OVERALL STABILITY 4.1 External Forces Acting on a Vessel In Chapter - - PowerPoint PPT Presentation

OVERALL STABILITY

4.1 External Forces Acting on a Vessel

In Chapter 4 we will study five areas:

• 1. The concept of a ship’s Righting Moment (RM), the

chief measure of stability.

• 2. KG and TCG changes and their effects on RM.
• 3. How Stability is effected by Damage to the Hull

using the “Added Weight” method.

• 4. Effects of a “Free Surface”.
• 5. Effects of Negative GM on ship stability.

4.2 Internal Righting Moment

EXTERNAL FORCES cause a vessel to heel. Recall Force x Distance = Moment – External Moment can be caused by wind pushing on one side of the vessel and water resisting the motion on the other side. – Each distributed force can be resolved into a resultant force vector. The wind acts above the waterline and the water resistance acts below the waterline.

Internal Righting Moment Righting Arm

G B F B

External upsetting force

Z WLf

Water resistance

MT

f f Ds CL

F

Internal Righting Moment

The two forces create a couple because they are equal in magnitude, opposite in direction, and not aligned. The couple causes rotation or heeling. The vessel will continue to rotate until it returns to Static Equilibrium (i.e. an Internal Moment is created which is equal in magnitude and opposite in direction). Giving M=0.

Internal Righting Moment

Internal Forces create a Righting Moment to counter the Upsetting Moment of the External Forces. The two internal forces are the weight of the vessel (s) and the resultant buoyant force (FB).

Internal Righting Moment

The perpendicular distance between the Weight and the Buoyancy Force vectors is defined as the RIGHTING ARM (GZ). The moment created by the resultant Weight and the resultant Force of Buoyancy is defined as the RIGHTING MOMENT (RM). It may be calculated by:

RM = GZ GZ F B

 =

s

Internal Righting Moment

Where: RM is the internal righting moment of the ship in ft-LT. s is displacement of the ship in LT. FB is the magnitude of the resultant buoyant force in LT. GZ is the righting arm in feet.

A ship in static equilibrium is affected by outside forces that will alter its state

• f equilibrium.

The forces of wind- and the opposing force of the water below the waterline- will cause an external moment couple about the ship’s center of flotation. MT

Wind

Water Resistance

MT B

Wind

Water Resistance

The ship reacts to this external moment couple by pivoting about F, causing a shift in the center of buoyancy. The center of buoyancy will shift because the submerged volume will change. Note that there is no change in weight or it’s distribution so there is NO change in the location of G!

MT B

Because the location of B changes, the location of where the FB is applied also

• changes. Because G does not move, the location of the Δs force does not change.

The displacement force and the buoyant for are no longer aligned. The heeling

• ver causes the creation of an internal moment couple.

FB s

MT B The external moment couple causes the creation of the internal moment couple to oppose it.

Wind

Water Resistance

As a result, the ship is now back into equilibrium, even as it heels over due to the wind force. s FB F

We are concerned with the created internal moment caused by the offsetting

• f the ship’s weight and the buoyant force.

The offset distance of the applied forces, GZ, is called the MOMENT ARM. The length of this moment arm is a function of the heeling angle, φ. MT

f

B FB Ds Z

Remember that a moment is created when a force acts at a

distance from a given point.

The created moment is called the internal

RM = GZDs = GZFB

In the case of the created internal moment couple, we have the two force, Ds and FB, acting over the distance GZ.

MT

f

B Z FB Ds This illustrates just one potential moment arm based upon one particular angle of φ. There are an infinite number of angles possible, therefore, an infinite number of moment arms that vary with the degree of heel, φ.

If we can plot the heeling angle f versus the created moment arm GZ, we can create the Intact Statical Stability Curve.

“Curve of Intact Statical Stability”

• r

“The Righting Arm Curve” – Shows the Heeling Angle () versus the righting arm (GZ). – Assumes the vessel is heeled over quasi- statically in calm water (i.e. external moments are applied in infinitely small steps).

4.3 Curve of Intact Statical Stability

This is a typical curve. Notice that it plots the angle of heel on the x-axis and the righting arm on the y-axis. The curve is in both the 1st and 3rd quadrants (the 3rd shows a heel to port). Typically only the curve showing a heel to starboard is shown as it is symmetrical.

