPLE
¢ The gate AB is 4.5 m long and 3 m wide into the
paper and hinged at B with a stop at A.
Neglecting the weight of the gate, compute the
water level h for which the gate will start to fall
44.5 kN
4.5 m
٥٨

EXAMPLE
¢ Gate AB is 5 m into the
paper and weights 30 kN
when submerged.
¢ It is hinged at B and
rests against a smooth
wall at A.
¢ determine the water
level h at the left which
will just cause the gate
to open
m
m
m
٥٩

EXAMPLE
¢ A tank has right triangular
gate near the bottom,
Compute:
¢ The hydrostatic force on the
gate
¢ the center of pressure on the
gate
٦٠

kNF
Ay
ApF
6.1569
)4*4(*)1*10*9810(
)sin(
=
=
=
=
αγ
Find: Force of block on gate
m
Ay
I
yycp
133.0
)4*4*10(
12/4*4 3
=
=
=−
F
kNF
kN
FF
FF
M
gb
gwgb
gbgw
378.104
6.1569
2
133.0
2
133.0
2133.0
0
,
,,
,,
=
=
=
−=
=∑
Fw,g
Fb,g
0.133
2
Example
٦٢

m
Ay
I
yycp
4641.0
)24*464.6(
12/6*4 3
=
==−
N
AyApF
000,318,1
)6*4(*)30cos33(*9810
)sin(
=
+=
== αγ
kNR
kN
FR
FR
M
A
A
A
05.557
1318)42265.0(
6
4641.03
)4641.03(6
0
=
=
−
=
−−=
=∑
F
RA
3-0.4641
6
Example
٦٣

Chapter 3 static forces on surfaces [compatibility mode]

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