- By
**ervin** - Follow User

- 118 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'Section 3.1' - ervin

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Section 3.1

Archimedes Principle:

The Buoyant Force on an object is equal to the weight of the volume of the water displaced by the object FB=rgV

Forces on a body in water

Distributed forces:

- Gravity: Distributed throughout volume of body based on mass density.
- Buoyancy: Distributed over wetted surface of body based on hydrostatic pressure
- Drag/Lift: Distributed over surface of body based on flow field when moving relative to medium

Box Shaped Barge:

Weight

G

Horizontal

components of

pressure force

are negated by

equal force on

opposite side of

barge.

B

A

FB

z

FB=PA;

P=rgz;

FB=rgzA;

V=zA;

FB=rgV

åF=0

Weight, Buoyancy, Drag and Lift forces all sum to

zero in each dimension

åM=0

All forces in each dimension are colinear and cancel;

i.e. there are no separation of the action points of

forces such that couples or moments are generated.

Example Problem

A boxed shaped barge 50ft wide, 100ft long, and 15ft deep has 10ft of freeboard in sea water:

- What is its draft?
- What is the hydrostatic pressure (psi) acting on the barge’s keel?
- What is the magnitude (LT) of the total hydrostatic force acting on the barge’s keel?
- What is the weight (LT) of the water displaced by the barge?
- Assuming that the buoyant force acts through a single point, what is the location of that point in 3 dimensions?
- Assuming minimum freeboard is 5ft, how many ft³ of coal at 50lb/ft³ can we load on the barge in seawater?
- If we then take the barge into freshwater, what will the new draft be?

Example Answer

Draft=Depth-Freeboard=15ft-10ft=5ft

Phyd=rgz=64lb/ft³×5ft×[1ft²/144in²]=2.22psi

Fhyd=Phyd×A=2.22lb/in²×50ft×100ft×[144in²/ft²]×[1LT/2240lb]=714LT

w=rgV=64lb/ft³×50ft×100ft×5ft×[1LT/2240lb]=714LT

Center of Buoyancy=at amidships, on centerline, 2.5ft above keel

Example Answer

TPI=AWP[ft²]{LT/in}/420=50ft×100ft/420{LT/in}=11.9LT/in

Change in draft=10ft-5ft=5ft×[12in/ft]=60in

Change in weight=60in×11.9LT/in=714LT

V=w/rg=714LT/(50lb/ft³)×2240lb/LT=32,000ft³

Example Answer

Current draft=TSW=10ft

w=rgSWVSW=64lb/ft³×50ft×100ft×10ft=3,200,000lb

VFW=w/rgFW=3,200,000lb/62.4lb/ft³=51,280ft³

TFW=VFW/AWP=51,280ft³/(50ft×100ft)=10.26ft

Increased draft means reduced freeboard below minimum spec

Section 3.2

Center of Mass/Gravity

The weighted average over area or volume based on given distribution summed such that result is equivalent to the total force applied through a single point.

What can change the Center of Gravity?

- Add/subtract weight
- Move weight/change distribution

Notation:

G=Location of Center of Gravity for ship

g=Location of Center of Gravity for object

Ds= Displacement of ship (LT)

W = Magnitude of Gravitational Force/Weight of object (LT)

Bo

B

L

K

C

L

- So far we’ve looked at ships that are in STATIC EQUILIBRIUM:
- SFx = 0
- SFy = 0
- SFz = 0
- SMp = 0

Now let’s take a look at what happens when a weight is added

to disturb this equilibrium

A change in weight (either adding or removing it) will cause a change in

the location of G, the center of gravity of the ship

- A change in VCG (or KG)

- A change in the TCG

TCG

G1

KGnew

- g

Go

B

L

K

It also causes a change in the

longitudinal CG (LCG), but

we’ll discuss that later...