Measure of Overall Stability

Curve of Statical Stability

Angles of Inclination: (Degrees)

Righting Arm - GZ -(feet)

Range of Stability Maximum Righting Arm Angle of Maximum Righting Arm Dynamical Stability Slope is a measure of tenderness or stiffness.



Intact Statical Stability

0.5 1 1.5 2 2.5 3 3.5 4 4.5 25 50 75 85

Heeling Angle  Moment Arm GZ

The above chart plots the data presented in the text on p. 4-6 an 4-7.

Intact Statical Stability

0.5 1 1.5 2 2.5 3 3.5 4 4.5 25 50 75 85

Heeling Angle  Moment Arm GZ

With φ at 0 degrees, the moment arm is also is 0. The buoyant force and the ship’s weight are aligned. No moment is created.

Intact Statical Stability

0.5 1 1.5 2 2.5 3 3.5 4 4.5 25 50 75 85

Heeling Angle  Moment Arm GZ

As the angle of heel increases, the moment arm also increases. At 25 degrees, shown here, GZ is 2.5ft.

Intact Statical Stability

0.5 1 1.5 2 2.5 3 3.5 4 4.5 25 50 75 85

Heeling Angle  Moment Arm GZ

As the angle increases, the moment arm increases to a maximum… here it is 4ft. As φ increases beyond this point the moment arm begins to decrease and the ship becomes in danger of capsizing…

Intact Statical Stability

0.5 1 1.5 2 2.5 3 3.5 4 4.5 25 50 75 85

Heeling Angle  Moment Arm GZ

...Remember, the internal moment couple created here is in response to the external couple created by outside forces. At GZ max the ship is creating its maximum internal moment. If the external moment is greater than the internal moment, then the ship will continue to heel over until capsized.

Intact Statical Stability

0.5 1 1.5 2 2.5 3 3.5 4 4.5 25 50 75 85

Heeling Angle  Moment Arm GZ

The angle of heel continues to increase, but the moment arm GZ, and thus the internal moment couple, decreases.

Intact Statical Stability

0.5 1 1.5 2 2.5 3 3.5 4 4.5 25 50 75 85

Heeling Angle  Moment Arm GZ

The angle has now increased to the point that G and B are now aligned again, but not in a good way. GZ is now at 0 and no internal moment couple is

• present. Beyond this point the ship is officially capsized, unable to right itself.

Curve of Intact Statical Stability Caveats!

Predictions made by the Curves of Intact Statical Stability are not accurate for dynamic seaways because additional external forces and momentum are not included in the analysis. ”Added Mass” However, it is a simple, useful tool for comparison and has been used to develop both intact and damaged stability criterion for the US Navy.

Curve of Intact Statical Stability

Typical Curve of Intact Statical Stability – Vessel is upright when no external forces are applied and the Center of Gravity is assumed on the centerline. (Hydrostatics) – As an external force is applied, the vessel heels over causing the Center of Buoyancy to move off the centerline. The Righting Arm (GZ) is no longer zero.

Curve of Intact Statical Stability

Typical Curve of Intact Statical Stability (cont.) – As the angle of heel increases, the Center of Buoyancy moves farther and farther outboard (increasing the Righting Arm). – The max Righting Arm will happen when the Center of Buoyancy is the furthest from the CG. This is max stability. – If the vessel continues to heel, the Center of Buoyancy will move back towards the CG and the Righting Arm will decrease.

Curve of Intact Statical Stability

Typical Curve of Intact Statical Stability (cont.) – Since stability is a function of displacement, there is a different curve for each displacement and KG. These are called the Cross Curves.

For all ships, there exists the CROSS CURVES OF STABILITY. Like the Curves of Form, they are a series of curves presented on a common axis.

• The x-axis is the ship’s displacement, Δs, in LT
• The y-axis is the righting arm, GZ, in ft
• A series of curves are presented, each representing a different angle of heel f

By plotting the data from the Cross Curves of Stability for a given displacement, you can create an Intact Statical Stability Curve.

