overview chapter 3 - buoyancy versus gravity = stability (see chapter objectives in text) builds on...

Overview

•Chapter 3 - Buoyancy versus gravity = stability

•(see Chapter Objectives in text)

•Builds on Chapters 1 and 2

•6-week exam is Chapters 1-3!

HYDROSTATICSReview (3.1)

• Archimedes Principle:

– “An object partially or fully submerged in a fluid will experience a resultant vertical force

equal in magnitude to the weight of the volume of fluid displaced by the object.”

– This force is called the “buoyant force” or the “force of buoyancy”(FB).

HYDROSTATICS Review (3.1)

Mathematical Equation:

Where. . .

FB is the magnitude of the resultant buoyant force in lb, is the density of the fluid in lb s2 / ft4 ,

g is the magnitude of the acceleration of gravity normally taken to be 32.17 ft / s2 .

is the volume of fluid displaced by the object in ft3 .

BF g weight

Hydrostatics

The forces lead to translations:

• Heave• Surge• Sway

• The moments lead to rotations:

• Roll• Pitch• Yaw

Vessel Degrees of FreedomAnd Static Equilibrium

HYDROSTATICSStatic Equilibrium : Forces and Moments

(3.1.2.1-2)• Sum of the Resultant Forces:

• Sum of the Moments about a reference point:

• Static equilibrium must consist of both conditions!

0 0 0Mx My Mz

BF g weight 0 ?

0 ?

0 ?

Fx

Fy

Fz

HYDROSTATICSStatic Equilibrium : Stability

t

B

Is this boat in static equilibrium?What are the component forces and moments?

Are they internal or external?

HYDROSTATICSStatic Equilibrium (3.1.1.2)

G

B

LW

Port Starboard

DistibutedHydrostaticForces

ResultantWeight, s

Resultant Vertical Buoyant ForceFB

Atmospheric Pressure

HydrostaticPressure

What is the hydrostatic pressure? F=p*A

Wave?

HYDROSTATICS

Static Equilibrium : Stability (3.2)

B

t

HYDROSTATICS

Changes in the Center of Gravity (3.2)

• The Center of Gravity (G) is the point at which all of the mass of the ship can be considered to be located (for most problems).

Terminology! UPPERCASE for ship; lowercase for a smaller weight.

• It is referenced vertically from the keel of the ship (KG or VCG or Kg).

• (1) Shifting, (2) adding, or (3) removing weight changes the location of the Center of Gravity.

HYDROSTATICSStatic Equilibrium : Stability

t

B

Where is the Center of Gravity?The Center of Buoyancy?

Are they vertically aligned? Why/Why not?

HYDROSTATICSChanges in the Center of Gravity (3.2.1.1)

C

B

L

L

WL

g

Gow Gf

K

• When weight is added to a ship, the CG will move in a straight line from its current position toward the cg of the weight being added. G0 to Gf. The distance is a ratio of the weight and disp.

• What happens to the Center of Buoyancy (and the ship)?

HYDROSTATICSChanges in the Center of Gravity (3.2.1.2)

• When weight is removed from a ship, G will move in a straight line from its current position away from the center of gravity of the weight being removed. G0 to Gf.

C

B

L

L

WL

g

Gow Gf

K

Changes in the Center of Gravity (3.2.1.3)

• When a small weight is shifted (but

not added or removed, CG will move parallel to the weight shift but a much smaller distance because it is only a small fraction of the total weight of the ship.

C

B

L

L

W L

go

gf

Go

Gfw

w

K

StarboardPort

HYDROSTATICS

HYDROSTATICSVertical Shift in the Center of Gravity (3.2.2.1)

Where: (note: some use the term “initial” for “old” and “final” for “new”

KGnew is the final vertical position of the center of gravity of the ship as referenced from the keel. KG’s are in “feet”.

KGold is the initial vertical position of the center of gravity of the ship as referenced from the keel.

old old addedweight addedweightnew

old addedweight

KG Kg wKG

w


And,

s new is the final displacement of the ship in LT. In this example, it is equal to the initial displacement plus or minus the weight added.

s old is the initial displacement of the ship in LT.

Kg added weight is the vertical position of the center of gravity of the weight being added as referenced from the keel. This line segment is a distance in feet.

w added weight is the weight of the weight to be added in LT.


• The first equation was for a weight addition or removal. What do we do for a weight shift? What is different

– Re-examine our first vertical shift equation. What changes?

old old addedweight addedweightnew

old addedweight

KG Kg wKG

w


• So, the final equation for vertical shifts is:

( )old old new oldnew

old

KG w Kg KgKG

w

• Example: A 150 pound person climbs in a 10 pound canoe and sits down. How much has KG shifted? KGold=0.5 ft Kg=?