C

L

G1

Go

B

L

K

C

L

When a weight is ADDED, the CG shifts TOWARD the added weight in line

with the CG of the ship and the cg of the weight

G0

B

L

C

L

When a weight is REMOVED, the CG shifts AWAY from the added weight

in line with the CG of the ship and the cg of the weight

- g

K

In the case of a weight SHIFT, the CG first shifts AWAY from the

removed weight….

- g

G2

G0

G1

- g

B

L

C

L

…and the TOWARDS the relocated weight

K

Let’s first consider a weight added directly over the centerline

This will cause the location of the CG to move TOWARD the weight ...

- g

G1

G0

KGold

KGnew

B

L

C

L

… Causing a change in the VERTICAL distance, or KG

K

KGnew =

+ waddx Kg

Dsold+ wadd

Use the concept of weighted averages to determine the new CG:

- g

G1

G0

B

L

K

C

L

Dsoldx KGold

KGnew =

+ (-w) x Kg

Dsold+ (-w)

It’s the same deal for removing a weight, only this time the weight is negative (i.e. removed):

- g

G0

G1

KGnew

KGold

B

L

K

C

L

L

In a relocation of a weight, look at it as SUBTRACTING one weight, and

ADDING another weight.

- g

G1

G0

Kg2

- g

KGnew

Kg1

KGold

B

L

K

w g2g1

- In this unique case, Dsnew and Dsoldand are the SAME THING!
- w1 and w2 are also the same thing!
- The weight has only moved, not been removed
- So we can rearrange the formula:

...This is ONLY for a single vertical weight shift!!

w g2g1

Where:

- GnewGoldis the distance between the old and new CG’s

- g2g1 is the distance between the old and new Cg locations
- of the relocated weight

...This relation will become important in the Inclining Experiment

+ Swix Kgi

KGnew =

Dsold +Swi

We can generalize the formula for vertical changes

in CG by the following:

- Given
- USS CURTS (FFG-38) floats on an even keel at a draft of 17ft
- KG = 19.5ft
- Lpp = 408ft
- It takes on 150LT of fresh water in a tank 6ft above the keel on the CL

- Find
- New vertical center of gravity (KG) after taking on water

G1

?

150

LT

19.5’

6’

G0

B

L

K

C

L

Step 1: Draw picture!

Step 2: Find Ds when floating at 17ft draft

- Go to curves of form for FFG in appendix
- Using curve 1, find the intersection w/ 17ft

Ds = 147 x 30LT

Ds = 4410LT

Step 3: Write the GENERAL Equation

Dsoldx KGold

KGnew =

4410LT x 19.5ft + 150LT x 6ft

4410LT + 150LT

KGnew =

86215.5 LT-ft + 900LT-ft

KGnew =

+ Swix Kgi

4560 LT

Dsnew

Step 4: Substitute in values into the general equation

KGnew =

4560 LT

KGnew =

19.10 ft

CHECK: Does this answer make sense?

YES! The CG shifts toward the added weight, lower than the original CG

Example Problem

A 688 Class Submarine is in port, pier side undergoing a maintenance period. The tender will be pulling periscopes tomorrow which requires the ship to maintain zero list, i.e. TCG=0ft.

The Engineering Dept needs to pump #2RFT dry to perform a tank inspection. What impact will this have on the sub’s TCG?

The sub has 10 MK 48 ADCAP torpedoes at 2LT each. How far and in which direction should these torpedoes be shifted to restore the sub’s TCG to zero?

Data: Do=6900LT Tcg#2RFT= -12ft (i.e. port of centerline);

TCGo=0ft Capacity#2RFT=5000galfw

Example Answer

w#2RFT=rgV

=5000gal×[1ft³/7.4805gal]×rgfw =668.4ft³×62.4lb/ft³×[1LT/2240lb] =18.62LT

Df= D0+Swa-Swr=6900LT-19LT=6881LT

TCGf =(TCG0D0+STcgawa-STcgrwr)/Df=(0ft×6900LT-[-12ft]×19LT)/6881LT =0.033ft (stbd of centerline) (Removed weight from port side)

Example Answer

TCGf=(TCG0D0+STcgawa-STcgrwr)/Df

0ft=(0.033ft×6881LT+dTcg×10torps×2LT/ torp)/6881LT

dTcg = -(0.033ft×6881LT)/20LT = -11.4ft (to port)

Shift 10 torpedoes each 11.4ft to port to compensate for the loss of weight on the port side of the sub.