In the Cross Curves of Stability, the data is presented assuming that:

KG = 0 (on the keel)

This is, of course, not realistic. It is done this way so that the curves may be generalized for all drafts. Once the curve data is recorded and plotted, a sine correction factor must be applied, shifting the KG to its correct position in order to get the

TRUE MOMENT RIGHTING ARM VALUE.

Cross Curves Example

Righting Arm (feet) Displacement (LT) 5 2.5 1000 2000 3000 10 degrees heel 30 degrees heel

At 2000 LT, the ship Has a RA of 2.5’ @10o Heel and 5’ @30o

Curve of Intact Statical Stability / “Righting Arm Curve”

Assumes: – Quasi-static conditions – Given Displacement – Given KG Cross Curves of Stability – Since MT moves as a function of φ, Righting Arms are calculated for each φ at regular intervals – Assumes a value of KG

4.4 Measure of Overall Stability

From the Curves of Intact Stability the following Measures of Overall Stability can be made: – Range of Stability – Maximum Righting Moment – Angle of Maximum Righting Moment – Dynamical Stability – Measure of Tenderness or Stiffness

Measure of Overall Stability

Range of Stability – The range of angles for which there exists a positive righting moment. – The greater the range of stability, the less likely the ship will capsize. – If the ship is heeled to any angle in the range

• f stability, the ship will exhibit an internal

righting moment that will right the ship if the external moment ceases.

Measure of Overall Stability

Maximum Righting Moment – The largest Static Moment the ship can produce. – Calculated by multiplying the displacement of the vessel times the maximum Righting Arm. – The larger the Maximum Righting Moment, the less likely the vessel is to capsize.

Measure of Overall Stability

Angle of Maximum Righting Arm – The angle of inclination where the maximum Righting Arm occurs. Beyond this angle, the Righting Arm decreases. – It is desirable to have a larger maximum angle so that at large angles of heel in a rolling ship the righting moment will continue to increase.

Measure of Overall Stability

Dynamical Stability: –

The work done by quasi-statically rolling the ship through its range of stability to the capsizing angle. – Can be calculated by the equation: . This is equal to the product of the ship’s displacement with the area under the Curve of Intact Statical Stability. – Not shown directly by the Curve of Intact Statical Stability. – Does not account for the actual dynamics, because it neglects the impact of waves and momentum.  D  d GZ

s

Measure of Overall Stability

Measure of “Tenderness” or “Stiffness” –

The initial slope of the intact statical stability curve indicates the rate at which a righting arm is developed as the ship is heeled over. This slope is GM! – A steep initial slope indicates the rapid development of a righting arm and the vessel is said to be stiff. Stiff vessels have short roll periods and react strongly to external heeling moments. – A small initial slope indicates the slower development of a righting arm and the vessel is said to be tender. Tender vessel have longer roll periods and react sluggishly to external heeling moments.

Example: Plot the Intact Statical Stability Curve for an FFG-7 displacing 5000LT Step #1. From the Cross Curves of Form, find the 5000LT displacement value

• n the x-axis.

Step #2. Record the righting arm value for each curve, from φ = 0 to 80 degrees Step #3. Draw the curve, using φ as x-axis, and GZ as y-axis

Intact Statical Stability, FFG-7 5 10 15 20 25 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 Heeling Angle Moment arm GZ

f GZ 0.00 5 2.00 10 3.80 15 5.80 20 7.75 25 9.75 30 11.75 35 13.30 40 14.75 45 16.10 50 17.20 55 18.00 60 18.60 65 19.00 70 19.30 80 19.50

Intact Statical Stability Curve for FFG-7 @ s = 5000LT … But a correction must still be made!!

Example Problem

The Statical Stability curve applies to a ship with D=3600LT. The ship is being pulled sideways into a 10° list by a tug attached to the ship 10ft above the

• Waterline. How

much force is the tug applying to the tow line?

Curve of Intact Statical Stability

• 2
• 1

1 2 3 4 5

• 10

10 20 30 40 50 60 70 80 90 100 Angle of Heel (Degrees) Righting Arm (GZ)(ft)

Example Answer

RM=GZΔ=1.2ft×3600LT=4320ft-LT Upsetting Moment from Tug=FTug×10ft=RM (in static equilibrium) FTug=4320ft-LT/10ft=432LT

F=10°

FTug

10ft

Water Resistance G B D FB F

In the Cross Curves of Stability, the data is presented assuming that:

KG = 0 (on the keel)

This is, of course, not realistic. It is done this way so that the curves may be generalized for all drafts. Once the curve data is recorded and plotted, a sine correction factor must be applied,shifting the KG to its correct position in order to get the

TRUE MOMENT RIGHTING ARM VALUE.

