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Academic Note.Pressure on Submerged Body.Center of pressure. Buyoyancy and flotation.
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1
Hydrostatic Forces and stability
The Joola after capsizing (Senegal 2003)
2
Hydrostatics Force
h
γ
p = 0
p = 0
p = h
h
p = 0
p = 0
p = hγ
• The pressure on the bottom is uniform so the
resultant force acts through the centroid.
• The pressure of the sides increases with
decreasing depth. The force will not act through
the centroid of the surface.
• The centroid is the geometric mean position
center of the surface. The line of action (center
of pressure) weights the area integral by the
force applied through that area.
3
Definition of the centroid
The centroid gives a
definition of the mean
position of an area (vol-
ume). It is closely re-
lated to the center of
mass a body.
A
x
y
C
dx
dy
y
yc
xc
dAi
i
xi
One adds up position of x for all the little pieces
dAi of the Area, A to get average x position, xc .
The x and y -coordinates of the centroid are
evaluated mathematically as
xc =
∑
i xi dAi∑
i dAi=
∫∫
Ax dx dy
A
yc =
∑
i yi dAi∑
i dAi=
∫∫
Ay dx dy
A
4
First moment of Area
The 1st moments of ar-
eas are the average dis-
placement of an area
about an axis of rota-
tion. They are closely
related to the centroid.
A
x
y
C
dx
dy
y
yc
xc
dAi
i
xi
The first moment of area about the y -axis is
Qy =∑
i
xidAi =
∫∫
A
x dx dy
Qy =
∫∫
A
(x − xc) dx dy + xc
∫∫
A
dx dy
Qy = 0 + xc
∫∫
A
dx dy = xcA
The first moment of area about the x -axis is
Qx =
∫∫
A
y dx dy = ycA
The first moments of area have units of m3 .
5
Second moment of Area
The 2nd moments of
areas are the average
(displacement)2 of an
area about an axis of ro-
tation. Has units of m4
A
x
y
C
dx
dy
y
yc
xc
dAi
i
xi
The second moment of area about the x -axis is
Ix =∑
i
y2i dAi =
∫∫
A
y2dx dy
It is sometimes called the moment of inertia of the
area. The second moment of inertia is always
positive since y2 > 0 .
The second moment of area about the y -axis is
Iy =∑
i
x2i dAi =
∫∫
A
x2dx dy
The product of inertia about an xy coordinate axes
Ixy =∑
i
xiyidAi =
∫∫
A
xydx dy
6
Parallel axis theorem
Working out the sec-
ond moments would be
troublesome as the axes
of rotations moved but
for the parallel axes
theorem. The mo-
ments of many objects
through their centroids
are known.
A
x
y
C
dx
dy
y
yc
xc
dAi
i
xi
The second moment of areas are
Ix = Ixc + y2cA
Iy = Iyc + x2cA
Ixy = Ixyc + xcycA
One writes down second moment through centroid,
then determines distance of centroid to axis of
rotation, and finally apply the parallel axis theorem.
7
Submerged inclined plane
Want to work out forces on inclined surface.
• The x -axis points out of page.
• The distance down incline is y . Depth is h .
• p = γh (gauge pressure).
8
Inclined plane
To determine net
force, need to add up
all contributions over
each small piece of the
area.
dF = p dA = γh dA one piece of A
FR =
∫∫
A
γh dA adding up
FR =
∫∫
A
γy sin θ dA h = y sin θ
FR = γ sin θ
∫∫
A
y dA if γ and θ constant
Now the integral over y is the first moment of Area,∫∫
A
y dA = ycA
9
Inclined plane
∫∫
A
y dA = ycA
Since F = γ sin θ∫∫
Ay dA
FR = γAyc sin θ = γhcA hc = yc sin θ
The net force on the plane depends on the depth of
the plate centroid below the surface. The net force is
the area multiplied by the pressure at the centroid.
10
Center of pressure
To determine the cen-
ter of pressure, add up
all contributions to the
force over each small
piece of the area, but
multiplied by y .
The center of pressure is essentially a weighted
average.
〈y〉 =
∑
i yiδFi∑
i δFi
yR =
∫∫
Ay dF
∫∫
AdF
yR =
∫∫
Ayp dA
∫∫
Ap dA
yR =
∫∫
Ayγy sin θ dA
∫∫
γ sin θy dA
yR =
∫∫
Ay2 dA
∫∫
y dA=
∫∫
Ay2dA
ycA
11
Center of pressure
yR =
∫∫
Ay2dA
ycA
The y position of
the force is the 2nd
moment of the area
with respect to the x
axis. This is essen-
tially the moment of
inertia about the x -
axis.
yR =Ix
Ayc
Parallel axis theorem Ix = Ixc + Ay2c .
yR =Ixc + Ay2
c
Ayc=
Ixc
Ayc+ yc
The resultant force FR always passes below the
centroid since yR > yc .