CL

Section 3.3:What happens when “G” leaves the Centerline?Initial Condition:

G shifts:

Ship responds:

WL

WL

WL

D

D

D

G1

G0

G1

B0

B0

FB

FB

B1

FB

BL

F1

K

K

CL

CL

As the ship lists/trims, the shape of the

submerged volume changes moving

B outboard until it slides under G.

*Since the total weight of the ship has

not changed, the total submerged volume

remains constant, but its shape changes.

K

CL

CL

Ship responds

to opposite weight

shift:

Where the lines of action of the various

centers of buoyancy cross* is the Metacenter

M

F2(-)

F1(+)

WL

G2

WL

G0

B2

B1

FB

B2

FB

B0

BL

F2

K

CL

*Lines of action cross at a single point only for

“small” angles of inclination (<10º).

Shapes which impact KM:

M

M

WL

B2

B1

B0

WL

B1

B2

BL

B0

K

BL

CL

K

CL

Highly curved hull cross-section:

Little buoyant volume at large

lever arm: M is at/near center of

curvature

Very flat hull cross-section:

Large buoyant volume at large

lever arm: M is high

Distance from G to MT = Metacentric Height = Major player in stability calculations (+ keeps ship upright)

Distance from B to MT = Transverse Metacentric Radius

ML

MT

GMT

WL

BMT

G

KMT

B

KG

KB

BL

TCG/TCB (-)

TCG/TCB (+)

K

CL

Locations and Line Segments for Hydrostatic Calculations

Example Problem

G2

G1

Radius

=3ft

A rocking chair’s “skids” have a radiusof curvature of 3ft. The chair’s initial center of gravity is 2.5ft above the skids. A box is put on the seat which raises the combined center of gravity to 3ft above the skids. Another box is put on top of the first which raises the combined center of gravity to 3.5ft above the skids.

- For each of these conditions, when the chair is tipped 45°, show how the forces of gravity and support are spatially related and predict how the chair will react when released.
- What point in this scenario is analogous to a ship’s metacenter?

Example Answer

G1

G1: Support is outboard center of gravity creating a couple which returns the chair upright.

G2: Support is aligned with center of gravity eliminating any couple. The chair maintains position.

G3: Support is inboard center of gravity creating a couple which tips the chair over.

The center of curvature of the rocking chair’s “skids” correspond to a ship’s metacenter.

Support

G2

Support

G3

Support

Section 3.4:Angle of List for Small Angles after Transverse Weight Shift

For a given transverse weight shift, what is the corresponding change in list angle?

g0

t

MT

gf

F

G0

D

WL

Gf

B0

Bf

FB

CL

BL

Bo

B

L

K

C

L

- Up to now we’ve considered ship’s floating on an even keel
- …(no list or trim).
- The following points are noted:
- K, keel
- B, center of buoyancy
- G, center of gravity

One point of particular note remains….

…MT, or the Transverse Metacenter

Bo

B

L

K

C

L

The Transverse Metacenter (MT) represents a convenient point of reference for

small changes in the angle of inclination, F, (less than 10o)

MT

B

L

C

L

For small changes in inclination, the point MT is where the ship is assumed

to rotate.

MT

F

Go

Bo

K

...The MT is generally about 10-30ft above the keel

O

)

(

There is also a Longitudinal Metacenter, or ML...

ϑ

…usually in the magnitude of 100 to 1000ft above the keel

When the ship reacts to an off-center load (which

will change the ship’s CG),...

...the center of buoyancy will shift

until it is vertically aligned

with the new CG...