4.5 Effect of a Vertical Shift in the Center of Gravity on the Righting Arm

– Must Apply a Sine Correction if:

• Using the Curve of Intact Statical Stability to correct for G not being

located at K

• Correcting the Curve of Intact Statical Stability for vertical

movements of G

MT B The external moment couple causes the creation of the internal moment couple to oppose it.

Wind

Water Resistance

External Moment Couple = Internal Moment Couple

s FB

MT

f

B1 Z0

G0Z0 = Moment Arm

When the ship heels over, the center of buoyancy, B, shifts. The shift creates a distance or “moment arm”.

MT

f

B1 Z0 For values taken from the Cross Curves of Stability, G0 is at the keel...

G0Z0 = Moment Arm

This value is recorded as G0Z0, the Initial Moment Arm.

MT

f

B1 Z0

The KG value for the ship is given… this is the ACTUAL G position from the keel... KG forms a similar triangle that gives the value for the SINE correction

Sin f = opp hyp

• pp =correction factor

hyp = KG Sine Correction factor = KG Sin f

MT

f

B1 Z0 Sin Correction = KG Sin f

MT

f

B1 Z0 Zv

GvZv = G0Z0 - KG sinf

Effect of a Vertical Shift in the Center of Gravity on the Righting Arm

As KG rises the righting arm (GZ) decreases. This change in GZ can be found from: Where: – Gv is the final vertical location of the center of gravity. – G0 is the initial location of KG. – Typically, G0GV=KGfinal

GvZv = G0Z0 - G0Gvsin F

Effect of a Vertical Shift in the Center of Gravity on the Righting Arm

Sine Correction:

Go, Zo=initial locations Gv, Zv=final positions

B G MT W L B0 G Z Z F b

s

P

F F

GvZv = G

0Z0 - G0Gv sin F

D

Effect of Increased Displacement on the Righting Arm

A higher displacement should increase the Righting Moment as RM= Displacement * RA But, if the added weight is high, then the KG increase could cause a reduction in GZ Weight added low down usually increases stability

Effect of a Vertical Shift in the Center of Gravity on the Righting Arm

4.6 Stability Change for Transverse Shift in CG So far we have only considered the case where the Center of Gravity is on the centerline (TCG=0). The center of gravity may be moved off the centerline by weight additions, removals, or shifts such as cargo loading, ordnance firing, and movement of crew.

Stability Change for Transverse Shift in the CG

F

Gt

MT

W1

Gv

Port Starboard

B

1

Gv ZV Zt L1

Gv G t cos

Gt

Fb

F F F F

Ds

Zv GT f

The red line indicates the COSINE Correction factor for a transverse change in G.

ZT

Note that GvGT is the TCG value and is the hypotenuse of this correction triangle...

Cosine Corr = GvGT cosf

Zv GT f The final moment arm, GTZT, is the correct moment arm ZT

GTZT = GvZv - GvGT cosf

Stability Change for Transverse Shift in the CG The new righting arm created by a shift in TCG may be computed at each angle from the Cosine Correction:

GtZt= GvZv -GvGtcos F

...Typically, GVGt=TCGfinal

Stability Change for Transverse Shift in the CG The new righting arm (GtZt) created due to the shift in the transverse center of gravity is either shorter or longer than the righting arm created if TCG=0. The range of stability has decreased on the side that the transverse center of gravity has shifted to but has increased on the side it shifted from.