• Last Comments:

– The general equation covers all cases for a change in KG. This is the equation you should apply to the exams!

HYDROSTATICSTransverse Shift in the Center of Gravity

(3.2.3)

• Shifts “side to side” of the Center of Gravity.

– Starboard is positive and port is negative!

• As in Vertical case, the Transverse movement of “G” may be caused by either (1) addition, (2) removal, or (3) shifting of weights.

HYDROSTATICSTransverse Shift in Center of Gravity (3.2.3)

• Results in a “List” on the Vessel.

– “List” occurs when a vessel is in static equilibrium and down by either the port or starboard side. No external forces are required to maintain this condition and it is permanent unless the Center of Gravity changes.

– “List” is different from “heeling”. Heeling occurs because an external couple is acting on the vessel. Heeling is a more temporary condition.


Example (Listing or Heeling?)


• The Transverse Center of Gravity is referenced in the transverse (athwartships) direction from the centerline of the ship and is labeled TCG.

• The equation used for a transverse shift in the Center of Gravity is the same as was used for the vertical shift! (With some changes in the notation.)

HYDROSTATICSTransverse Shift in Center of Gravity (3.2.3.4)

• Remember a weight shift is just like removing a weight from its original location and adding it to its final location. So for just a weight shift, the generalized equation simplifies to:


old

TCG w TCg TCgTCG

w

• Example: Your 100 LT ship is initially upright. You pump 5 LT of water from a point 15 ft starboard of centerline to 10 ft port of centerline. What is the new TCG? (We will use that answer later to find the angle of heel.”

HYDROSTATICS


old

TCG w TCg TCgTCG

w

Vertical and Transverse Changes in “G”The Key Equations!


old

KG w Kg KgKG

w

When faced with a change in weight (add, sub or move), first sketch it, then solve KG, then solve

TCG!

HYDROSTATICSMetacenter (3.3)

• A reference point for hydrostatic calculations for small angles of roll (less than 10 degrees) or pitch (less than five degrees).• Defined as the intersection of the buoyancy forces and the ship centerline.

HYDROSTATICSMetacenter (3.3)

•The higher the metacenter, the more stable the ship is! •There is a different metacenter for ship pitching in

the longitudinal direction and ships rolling in the transverse direction.

•BMT is for roll, BML for pitch. Which is higher?•If the subscript is omitted, it means BMT.

HYDROSTATICSMetacentric Radius (3.3.1.1)

• The distance from the Metacenter to the Center of Buoyancy is defined as the Metacentric Radius (BM).

Zero pt.

B

MT

K

TIBM

32

3TI y dx

Quick Review• Finding KMT from the Curves of Form

• For a draft of 10 ft…

• Curve 8• Gen’l Scale =

192• 192*0.06 ft

• KMT=11.5 ft

HYDROSTATICSMetacentric Height (3.3.1.2)

• The distance between the Center of Gravity (G) and the Metacenter (M) is defined as the Metacentric Height (GM).

Zero pt.

GMT

HYDROSTATICSMetacentric Height (3.3.1.2)

• Why is GM important?

– If G is below M, then GM is said to be positive. The ship does not want to capsize. This is GOOD!

– If G coincides with M, then GM is said to be zero. A vessel would stay heeled. This is not very good.

– If G is above M, the GM is said to be negative. The ship will tip over. This is REALLY BAD!

Good (positive) GM!The ship wants to roll upright.

Zero pt.

GM

T

B

Bad (negative) GM!The ship wants to roll over. G is either too high or M is too low!

Zero pt.

GM

T

B


• B and M are functions of the hull shape and are generally constant over the life of the ship. G is based on the weights and changes constantly.

•To be safe at sea, we need to find the ship’s KG to make sure it is sufficiently below M!

KG = KM - GMWhere,

• KM is shown on the Curves of Form.

• GM is found from both calculations and by an Inclining Experiement


•KM=KB + BM where:

– KB is found by numerical integration but for most vessels is between 40-50% of the draft– BM is found by:

IT is the “Transverse Moment of Area of the Waterplane” and has the units of ft4

For a box-shaped barge it simplifies to:

TT

IBM

3

12T

LBI

32

3TI y dx

Example stability check

• You have just bought a 30-foot long floating dock, made some modifications and will now put it in the water. It is 6 ft wide and 2.5 feet deep. KG=2 ft and it has a 1 ft draft. Will it be stable? (eg Find GM and determine if it is positive!)