G1

…G1 can be assumed to move

PERPENDICULAR from the CL

B1

B

L

C

L

FB

MT

F

Go

Bo

Remember, this is only for listing

of 10o or less

The short leg is G0G1

The long leg is G0MT

The hypotenuse is G1MT

tan F = G0G1

G0MT

Look at the right triangle formed by this

shifting…

F

Go

G1

SO….

Bo

B1

C

L

(tan F = opp/adj… remember?)

FB

G0MT

MT

Go

KMT

- From the Curves of Form you can get KMT
- The Vertical Center of Gravity is KG0
- G0MT =KMT - KG0

Bo

KG

C

L

K

With that fact understood, we can now determine the ANGLE OF LIST

of a vessel due to a change in loading.

How?

- G0G1is the change in the transverse Center of Gravity

The USS Simpson (FFG-56) floats on an even keel at a 16ft draft. The

KG is 20ft above the keel. After 1 week, 50LT of fuel has been used from

a tank 11ft to port and 15ft above the keel.

Find the angle of list after the fuel has been used.

Step 1: Find the ship’s displacement

From the curves of form, curve #1, 16ft draft crosses at 132

132 x 30LT = 3960LT

Ds1

KG1 = 3960LT x 20ft - (50LT x 15ft)

(3960 - 50)LT

KG1 = 78,450LT-ft

3910LT

Step 2: Find the new vertical CG (KG)

KG1 = 20.06ft

Ds1

TCG1 = 3960LT x 0ft - (50LT x -11ft)

(3960 - 50)LT

TCG1 = 0 - (-550LT-ft)

3910LT

Step 3: Find the Transverse CG (TCG)

(minus because it’s to port)

TCG1 = 0.141ft

(shifts to starboard, away from

removed weight)

G0G1 is the change in the Transverse CG:

- G0 = 0 (on the centerline)
- G1 = .141ft

G0G1 = .141ft

- G0MT =KMT - KG0
- KMT from curves is 113 x .2ft = 22.6ft
- KG0 = 20ft

G0MT = 2.6ft

Step 4: Define lengths of G0G1 and G0MT

tan f = opposite

adjacent

2.6ft

tan f = G0G1

G0MT

tan f = .141ft

2.6ft

.14ft

Step 5: (Almost there!) Find tan f:

F

Go

G1

tan f = 0.0541

atan 0.0541 = f

3.10o = f

C

L

Section 3.5The Inclining Experiment

In the previous section, we derived the relationship between a shift in weight and the resultant list/trim angle:

tan(F) = wt/(DG0MT)

- w,t are the weight and distance moved – usually known
- The location of MT and the magnitude of D are properties of the hull shape read from the Curves of Form for the appropriate draft (T).
- How do we find the location of G0?

How do we find the location of G0?

We determine it experimentally after new construction for a class or any major permanent complex weight redistributions for a given ship (alteration/conversion).

Inclining Experiment Procedure:

1. Configure the ship in a “light” condition

2. Bring on large weights (~2% of Dship), move to known distances port and starboard of centerline and measure tan(F) using “plum bob”. Measure & record Dincl using draft and Curves of Form.

3. Plot wt vs. tan(F); divide slope by Dincl to get GinclMT

4. Calculate KGincl = KMT(from Curve of Form)–GinclMT

5. KG0=KGlight=(KGinclDincl–Kginclwtswinclwts)/(Dincl–winclwts)

Inclining Experiment Tools

-Plot:

-Plumb Bob:

Mast

dadj

tan(F)=dopp/dadj

F

Inclining Moment, wt (LT-ft)

Scale

dopp

Tangent of Inclining Angle (Tan[F])

So far we’ve established that the

angle of list can be found using

the right triangle identified here:

F

The short leg is G0G1

The long leg is G0MT

The hypotenuse is G1MT

Go

G1

SO….

Bo

tan F = G0G1

G0MT

B1

C

L

FB

...And so we can find the angle of list

F

Up to now, however, G0MT has

been given based upon a KG that

has been provided.