GTZT = G0Z0 - KG sinf - GvGT cosf

Combining both the vertical and horizontal corrections by substituting for GvZv you can get a final general formula for determining moment arms:

Example Curves

With Cosine Correction Statical Stability Curve and Corrections

• 15
• 10
• 5

5 10 15

• 1

4

• 1

3

• 1

2

• 1

1

• 1
• 9
• 8
• 7
• 6
• 5
• 4
• 3
• 2
• 1

1 2 3 4 5 6 7 8 9 1 1 1 1 2 1 3 1 4 Angle of Heel (degrees) Righting Arm (GZ)(ft) Righting Arm from Cross Curves With Sine Correction With Cosine Correction

Various Righting Arm Conditions

G B FB D Z F=0 G B FB D F=30 G B FB D F=RAMax G B FB D F= >Capsize Angle G B FB D F=Capsize Angle Z

Example Problem

FFG-7, with draft 13.5ft , which would

• therwise be on an even keel, is heeling

15° to starboard in a gale. KG is 17ft. What is the Righting Moment?

Example Answer

• RM=GZΔ
• GVZV=G0Z0-G0GVsinF
• From Curves of Form:

Draft of 13.5ft-> Δ =100×30LT=3000LT

• From Cross Curves for Δ =3000LT, φ=15°,

G0Z0=6ft

• G0GV=KG=17ft
• GVZV=6ft-17ft×sin(15°)=1.6ft
• RM=1.6ft×3000LT=4800ft-LT

Example Problem

• FFG-7, with draft 13.5ft, which would
• therwise be on an even keel, is heeling

15° to port in a gale. KG is 17ft. While in this condition, 50LTs of unsecured stores shift from 20 ft starboard of centerline to 20ft port of centerline.

• What is the Righting Moment?
• What would the Righting Moment be if

the weight had shifted the other way?

Example Answer

• RM=GZ Δ
• GfZf=G0Z0-G0GVsin φ -GVGtcosφ
• From Curves of Form:

Draft of 13.5ft-> Δ =100×30LT=3000LT

• From Cross Curves for Δ =3000LT, φ =(-)15°, G0Z0=(-)6ft
• G0GV=KG=17ft
• GVGt= TCGf=(TCG0 Δ 0+STcgawa-STcgrwr)/ Δ f

=(0ft×3000LT+(-)40ft×50LT)/3000LT=(-).67ft

• GfZf=(-)6ft-17ft×sin(-15°)-(-.67ft)×cos(-15°)=(-).953ft
• RM=(-).953ft×3000LT=(-)2860ft-LT vice (-)4800ft-LT in the case

without the weight shift (~40% reduction)

• GfZf=(-)6ft-17ft×sin(-15°)-(+.67ft)×cos(-15°)=(-)2.247ft
• RM=(-)2.247ft×3000LT=(-)6741ft-LT vice (-)4800ft-LT in the

case without the weight shift (~40% improvement)

4.7 How Does a Ship Sink?

• 3 of the 6 Degrees of Freedom can sink a ship
• Foundering(heave): Ship fills up with water from the

bottom up and simply sinks on a relatively even keel (Loss of Buoyancy)

• Overwhelm and Capsize: Ship still floats and is stable

but has insufficient Righting Arm and Dynamical Stability for weather and sea conditions, which eventually roll the ship past range of stability

• Plunging: Pitch angle becomes excessive causing

the ship to sink bow or stern first

• Progressive Flooding: Excessive list and/or trim angle

and/or failure of bulkheads adjacent to flooded compartments resulting in one or more of the loss modes above.

Damage Stability

“Flooding” - Water ingression such that the vessel has sinkage and trim but no list. May be intentional. “Damage” - Water ingression such that the vessel has sinkage, trim and list.

When a vessel is damaged, creating a a gap or hole in the hull, water will breech the ship. This results in:

• Increase in draft
• Change in trim
• Permanent angle of list

The result of this flooding can be determined two ways:

• Lost Buoyancy Method
• Added Weight Method

Consider a vessel that has been damaged such that a portion of the bottom is now open to the sea... The vessel’s draft will increase because an amount of the buoyancy was lost...

Lost buoyancy considers the amount of buoyancy “lost” as a result of the

hole, and determines the value based upon the change in parallel sinkage that results. The change in the draft reflects the amount of buoyancy lost. The ship sinks until the available submerged volume is again equal to the ship’s displacement

Original draft

Lost Buoyancy Method

Analyzes damage by changes in buoyancy versus changes in the Center of Gravity. Premise is that the ship’s CG does not move. Since weight does not change, total buoyant volume must also be constant. Therefore, the ship makes up any lost buoyancy volume from damage by listing, trimming, and draft changes.

