Hydrostatics & stabilityNomenclature

Talking with Naval Architects

Coordinates

Heel= static roll Trim = static pitch

BowStern

Keel(K)

Nomenclature

• The bow is pointing forward,• The stern is facing aft (backwards)• The starboard side is on your right as you

face forward,• The port side is on your left as you face

forward

Archimedes Principle

• The buoyancy of an object is equal to the weight of displaced fluid

Buoyancy ForceStatic Pressure

P=ρgz B

T

F = P*A = [ρgT]*BL = ρg*TBL = ρg*(Volume)

z

Tip: All fluid forces of importance for offshore structures are pressure forces!

Displacement

LBT=∇L

BT

BLBTC=∇L

TB

L = LengthB = Beam (width)T = DraftCB = Block coefficient∇= Displacement (Volume) ∆= Displacement (Buoyancy Force)γw = Density of fluid = ρg

∆ = γwLBT

∆ = γwLBTCB

30 m

100m

4000 t

T ?γw =1.025 t/m3

Seawater

∆= γwLBT = 4000 t

T = 4000 t / [γwLB] = 4000 t / [1.025 t/m3 * 100 m * 30 m] =

1.3 m Draft

Example 1: Compute draft (T) for prismatic barge, 100 m long, 30 m wide, weight 4000 t

Hydrostatic Stiffness - HeaveW

W + δW

δTT

K = δW/δT = γwAwp

Awp = LB (Waterplane Area)

Hydrostatic Stiffness – Roll/Pitch

Pontoon Cross Section10 m

5 m

60 m typ

10 m Dia typ

25 m

Example 2: Semi-Submersible Weighs 3000 t. How much payload can it carry? What is its vertical stiffness? Seawater is 1.025 t/m3.

Three Rules of Floating Platform Design

First Rule of Floaters: Buoyancy must equal weight plus any external vertical forces.

Weight includes:

1. Hull Steel Weight

2. Hull Outfitting

3. Topsides Payload (fixed and variable)

4. Topsides Structure

5. Ballast in Hull (fixed and variable)

External Vertical Forces Include

1. Component of Mooring Load

2. Component of Riser Load

3. Suspended weights (e.g. from crane)

2nd Rule of Floaters: The weight shall be positioned such that the hull will not tip over!

Typhoon TLP after Hurricane Rita

3rd Rule of Floaters: There should be enough Reserve Buoyancy to maintain balance and stability even with tanks flooded!

P-36 After Explosion in Column – Not Enough Reserve Buoyancy

Thunderhorse Semi after Hurricane Dennis – Reserve Buoyancy in Deck saved it from sinking!

Transverse Stability

W B B

W

Case 1: Positively stable Case 2: Negatively stable or unstable

Stability = Tendency to return to a previous condition when perturbed!

The metacenter is the point of intersection between the action of the buoyant force and the centerline of

the vessel.

W

L

W1 L1

G

B B1

M

K

z

Lost Buoyancy

Gained Buoyancy

Metacentric Height is the distance from the center of gravity to the

metacenter.

W

L

W1 L1

G

B B1

M

K

z

K – keel pointG – point of action of weight, i.e. center of gravityB – point of action of buoyancy, i.e. center of buoyancy; The position of B

shifts with the amount of heel (B to B1)M – Point of intersection of line of buoyancy and centerline, i.e. MetacenterGM – Distance between G and M, i.e. metacentric height.KB - Distance from Keel Point to BKG- Distance from Keel Point to G

“GM” is the metacentric height

and must be positive!

Metacentric Height, GMW

L

W1 L1

G

B B1

M

K

z

KGBMKBGM −+=

BM

KB

KG

B

M

TB

LBTLBIBM xx

1212/ 23

==∇

=

L

BT

Different “B”

“BM” is a function of the waterplane inertia (moment of square of the distance from the axis)

Metacentric Height (GM) and Restoring Moment

W

L

W1 L1

GB B1

M

K

z

θ

Μ θ = GZ*B ~ GM*B* sin(θ) (small angles)

M θ

GZ is the “righting arm”

Example Restoring Moment Curve

• Plot GZ (righting arm) vs. angle of heel– Ship’s G does not change as angle changes (not

true if tanks are partially full)– Ship’s B always at center of underwater portion of

hull– Ship’s underwater portion of hull changes as heel

angle changes– GZ changes as angle changes

B

G

M

K

T

B

GM=T/2 + B2/12T - KGStable if GM > 0

KG<T/2 +B2/12T

Substitute T = W/(γwLB)

WLB

LBWKG w

w 122

3γγ

+≤

0

10

20

30

40

50

60

70

0 5000 10000 15000 20000

Weight, t

Dra

ft (T

), K

Gm

ax -

m

T

Kgmax

L=100 mB = 30 m

Max KG for Payload is required in operating manual for vessels.