Go

G1

We’ll now see how KG can be

found by determining G0MT

Bo

B1

C

L

FB

This is done by the Inclining Experiment

By using a known weight and placing it at a known distance

an angle of list can be measured

By repeating this process - port and starboard- we can

graph the relationship between the moment created by the

weight and the angle of inclination

This will allow an average inclined KG to be determined,

and from that a KG for the ship in an condition of no list or

trim can be established

G0MT

In earlier discussions an equation was derived for a shift in of

a single weight:

Ds G0G1 = w g0g1

…where g0g1 was the distance that the weight was shifted. Let’s call that distance “t”. Sooo,...

Ds G0G1 = wt

And re-look at the equation for the angle of list:

G0MT

Ds G0G1 = wt

G0G1 = wt

Ds

G0MT tan F = G0G1

wt

Ds

G0MT tan F =

Note that the common term in both equations is G0G1. So

let’s isolate it in each equation:

tan FDs

G0MT =

wt

Ds

G0MT tan F =

That’s nice,… but not nice enough... One more rearrangement

and we’ll have what we really want, G0MT:

tan FDs

G0MT =

- Let’s review what we know:
- “w” is a known weight that is relocated
- “t” is the distance the weight is moved
- “tan f” is the angle created by the weight shift
- “Ds” is the displacement of the ship

This will be the formula that governs the Inclining Experiment

tan FDs

G0MT =

- In the Inclining Experiment:
- The distance “t” is varied, changing the angle of list, tan f
- “w” and Ds will remain constant
- By varying t, thus varying the created moment of wt,

the angle of inclination will change

By plotting wt versus tan f, you can determine the average G0MT

tan FDs

G0MT =

DWt

Dtan f

Dy

Dx

=

(slope of wt vstan F curve)

Ds

AverageG0MT =

Remember, slope is Dy/Dx:

Or...

So...

Ds

AverageG0MT =

When you vary the distance t, and thus the moment, you’ll vary the inclination

angle. The result is plotted in an example here:

The slope of the “best fit” line, Dy/Dx, when divided by the displacement,

will give the average G0MT distance:

Having found the AverageG0MT, you can find the KG when the

ship is loaded with the inclining weight:

KG = KMT - G0MT

KG light = KG inclinedxDs old - Kg x w

Ds new

The problem now degenerates to a simple “’change in vertical

center of gravity, KG, equation:

KG light, the KG of the ship with considering the ship’s weight only-

no crew, stores, fuel, etc.- is what we wanted!!

- Using a known weight and a measured distance, a moment is created
- The moment creates a list that can be measured
- By repeating the process with the same weight over different distances and plotting the results, the average G0MT can be found
- Once G0MT is found, you can find KG of the light ship

Example Problem

- The USS OHIO has just completed her Overhaul and Conversion from an SSBN to an SSGN and Special Operations Forces platform. She is pierside performing a required Inclining Experiment. Dlightship=18700LT; KMT=21ft. The inclining gear weighs 400LTs and is centered 47ft above the keel. 375LTs is moved to the following transverse distances resulting in the corresponding list angles.Distance to Starboard(ft) List Angle(°) -50 -12.8 -25 -6.4 0 0 25 6.5 50 12.7

What is KGlight?

Example Answer

Multiply transverse distances by 375LT to get inclining moment. Take tangent of list angle and plot the two derived sets of data against one another:

Example Answer

Slope=(18750-[-18750]LT-ft)/(.225-[-.227]) =83000LT-ft

GMTincl=slope/Dincl=83000LT-ft/19100LT =4.35ft

KGinc l=KMT-GMTincl=21ft-4.35ft=16.65ft

KGlight =(KGinclDincl-Kgwtswwts)/Dlight =(16.65ft×19100LT-47ft×400LT)/18700LT =16ft

Section 3.6 Longitudinal Changes

Tm=(Taft+Tfwd)/2

Trim=Taft - Tfwd

- If ship is “trimmed by the stern”,
- Bow is up
- Taft> Tfwd
- Trim is (+)

DWL

WL

Taft

Tfwd

F

Ap

Fp

daft

dfwd

The change in draft will be evident in a change of draft

forward...