The Added Weight Method considers the resulting flooding as though it was a weight added to the ship. This is the method that will be used in this course. A flooded compartment does not fill completely with water, however. Compartments contain equipment, furniture, structural components, and cargo. A correction factor must be added to the volume of the compartment to accurately reflect conditions...

Original draft

This correction factor is called: PERMEABILITY = AVAILABLE VOLUME TOTAL VOLUME Some typical factors are: Watertight compartment (warship) 97% Watertight compartment (merchant) 95% Accommodation spaces 95% Machinery spaces 85% Dry cargo spaces 70% Bunkers, stores, cargo holds 60%

Added Weight Method “The One We Will Use”

Damaged Ship Modeled as Undamaged But with Water- Filled Spaces. Average Distances of Space from Keel, Midships, and Centerline Known & Water Density Known. Therefore Can Solve for Shifts in “G” as a Weight Addition Problem

Added Weight Method

Independently Solve for Damaged Condition – KG – TCG – Draft and Trim Must know compartment contents to find Total of (Water Weight) Added. This involves a “Permeability” factor.

Permeability

Compartments are rarely 100% flooded during damage, due to trapped air, equipment, etc. – Ratio of volume occupied by water to the total gross volume is defined as “permeability”. Permeability = Volume Available for Flooding Total Gross Volume – Permeability is always < or = to 100%!

Damage Stability Design Criteria

Guiding rules for vessel design. Note that criteria used in static analysis will neglect the impact of dynamic forces such as wind and waves.

Damage Stability Design Criteria

Three Main Criteria

• “MARGIN LINE”
• “LIST”
• “EXTENT OF DAMAGE TO HULL”

MARGIN LINE LIMIT

• Highest permissible location of any

damaged waterplane.

• Must be at least 3 inches (0.075 m) below

top of the bulkhead deck at the side.

LIST LIMIT

• Heel by damage  20 degrees.
• Naval machinery to operate indefinitely at a

permanent list  15 degrees (most will function up to ~25 degrees for a few hours).

• Assumes personnel can continue damage control

efforts effectively at a permanent list of 20 degrees.

• Ship must possess adequate stability against

weather to be towed when at 20 degree list.

EXTENT OF DAMAGE TO THE HULL LIMIT

•  100 ft LOA: must withstand flooding in one

space.

• 100 - 300 ft LOA: flooding in two adjacent

compartments.

• Warships, troop transports and hospital ships over

300 ft LOA: hull opening up to 15 % of Lpp.

• Others  300 ft: hull opening up to 12.5% of Lpp.

Foundering and Plunging A vessel as result of “damage” or other events can be lost several ways: Insufficient transverse stability. It rolls over. – (Could be static or dynamic.) Insufficient longitudinal stability. “Plunging” If insufficient buoyancy. It sinks. “Foundering”

Example Problem

An FFG-7 with a draft of 13.5ft and a KG of 19ft on an even keel inport sails into the North Atlantic during Winter.

– While there, topside becomes coated with a 6in thick coating of ice

• f density of 55lb/ft³. The topside area covered is 20,500ft² and

has a Kg of 40ft. – In this condition, a space heater in CIC shorts generating a fire which is only extinguished by completely filling the 97% permeable 40ft×40ft×10ft space with firefighting (sea) water. The space is centered 45ft above the keel and 2.5ft port of centerline.

What is the Righting Moment for a 15° port list and how could the resulting problem have been prevented?

Example Answer

• wice=ρgV=55lb/ft³×.5ft×20,500ft²×1LT/2240lb =252LT(@Kg=40ft)
• wffwater= ρ gV=64lb/ft³×40ft×40ft×10ft×.97×1LT/2240lb

=443LT(@Kg=45ft)

• Δ(Curves of Form[T=13.5ft])=100×30LT=3000LT
• KGf=(KG0 Δ 0+Kgawa-Kgrwr)/ Δ f
• KGf=(19ft×3000LT+40ft×252LT+45ft×443LT)

/(3000LT+252LT+443LT)

• KGf=23.5ft
• TCGf=(TCG0 Δ 0+Tcgawa-Tcgrwr)/ Δ f
• TCGf=(0ft×3000LT+0ft×252LT+(-)2.5ft×443LT) /(3695LT)
• TCGf=(-)0.3ft
• G0Z0(Cross Curves[Δ =3695LT; φ =(-)15°])=(-)6ft
• GfZf=G0Z0-KGfsinφ-TCGfcosF
• GfZf=(-)6ft-23.5ft×sin(-15°)-(-.3ft)×cos(-15°)=(+).372ft

Example Answer

• R.M.=Δ×GfZf=3695LT×0.372ft=(+)1375ft-LT to port for a

port list: The ship capsizes!