12
Center of Pressure, x
xR =
∫∫
AxydA
ycA
The mean x-position
of the force can be
determined by similar
technique. This is just
the product of inertia
for the coordinate sys-
tem.
xR =
∫∫
AxydA
ycA=
Ixy
ycA
xR =Ixyc + xcycA
ycA
xR =Ixyc
ycA+ xc
13
Geometric properties for shapes
Rectangle
A = ba
Ixc = ba3/12
Iyc = ab3/12
Ixyc = 0
x
y
c
a/2
b/2 b/2
a/2
Circle
A = πR2
Ixc = Iyc = πR4/4
Ixyc = 0
x
y
R
c
Half-circle
A = πR2/2
Ixc = 0.1098R4
Iyc = 0.3927R4
Ixyc = 0
xc
R R
y 4R3π
14
More shapes
Triangle
A = ba/2
Ixc = ba3/36
Ixyc = ba2(b − 2d)/72
c
y
a
d
x
(b+d)/3
a/3
b
Quarter Circle
A = πR2/4
Ixc = Iyc = 0.05488R4
Ixyc = −0.01647R4
x
yR
3π4R
c
Ixyc is only non-zero if the shape does not have a
bilateral symmetry.
15
Worked example
Determine the re-
sultant force on
the plate and the
reaction at the step. Inlet
3.0 m
step
1.0 m
1.2 m
ρ = 1000 kg/m3
Draw FBD for plate.
Ignore the weight force.
W = 0 , Hy = 0
W
H
H
rF
x
y
stepF
16
Pipe-inlet
Determine resultant force
Fr = ρghcA
A =1
2ab =
1
21.2 × 1.0 = 0.60 m2
hc = 3.0 +a
3= 3.0 +
1
31.0 = 3.333 m
Fr = 1000 × 9.810 × 0.600 × (3 + 0.333) = 19.6 kN
Ixc =1.2 × 1.03
36= 0.0333 m4
Now for center of pressure
yr =Ixc
Ayc+ yc
yr =0.0333
0.60 × 3.333+ 3.333
yr = 3.349 m
17
Pipe-inlet: Step reaction
Net torque about hinge
must be zero.
W
H
Hx
y
stepF=19.6 kN
= 0
= 0
0.349 m
Fr
Fstep × 1.0 = Fr(0.349)
Fstep = 19.6 × 103(0.349)
Fstep = 6.85 kN
The force on the hinge determined from
Fr − Fstep + Hx = 0 .
Hx = Fstep − Fr = 6.85 − 19.6 = −12.7kN
The force on the hinge acts to the left (opposes Fr )
18
Pressure Prism
This is a intuitive recipe for determining the force on
submerged surfaces. Useful for surfaces that are
rectangular in shape.
• Gauge pressure is zero at top and γh at bottom.
• Pressure variation with h is linear.
• Average pressure 〈p〉 = γh/2
• Resultant force Fr = 〈p〉A = γ(hA/2)
• Volume of pressure prism (= γhA/2) .
• The center of pressure passes through the
centroid of the pressure prism.
19
Pressure Prism
The pressure prism can be regarded as arising from 2
parts. Let w be width of surface into page. Force
due to rectangle ABDE .
F1 = (γh1)(h2 − h1)w
Force due to triangle BCD.
F2 =
(
1
2γ(h2 − h1)
)
(h2 − h1)w
The resultant force is FR = F1 + F2
20
Pressure Prism
Determination of Center of Pressure done from
moments of the forces.
FRyR = F1y1 + F2y2
⇒ yR =F1y1 + F2y2
FR
Moment of rectangular part about AB level is 1/2
distance apex to base, i.e. y1 = 12(h2 − h1) .
Moment of triangular part is 2/3 distance apex to
base, i.e. y2 = 23(h2 − h1) .
21
The tank problem
Want to determine
the force on the cover
plate.
Will also determine
the center of pressure.
Air
gp =50 kPa
Water 2.0 m
0.6 m
0.6
Useful hint: While the air pressure inside the tank
needs to be taken into consideration, the impact of
atmospheric pressure can be ignored (the tank
pressure is a gauge pressure).