HYDROSTATICSCalculating Angle of List (3.4)

• As a weight shifts across the deck of a vessel, the vessel lists (or “inclines”. How can we predict

the angle of inclination (list)?

• Derivation of Equation

– Draw two vessels, one upright and one listing. Show a weight moving, along with the CG and B.

Calculating Angle of List (3.4)

HYDROSTATICSCalculating Angle of List (3.4.2)

• The weight is shifted causing a shift in the Center of Gravity.

• A moment is created causing the vessel to incline.

• The underwater shape of the hull changes causing the Center of Buoyancy (B) to move until it is in line with the Center of Gravity (G) and the vessel is back in static equilibrium.


• From the geometry and then some substitution, we get:

Zero pt.

tan

tanO F OG G G M

GM w t

W

t

G

M

B


• This equation only works for small angles because it assumes that the Metacenter does not move!

•Note that for small angles, tan = sin! So you can calculate GM from either along the old or new inclined axis.

•Example: You move a 1 LT weight 25 feet to starboard on your 100 LT ship and it lists 2 degrees. What is GM? How would you find KG?

HYDROSTATICSInclining Experiment (3.5)

• Uses small-angle hydrostatics to find the vertical center of gravity (KG) of a ship.

• Process:– A weight is moved a transverse distance, causing a shift in the TCG, and resulting in measurable inclination (list).

HYDROSTATICSInclining Experiment (3.5)

Navy 44 Incline Experiment

HYDROSTATICSInclining Experiment (3.5.1)

• Solving the Angle of Heel equation for the metacentric height (GM), we find:

• The easiest way to do this experiment is to use one set of weights at one distance off centerline. Alas, this would have significant experimental errors, so we measure the inclination with different weights and different positions.

tan

w tGM


• We then plot the data on a graph where the y-axis is the Inclining Moment (wt) and the x- axis is the Tangent of the inclining angle (Tan ).

• The average value of GM can be found from the slope of the line. We can see that:

1 1 1

tan

w t rise yGM slope

run x


• Recall: We want to find the Center of Gravity which can be found by the equation:

KG=KM-GM

• KM is found from the Curves of Form

• GM is found from the Inclining Experiment


• Removing the Inclining Apparatus we must recalculate KG. This is done as a weight removal problem:


old

KG w Kg KgKG

w


•Shipboard Considerations:– No initial list.– Minimum trim.– Dry bilges.– Liquid fuel and oil to be in accordance with the

Shipyard Memo.– Sluice valves closed.– All consumables are to be inventoried.

• Minimum number of personnel remain onboard.• See the example in your text!

HYDROSTATICSLongitudinal Changes in the Center of Gravity

(3.6)• A longitudinal shift in the CG will result in the

vessel having some trim.

• Trim is the difference between the forward and aft drafts, Tf and Ta. It may be calculated by:

T r im = T a f t - T f w dThe Mean Draft is:

Ex. A ship has a draft of 15’ fwd and 16’ aft. Trim = 1 ft


(3.6)

• A vessel is trimmed by the bow when the bow has a deeper draft. This is indicated by a negative trim.

• A vessel is trimmed by the stern when the stern has a deeper draft. This is indicated by a positive trim.

• What is the point which the vessel trims about?


(3.6)

• What happens when a weight is shifted forward or aft?

– The vessel goes down by the bow or stern depending on the direction of the weight

shift.

– Note that the change in trim is independent of the original location of the weight. (i.e. It only matters whether the weight moves forward or aft)


“The Trim Problem” (3.6)

• Draw a picture of what is happening when a vessel trims due to a weight shift:


(3.6)• As the weight shifts forward, a new operating waterline is

created and the draft decreases aft and increases forward.

Taft

Tfwd


(3.6)• We now have two similar triangles and will draw a

third which represents the change in trim.

•Recall: Trim = Taft -Tfwd

•So the total Trim with a change in trim is:

•And with no initial trim, then the change in TRIM is:

aft aft fwd fwdTRIM T T T T

aft fwdTRIM T T


(3.6)• To calculate the final drafts we will need to find:

•Where MT1 is from the Curves of Form (2.10) •We use similar triangles (ratios) to find the change in draft due to the weight shift.

1 aft fwd

w lTRIM T T

MT

aft fwd

aft fwd

T T TRIM

d d LPP


(3.6)

• Example: You have a 1000 x 200 x 90 foot tanker (100,000 LT) with F at Stn 6. It has zero TRIM. You move 1000 LT of oil 450 ft aft. What is the new draft at the stern? MT1~ IL/420L=150,000 FTLT/in

overview chapter 3 - buoyancy versus gravity = stability (see chapter objectives in text) builds on...

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