• KMt(Curves of Form[Δ=3695LT;T=15.25ft])=112*.2ft=22.4ft

(GMt=KMt-KGf=22.4ft-23.5ft=(-)1.1ft; Stable?)

• Center of Gravity is above Metacenter; ship rolls to port

due to offset of flooded compartment and capsizes.

• Prevent by keeping topside clear of ice and dewatering fire

spaces as soon as possible.

4.8 Free Surface Correction (Small Angles of Heel)

Free Surface - A “fluid” that moves freely. Fluid Shift is a weight and causes the CG to shift in both the vertical and horizontal directions. – Vertical shift is small for small angles and is usually ignored. – Horizontal shift always causes a reduction in the righting arm (GZ).

Free Surface Correction

Free Surface Correction (FSC) The distance the center of gravity would have to rise to cause a reduction in the righting arm equivalent to that caused by the actual transverse shift. "Virtual" center of gravity (Gv) The effective position of this new VCG. Effective Metacentric Height (GMeff) The distance from the virtual center of gravity (Gv) to the metacenter. Note: Dynamic effects are neglected.

Free Surface Effect

Static effects for small angles (F<=7°)

– Effective “g” for tank is above tank g analogous to relationship between M and B

B0 Bf MT g0 gf geff

Free Surface Correction The Big Picture

.

WL WL 1 MT B1 B G Z Z1 G1 K Gv GM eff FSC

g1 g

F

F

F

Free Surface Effect

The new, effective VCG is Gv, so a sine correction is applied to get the statical stability curve

G1Z1 = GtZt - GGv sin f

• r

G1Z1 = GtZt - FSC sin f

Free Surface Correction

The free surface correction to GM for small angle hydrostatics is: where: t

is the density of the fluid in the tank in lb s2/ft4

s

is the density of the water the ship is floating in lb s2/ft4 it is the transverse moment of area of the tank's free surface area in ft4 . s is the underwater volume of the ship in ft3.

FSC =

t s it s

Free Surface Correction

it is calculated for a rectangular tank as:

The dimensions are for the free surface!

it = (L) (B)3 12

L B C L Tank Y X

Effect on Ship “G” and Stability

GZeff=G0Z0-G0Gvsinφ-GvGtcos φ -FSCsin φ

– Calculation of KG, etc. is already accounted for in this equation – Free Surface Correction (FSC) already accounts for size of ship.

GMeff=GM-FSC=KM-KG-FSC

– A large FSC has exactly the same effects on list and stability as a higher KG.

How do we minimize adverse effects of free surface effect?

• Compartmentalization
• Pocketing (Keep tanks >95% full)
• Empty Tanks
• Compensated Fuel Oil Tanks
• Dewater quickly after a casualty -

flooding or fire

4.9 Metacentric Height

Recall that Overall Stability is measured by:

• Range of Stability
• Dynamical Stability
• Maximum righting moment
• The angle at which the maximum righting

moment occurs.

Initial Slope of the Curve of Intact Stability At small angles, a right triangle is formed between G, Z, and M. The righting arm may be computed: As   0, if the angle is given in radians the equation becomes:

GZ = GM sin



GZ __ = GM __ __ __

Initial Slope of the Curve of Intact Stability Metacentric height can then be found from the initial slope of the Curve of Intact Statical Stability:

GM = GZ sin GZ (radians) (for smallAngles) GM GZ (if

= 1 radian)

__



__ __ __ __



=

= 

To find the slope either: – Find the change in the y-axis over a given change in the x-axis. – Draw a straight line with the initial slope and read the value of GZ at an angle of 57.3 degrees (i.e. one radian).

