1
Hydrostatic Forces and Buoyancy
The Joola after capsizing (Senegal 2003)
2
Hydrostatics Force
p=0
p=0
h
h
p = γh
p =γ h
p=0
p=0
• 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 (volume). It is closely related to the center of
mass of a body.
y
dAi
xi
A
dx
dy
xc
C
yc
yi
x
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
RR
P
i xi dAi
A x dx dy
P
xc =
=
A
i dAi
RR
P
y dx dy
yi dAi
i
A
P
yc =
=
A
dA
i
i
4
First moment of Area
The 1st
moments of
areas are the average
displacement of an area
about an axis of rotation. They are closely
related to the centroid.
y
dAi
xi
A
dx
dy
xc
C
yc
yi
x
The first moment of area about the y -axis is
ZZ
X
Qy =
xi dAi =
x dx dy
A
i
Qy =
Qy
ZZ
(x − xc ) dx dy + xc
A
ZZ
= 0 + xc
dx dy = xc A
ZZ
dx dy
A
A
The first moment of area about the x -axis is
ZZ
Qx =
y dx dy = yc A
A
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 rotation. Has units of m4
y
dAi
xi
A
dx
dy
xc
C
yc
yi
x
The second moment of area about the x -axis is
ZZ
X
Ix =
yi2 dAi =
y 2 dx dy
i
A
It is sometimes called the moment of inertia of the
area. The second moment of inertia is always
positive since y 2 > 0 .
The second moment of area about the y -axis is
ZZ
X
x2i dAi =
x2 dx dy
Iy =
i
A
The product of inertia about an xy coordinate axes
ZZ
X
xi yi dAi =
Ixy =
xydx dy
i
A
6
Parallel axis theorem
Working out the second moments would be
y dAi dx
troublesome as the axes
xi
of rotations moved but
xc
for the parallel axes
theorem.
The moyi
ments of many objects
through their centroids
are known.
The second moment of areas are
A
dy
C
yc
x
Ix = Ixc + yc2 A
Iy = Iyc + x2c A
Ixy = Ixyc + xc yc A
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
FR
FR
FR
= p dA = γh dA one piece of A
ZZ
=
γh dA adding up
Z ZA
=
γy sin θ dA h = y sin θ
A
ZZ
= γ sin θ
y dA if γ and θ constant
A
Now the integral over y is the first moment of Area,
ZZ
y dA = yc A
A
9
Inclined plane
ZZ
y dA = yc A
A
Since F = γ sin θ
RR
Ay
dA
FR = γAyc sin θ = γhc A
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 center 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.
P
yi δFi
i
hyi = P
i δFi
RR
A y dF
RR
yR =
A dF
RR
A yp dA
RR
yR =
A p dA
RR
yγy sin θ dA
A
RR
yR =
γ sin θy dA
RR 2
RR 2
y dA
y dA
A
A
yR = RR
=
yc A
y dA
11
Center of pressure
yR =
RR
2 dA
y
A
yc A
The y
position of
the force is the 2nd
moment of the area
with respect to the x
axis. This is essentially the moment of
inertia about the x axis.
Ix
yR =
Ayc
Parallel axis theorem Ix = Ixc + Ayc2 .
Ixc
Ixc + Ayc2
=
+ yc
yR =
Ayc
Ayc
The resultant force FR always passes below the
centroid since yR > yc .
12
Center of Pressure, x
xR =
RR
A xydA
yc A
The mean x -position
of the force can be
determined by similar
technique. This is just
the product of inertia
for the coordinate system.
xR =
xR =
xR =
RR
A xydA
Ixy
=
yc A
yc A
Ixyc + xc yc A
yc A
Ixyc
+ xc
yc A
13
Geometric properties for shapes
Rectangle
A = ba
3
a/2
c
Ixc = ba /12
Iyc = ab3 /12
x
a/2
y
Ixyc = 0
b/2
b/2
Circle
A = πR2
R
Ixc = Iyc = πR4 /4
x
c
y
Ixyc = 0
Half-circle
A = πR2 /2
Ixc = 0.1098R
Iyc = 0.3927R4
Ixyc = 0
c
4
y
R
x
R
4R
3π
14
More shapes
Triangle
d
A = ba/2
a
c
3
Ixc = ba /36
y
(b+d)/3
b
Ixyc = ba2 (b − 2d)/72
a/3
4R
3π
Quarter Circle
A = πR2 /4
c
Ixc = Iyc = 0.05488R4
Ixyc = −0.01647R4
x
R
x
y
Ixyc is only non-zero if the shape does not have a
bilateral symmetry.