…and aft.

dTaft

w

_

.

dTrim

dTfwd

The difference between the fore and aft drafts is the change in trim:

Trim = dTaft - dTfwd

The ship will pivot about the center of flotation, F.

F

Graphically, it looks like this. First, the ship is represented with a line

representing its initial state:

FP

AP

)(

O

_

.

F

As weight is added, the the ship rotates about F:

)(

O

_

.

FP

AP

F

w

You can simulate this on your paper by turning the sheet in the direction that the

bow or stern would sink because of the added weight, then drawing a line to

represent the new position.

The changes in draft can now be read directly…

dTfwd

dTrim

)(

O

_

.

w

F

dTaft

Now, rotate the sheet so that the line drawn becomes level and acts as

the new waterline:

)(

O

_

.

AP

w

FP

F

dTaft is below the WL, so it’s subtracted. dTfwd is above the waterline, so it’s added to the draft.

1. Change in draft due to the parallel sinkage of the vessel due to the added weight, “w”:

dTPS = w

TPI

2. Change in draft due to the moment created by the added weight

at a distance from F, or “wl”:

dTrim = wl

MT1”

There are two aspects of draft to consider when finding the change in draft:

These two measurements- change due to parallel sinkage and

change in trim due to moment- when added with the initial

draft will give you the TOTAL draft, forward and aft:

dTrim = wl

MT1”

dTPS = w

TPI

Tfwd new = Tfwdold +/- dTPS +/- dTrim

AND

Taft new = Taftold +/- dTPS +/- dTrim

Let’s consider change due to the parallel sinkage of the vessel first:

dTPS = w

TPI

- TPI, Tons Per Inch Immersion is a geometric function of the vessel at
- a given draft and is taken from the Curves of Form
- The added weight, w, will cause the vessel to “sink” a small distance for
- the length of the entire vessel
- We assume that the weight is applied at F! This assures that the sinkage
- is uniform over the length of the ship

Now consider the change in trim due to the created moment of the added

weight:

dTrim = wl

MT1”

MT1”, or the Moment to Trim 1”, is also from the Curves of Form

The weight, w, at a distance, l, from the center of flotation, F, creates

a moment that causes the ship to rotate about F

This rotation causes one end to sink and the other end to rise

The degree of rise or fall depends on the location of F with regard to

the entire length of the ship as given by Lpp

MT1”

The value for dTrim will be for the entire length of the ship:

l

dTfwd

dTrim

)(

O

_

.

w

F

dTaft

Lpp

...Now we need to find how much of the trim is aft and how

much is forward!

To find the trim distribution, consider the similar triangles formed below:

l

dTfwd

dTrim

)(

O

_

.

w

F

dTaft

daft

dfwd

Lpp

The largest triangle shows the TOTAL change in trim, dTrim

The hatched green triangle shows the forward trim dTfwd

The hatched yellow area triangle shows the aft trim, dTaft

dTaft

dTfwd

Lpp

dfwd

daft

=

=

l

dTfwd

dTrim

)(

O

_

.

w

F

dTaft

daft

dfwd

Lpp

For these similar triangles there is a ratio aspect that relates to each:

(The short leg divided by the long leg of the triangle!)

Knowing how to find the change in draft from both parallel sinkage and from the induced moment, you can now find the total draft change, fore and aft:

Tfwd new = Tfwdold +/- dTPS +/-dT

AND

Taft new = Taftold +/- dTPS +/-dT

Calculating Draft Changes

Procedure:

- Calculate impact of weight addition/removal to mean draft using TPI.
- Calculate impact of weight addition/removal to trim at given distance from center of floatation.
- Calculate trim effect on fwd and aft drafts separately.
- Separately add mean draft impact to trim effects to determine final drafts fwd and aft.