22
Tank Problem, pressure prism
airp
airp
pwater
γ
x 2.0γ
2.0 m
0.6 m
ABC
x0.6
The over-pressure due to air in the tank, 50 kPa is
constant with depth. Pressures and forces due to
(A+B) and C are
FA+B = (50, 000 + 2.0 × 9, 800) × 0.62 = 25, 060 N
FC = ( 1
20.6 × 9, 800) × 0.62 = 1060 N
Total force is FR = 26, 100 N . The center of
pressure is
yR =25060 × 0.3 + 1060 × 0.400
26100= 0.304 m
23
Force on curved surface
Curved surfaces occur in many
structures, e.g. dams and cross
sections of circular pipes.
The loads on the surface are all due
to pressure forces. Look at forces
acting on wedge of water ABC.
Weight force W due to
weight of wedge of water.
Pressure forces F1, F2 due
to water above and from
left.
Reaction Forces FH , FV due
to wall of tank.
24
Curved surface: Free body diagram
The weight force W passes
through the center of gravity
of the wedge.
For static equilibrium,
F1 + W = FV
F2 = FH
Also F2 is co-linear with FH and FV is co-linear
with the resultant of F1 + W .
25
Curved Surface, example
Determine the resultant force on the curved part of
the base and also determine its line of action.
The bottom corner of
the tank is a circle of
radius 2.0 m .
The tank length (out of
page) is 8.0 m .
2.0 m
2.0 m
2.0 m
The centroid of the
quarter circle wedge is
xc =4R
3πxc = 0.84883 m
It is 0.84883 m from
the left boundary of
quarter circle.
2.0 m
2.0 m
2.0 m
F
W
1
26
Curved Surface, Vertical
2.0 m
2.0 m
2.0 m
F
W
1
F1 = γ2.0(2.0 × 8.0)
F1 = 313.6 kN
W = γ1
4π(2.0)2 × 8.0
W = 246.3 kN
The total vertical force is
246.3+313.6 = 559.9 kN
F1 line of action 1.0 m from wall.
W line of action 2.0− 0.84883 = 1.151 m from wall.
Line of action for F1 and W .
xR =313.6 × 1.0 + 246.3 × 1.151
559.9xR = 1.066 m
27
Curved Surface, Horizontal
2.0 m
2.0 m
2.0 m
Rectangle
F� = γ2.0(2.0 × 8.0)
F� = 313.6 kN
F△ = γ1
22.0(2.0 × 8.0)
F△ = 156.8 kN
The net horizontal force is 313.6 + 156.8 = 470.4 kN
F� line of action 1.0 m below 2.0 m line.
F△ line of action 1.33 m below 2.0 m line.
Line of action for horizontal force.
yR =313.6 × 1.0 + 156.8 × 1.333
470.4yR = 1.11 m
28
Curved Surface, Summary
2.0 m
2.0 m
2.0 m1.11 m
1.06 m559.9 kN
470.4 kN
731.3 kN
The net force is
FR =√
559.92 + 470.42 = 731.3 kN
29
Buoyancy: Archimedes principle
When a body is wholly or partially immersed in a
fluid there is an upward buoyancy force equal to the
weight force of the fluid displaced by the body.
hA
p
p pxx
1
2p = p + h
1γ
Consider pressure forces on a rectangular slab
Fnet:pressure = p2A − p1A
Fnet:pressure = (p1 + γh)A − p1A
FB = γhA = γV
The buoyancy force arises as a result of higher
pressure on the bottom surface.
30
Archimedes principle: Proof
Any body can be decomposed into a number of very
small slabs. The buoyancy force on each slab is just
δFi = γδVi .
iδV
Therefore the proof for a rectangular slab can be
generalized to a body of arbitrary shape.
FB =∑
i
δFi = γV.
Buoyancy force does not depend on the density of
the submerged object. The buoyancy force only
depends on the density of the fluid and the
volume(shape) of the submerged object.
31
Archimedes principle
The Buoyancy force of a submerged body passes
through its centroid. Called the center of buoyancy.
W = mg
BF = Vγ
CGC
The buoyancy force for a partially submerged object
passes through the centroid of the displaced volume,
V ′ .
W = mg
CGC
F = V’γB
The weight force passes through the center of gravity
and does not always pass through the centroid.
32
Stability
The stability of a body depend on what happens
when it is displaced from the equilibrium position.
NeutralStable Unstable
• The equilibrium is stable if the forces acting on
the object act to return it to its equilibrium
position.
• The equilibrium is unstable if the forces acting
on the object act to send it away from its
equilibrium position.
• The equilibrium is neutral if there are no net
forces acting on the object to return it or
remove it from equilibrium.