15
Worked example
Determine the resultant force on
the plate and the
reaction at the step.
3.0 m
1.2 m
Inlet
1.0 m
step
ρ = 1000 kg/m3
Draw FBD for plate.
The weight force can
be ignored during the
calculation and W = Hy
Hy
Hx
Fr
Fstep
W
16
Pipe-inlet
Determine resultant force
Fr = ρghc A
1
1
A = ab = 1.2 × 1.0 = 0.60 m2
2
2
1
a
hc = 3.0 + = 3.0 + 1.0 = 3.333 m
3
3
Fr = 1000 × 9.810 × 0.600 × (3 + 0.333) = 19.6 kN
Ixc
1.2 × 1.03
=
= 0.0333 m4
36
Now for center of pressure
yr
yr
yr
Ixc
+ yc
=
Ayc
0.0333
+ 3.333
=
0.60 × 3.333
= 0.016 + 3.333
yr = 3.349 m
17
Pipe-inlet: Step reaction
Hy = 0
Net torque about hinge
must be zero.
0.349 m
Fr =19.6 kN
Hx
Fstep
W= 0
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 hpi = γh/2
• Resultant force Fr = hpiA = 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.
1
F2 =
γ(h2 − h1 ) (h2 − h1 )w
2
The resultant force is FR = F1 + F2
20
Pressure Prism
Determination of Center of Pressure done from
moments of the forces.
FR y R = F1 y 1 + F2 y 2
F1 y 1 + F2 y 2
⇒ yR =
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 of distance from
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.
pg = 50 kPa
Air
Water
2.0 m
0.6
0.6 m
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
pair
γ x 2.0
B
2.0 m
A
C
γ x0.6
0.6 m
pair
pwater
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
= ( 12 0.6 × 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
= 0.304 m
26100
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
4R
3π
= 0.84883 m
F1
2.0 m
xc =
xc
2.0 m
It is 0.84883 m from
the left boundary of
quarter circle.
W
2.0 m
26
Curved Surface, Vertical
F1 = γ2.0(2.0 × 8.0)
F1 = 313.6 kN
F1
2.0 m
W
2.0 m
W
1
= γ π(2.0)2 × 8.0
4
= 246.3 kN
W
2.0 m
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
xR
313.6 × 1.0 + 246.3 × 1.151
=
559.9
= 1.066 m
27
Curved Surface, Horizontal
Rectangle
F = γ2.0(2.0 × 8.0)
2.0 m
2.0 m
2.0 m
F = 313.6 kN
F△
F△
1
= γ 2.0(2.0 × 8.0)
2
= 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
yR
313.6 × 1.0 + 156.8 × 1.333
=
470.4
= 1.11 m
28
Curved Surface, Summary
2.0 m
2.0 m
2.0 m
1.11 m
470.4 kN
731.3 kN
1.06 m
559.9 kN
The net force is
p
FR = 559.92 + 470.42 = 731.3 kN
Static balance and Action and Reaction can be
applied
FPressurePlusWeightOnWedge + FTankOnWedge = 0
FTankOnWedge + FWedgeOnTank = 0
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.
p1
A
h
px
px
p = p1 +γ h
2
Consider pressure forces on a rectangular slab
Fnet:pressure = p2 A − p1 A
Fnet:pressure = (p1 + γh)A − p1 A
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 .
δVi
Therefore the proof for a rectangular slab can be
generalized to a body of arbitrary shape.
X
δFi = γV.
FB =
i
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.
FB = γ V
C
CG
W = mg
The buoyancy force for a partially submerged object
passes through the centroid of the displaced volume,
V′ .
FB = γ V’
C
CG
W = mg
The weight force passes through the center of gravity
and does not always pass through the centroid.