Initial Slope of the Curve of Intact Stability

Metacentric Height

LET’S EXAMINE EACH GM CONDITION – GM Positive (G < M) – GM Zero (Neutral Stability) (G = M) – GM Negative (G > M)

Metacentric Height

Positive Stability

Metacentric Height

Neutral Stability

Metacentric Height

Negative Stability

Metacentric Height

SUMMARIZING GM CONDITIONS – GM Positive = Positive Stability (M > G) – GM Zero = Neutral Stability (M = G) – GM Negative = Negative Stability (M < G) Metacentric Height only a good indicator of stability over small angles. GM is initial slope of Curve Intact Stability

Stability Status

Weight Margin Stability Margin Adequate Adequate Inadequate Inadequate Status 1 Status 2 Status 3 Status 4

Example Problem

An FFG-7 with a draft of 13.5ft and a KG of 17ft on an even keel inport goes to sea.

A space heater in CIC shorts generating a fire which is extinguished by completely filling the 97% permeable 40ft×40ft×10ft space with firefighting (sea) water. The flooded volume is centered 45ft above the keel and 2.5ft starboard of centerline.

• 1. What is the equilibrium list angle in this condition?
• 2. A large wave hits and forces the ship to a temporary 15°

starboard list. What is the Righting Moment?

• 3. The CIC overhead, weakened by the heat of the fire, is blown
• ff by a gale force wind, making the compartment now a free
• surface. What is the equilibrium list angle and Righting Moment

for a temporary 15° starboard list?

Example Answer

• wffwater=ρgV=64lb/ft³×40ft×40ft×10ft×.97

×1LT/2240lb =443LT(@Kg=45ft)

• Δ(Curves of Form[T=13.5ft]) =100×30LT=3000LT
• KGf=(KG0 Δ 0+Kgawa-Kgrwr)/ Δ f
• KGf=(17ft×3000LT+45ft×443LT) /(3000LT+443LT)
• KGf=20.6ft
• TCGf=(TCG0 Δ 0+Tcgawa-Tcgrwr)/ Δ f
• TCGf=(0ft×3000LT+2.5ft×443LT) /(3443LT)
• TCGf=0.32ft

Example Answer

• KMt(Curves of Form[Δ =3443LT;T=14.6ft]) =114×.2ft=22.8ft
• GMt=KMt-KGf=22.8ft-20.6ft=2.2ft
• tan(φ)=TCGf/GMt=0.32ft/2.2ft; φ =8.3°
• G0Z0(Cross Curves[Δ =3443LT; φ =15°])=6ft
• GfZf=G0Z0-KGfsin φ -TCGfcos φ
• GfZf=6ft-20.6ft×sin(15°)-(.32ft)×cos(15°)=0.36ft
• R.M.= Δ ×GfZf=3443LT×0.36ft=1240ft-LT

Example Answer

CIC Overhead Blown Off:

• it=lb³/12=40ft×(40ft)³/12=213,333ft4
• VS=Δ/(ρg)=3443LT×2240lb/LT/(64lb/ft³)=120,505ft³
• FSC=(ρtit)/(ρSVS)=it/VS=1.77ft (ρt=ρS)
• GMeff=KMt-KG-FSC=22.8ft-20.6ft-1.77ft=0.43ft
• tan(φ)=TCGf/GMeff=0.32ft/0.43ft; f=36.7°(vice 8.3°)
• GfZf=G0Z0-KGfsinφ-TCGfcos φ -FSCsin φ
• GfZf=6ft-20.6ft×sin(15°)-(.32ft)×cos(15°)-1.77ft×sin(15°)=(-)0.1ft
• R.M.= Δ ×GfZf=3443LT×(-)0.1ft=(-)344.3ft-LT(vice +1240ft-LT)
• At 15°, ship lists starboard but wants to list further starboard to

reach 36.7°

Example Problem

Below are body plan views of three proposals for outriggers. Sketch the respective curves of Intact Statical Stability and comment on the stability and ride characteristics for each option.

G G G

Example Answer

F GZ Statical Stability F GZ Statical Stability F GZ Statical Stability

Positively stable: Very stiff roll characteristic Unstable at Zero List: Very likely to Loll and settle

• n one of the outriggers

Neutrally Stable: until deck hits water Very tender roll characteristic G G G M M M