Example: Maximum KG as function of total weight for the 100 x 30 m barge.

KGBMKBGM −+=

A. A rectangular barge is 80 m long by 20 m wide. It weighs 5000 tonnes with its cargo. What is it’s draft if the density of water is 1.025 t/m3?

B. What is the maximum value of KG for this barge for positive stability, i.e. GM>0?

C. Man is in rowboat on lake. There is an anchor in the boat. He throws the anchor overboard. Does the level of the lake:

• Raise• Go down• Stay the same?

D. A glass of water has an ice cube. When the ice melts is the level of water.

• Higher• Lower• The Same?

Hint: The density of water is greater than the density of ice.

E. A glass of Scotch has an ice cube. When the ice melts, is the level• Higher• Lower• The Same?

Hint: The density of Scotch is less than the density of ice.

hVw

Vboat = (Wboat + Wanchor)/γw

h = (Vboat + Vw)/(WL) = 1/(WL)*(Wboat/ γw+Wanchor/γw+Vw)

W

hVw

Vboat = Wboat /γw

h2 = (Vboat + Vanchor +Vw)/(WL) = 1/(WL)*(Wboat/ γw+Wanchor/γa+Wanchor/γw+Vw)

W

Vanchor = Wanchor/γa

h1 = (Vboat + Vw)/(WL) = 1/(WL)*(Wboat/ γw+Wanchor/γw+Vw)

Since γa>γw

h2 < h1…. The level goes down!

h Vw

Vice = Wice /γw

W

h Vw

Vicewater = Wice /γw

W

The ice cube and the melted ice cube weigh the same!.... H is the same…

Ice cube in water

h1 Vw

Vice = Wice /γice

W

h2 Vw

Vicewater = Wice /γw

W

Since γice < γw

Vice > Vicewater

h1 > h2

Ice cube in scotch

Class Rules for Stability

Overturning Moment

6 m

Wind Force

Reaction force from:

a) Centroid of Drag or

b) Moorings (use worst)

Wind Load Calculations

Wind Force

1

23

4

5

Select different block areas based on elevation, shape

z2

z1

z4

z5

Tip: For wind calcluations use the waterline as z=0 .. Convert to the keel reference for stability!!!

6 m3.6 m

z3

Force in normalized by Ur2

Centroid based on 1st moment of forces

Zc_wind

Wind Force Coefficient

• Cw = Wind Force Divided by Velocity (at 10 m) Squared

• Units: – tonne/(m/sec)2

– Kips/(ft/sec)2

– kN/(m/sec)2

Overturning Moment

6 m

Wind Force = 898*Ur2 (N)

Reaction force from:

Moorings at keel (assume worst)

2.4 + 16.5 = 18.9 Moment Arm

2.4 m draft

• Watertight Integrity– Compartments and fittings designed for

hydrostatic head (minimum 20 ft) of surrounding structure.

• Weathertight Integrity– Under design sea conditions water will not

enter the structure (e.g. typically limited by vent openings)

0

2000

4000

6000

8000

10000

12000

14000

16000

18000

0 10 20 30 40 50 60 70 80

Theta (deg)

Mom

ent (

t-m)

Righting Moment KG=5Overturning Moment

Min extent of weathertight integrity (assumed when

deck imerses!)

1st Intercept

2nd InterceptB

A

C

KG = 5 GM = 9.9

KG = 10 GM = 4.9

Intact Stability Example

KG 5 10Area A 88500 20000Area B 72750 67500Area C 10500 22500A+B 161250 87500B+C 83250 90000Ratio (A+B)/(B+C) 1.94 0.97Min Reqd 1.4 1.4

OK NO!!

Even with positive GM the barge does not meet Class Rules for Stabiltiy:

Reduce KG or Reduce Weight.. Or? Add fixed ballast..

Increases Weight but lowers KG

0

1000

2000

3000

4000

5000

6000

7000

8000

9000

0 5 10 15 20 25 30 35 40

Theta (deg)

Mom

ent (

t-m)

Righting Moment KG=10Overturning Moment

A

B

C

Intact Stability Rules for Offshore Floating Platforms

• Must have a positive GM for all conditions• Survive overturning moment from 100 kt wind

(51.5 m/sec)– Waves not considered in class rules– Reaction from drag (free floating) or mooring,

whichever is worse• Inclining test is required for first unit of series• Righting moment curves and overturning

moment curves are required for all operating drafts

Damage Stability Rules

• Consider accidental floating of any compartment with external connections to the sea, or adjacent to the sea

• Mean heeling angle plus overturning due to 50 kt wind should not exceed limits of weathertight integrity.

Wind Tunnel Testing

Photos from Force Technology (www.force.dk)