33
Buoyant stability: submerged
The buoyant force acts through the centroid of the
object. The gravitational weight force acts through
the center of gravity.
• A completely submerged body is stable if the
center of gravity lies below the centroid.
• A completely submerged body is unstable if the
center of gravity above the centroid.
34
Buoyant stability: floating
• Even though the center of gravity lies above the
centroid the resultant torques is a restoring
torque.
• The centroid (of the displaced volume) can shift
as the body has an angular displacement. It is
the movement of the centroid to the right that
gives this body its stability.
35
Buoyant stability: floating
• In this case the body is unstable.
• The buoyancy force and gravitational force act
to create an overturning torque.
36
Buoyant stability: Technical
Consider a floating body given a small angular
deflection. The magnitude of the buoyancy force will
stay the same, (weight force does not change) but
the location of the centre of buoyancy changes.
θdΒ1
G
W
θd
FB
Β2
O
M
O = Waterline point about which boat rolls.
B1 = original buoyancy point.
B2 = buoyancy point after displacement.
The line of action of the original buoyancy force
(through the center of gravity) and new buoyancy
force intersect at the metacenter.
37
Buoyant stability: Terminology
Metacenter The metacenter M is the point of
intersection between the original line of action
and new line of action of the buoyancy force
G The center of gravity moves as mass is added or
removed, so the metacentric height d(GM)
changes as mass is added or removed.
d(GM) The metacentric height d(GM) is the
displacement of M from G . Negative if M
below G
dθ The angular displacement should be small, e.g.
less than 20o .
Positive Stability The ship is stable if d(GM) > 0
, i.e. the metacenter lies above the center of
gravity.
Negative Stability The ship is unstable if
d(GM) < 0 , i.e. the metacenter lies below the
center of gravity.
Neutral stability Occurs when GM = 0 .
38
The stability conditions
ΒΒ
G
M
FB
W1 2
d(GM) > 0
The torque applied to
the vessel is a restoring
torque
Β
G
1 Β2
FB
W
Md(GM) < 0
The torque applied to
the vessel is an over-
turning torque
39
Pitch and Roll
W
GΒ1θ
PITCH AXIS
y
x
M
B
dxdy
dzΒ2
ROLL AXIS
ROLL AXIS O
F
d(B1M) =Iy
V
V = displaced volume
Pitch and roll refer to the rotation about long and
narrow axes of the vessel.
40
Buoyant Stability Recipe
• One determines position of center of gravity
• Determine water level
• One determines center of buoyancy of displaced
volume
• Then d(BG) is known.
• Finally, d(BM) is evaluated with
d(B1M) = BM =Iy
V
• Stable if d(BM) − d(BG) > 0
• Unstable if d(BM) − d(BG) < 0
41
Stability example 1:Roll
A wooden pine block, spe-
cific gravity = 0.500 .
(out of page dimension =
2.0 m )
Is it stable?
1.2 m
1.0 m
y
Determine water-level.
FB = FG
ρwaterVsubmergedg = mblockg = ρblockVblockg
ρwater×1.0×y×2.0 = 0.50×ρwater×1.0×1.2×2.0
y = 0.50×1.2 = 0.60 m
Center of gravity is 0.60 m above base.
Center of buoyancy is 0.30 m above base.
42
Stability example 1: continued
d(KB) = 0.30 m
d(KG) = 0.60 m1.2 m
1.0 m
0.60 m
MG
K
B
The Metacentric height for roll is
d(BM) =Iy
V=
112
1.032.0
1.0 × 0.6 × 2.0
=1.0
12 × 0.60= 0.1389 m
The metacentric height is 0.3000 + 0.1389 = 0.4389
m above the keel.
Since M is below G the block is not stable.
43
Stability example 1: Pitch
d(KB) = 0.30 m
d(KG) = 0.60 m
Out of page length
= 1.00 m
1.2 m
0.60 m
2.0 m
MG
K
B
The Metacentric height for pitch is
d(BM) =Iy
V=
112
2.031.0
1.0 × 0.6 × 2.0
=4.0
12 × 0.60= 0.5556 m
The metacentric height is 0.3000 + 0.5556 = 0.8556
m above the keel.
Since M is above G the block is stable in pitch.
44
Comments about stability
What can be done to im-
prove stability?
1.2 m
1.0 m
0.60 m
GM
K
B
Stability is affected by position of M, B and G .
d(BM) =Iy
V
• Increase width. Makes Iy larger. (Catamaran).
• Lower position of G . Add ballast (near the
bottom). This also tends to raise B from the
keel, and raises the position of M .