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 .