Lpp = 101.7 ft Draft = (10.5 + 10.1)/2 = 10.3ft

amidships = 50.85 ft Ds = 2LT x 205 = 410LT

LCF = 55.8 ft from FP, or 4.95 ft aft of amidships

DRAW A PICTURE!

Dfwd = 55.8

Daft = 45.9

19.95

)(

dTfwd

O

_

.

10LT

dTrim

dTaft

F

101.7

Example:

The YP floats at a draft 10.5 ft aft and 10.1ft forward. A load of 10LT is placed

15ft forward of amidships. Find the final forward and aft drafts.

Daft = 45.9

19.95

)(

dTfwd

O

_

.

10LT

dTrim

dTaft

F

101.7

Step 1: Find change due to parallel sinkage

dTPS = w

TPI

dTPS = 10LT

235 x .02LT/in

dTPS = 2.13in

Daft = 45.9

19.95

)(

dTfwd

O

_

.

10LT

dTrim

dTaft

F

101.7

Step 2: Find change due to moment

dTrim = wl MT1”

dTrim = 10LT x 19.95ft 252.5 x .141 LT-ft/in

dTrim = 5.60in

dTaft

dTfwd

=

=

Dfwd = 55.8

Daft = 45.9

Lpp

daft

dfwd

19.95

dTaft

dTfwd

4.21in

=

=

)(

dTfwd

O

_

.

101.7ft

45.9ft

55.8ft

10LT

dTrim

dTaft

F

101.7

Step 3: Divide the dTrim based on similar triangles

5.60in

dTaft

2.53 in

=

45.9ft x

=

101.7ft

5.60in

3.07 in

=

dTfwd

55.8ft x

=

101.7ft

Daft = 45.9

19.95

)(

dTfwd

O

_

.

10LT

dTrim

dTaft

F

101.7

Step 4: Sum the changes in draft fore and aft

Forward:

Tfwd new = Tfwdold +/- dTPS +/-dTmoment

Tfwd new = 10.1ft + (2.13in + 3.07in) x (1ft/12in)

Tfwd new = 10.1ft + .43ft

Tfwd new = 10.53ft

Daft = 45.9

19.95

)(

dTfwd

O

_

.

10LT

dTrim

dTaft

F

101.7

Step 4: Sum the changes in draft fore and aft

Aft:

Taft new = Taftold +/- dTPS +/-dTmoment

Taft new = 10.5ft + (2.13in - 2.53in) x (1ft/12in)

Taft new = 10.5ft - .033ft

Taft new = 10.467ft

Background Lab 2

Lab Objectives

- Reinforce students’ understanding of Archimedes Principle
- Reinforce student’s concept of static equilibrium
- Reinforce student’s concept of the center of floatation

Background Lab 2

Concepts/Principles:

- Archimedes Principle
- Static Equilibrium
- Center of Floatation
- Simpson’s First Rule
- Interpolation
- Hydrostatic Force
- TPI
- MT1”

Background Lab 2

Terminology

- Displacement
- Buoyant Force

Equations

- D=rgÑ=FB

General Safety

- Immediately clean up any water spilled to avoid fall hazard

Apparatus

Equipment

- Floating bodies
- Tanks with weirs and spillways
- Buckets
- Scale
- Rulers
- 5 lb weights

Procedures for taking measurements

- Record results measurements of models and weighing of buckets

Data Collection/Reduction

Data to be collected & Expected results

- These should be equal
- Weight of model
- Weight of water
- Calculated water volume displaced
- Hydrostatic Force
- Longitudinal Center of Floatation (LCF)

Sources of error

- Measurements
- Insufficient drip time

Section 3.8: Dry Docking

How is the ship’s weight shared between docking blocks and buoyant force?

Requirements for Static Equilibrium still apply:

- SF=0; SM=0
- SFV=(-)D+FB+Fblocks=0
- FB=rgÑS
- D =rgÑS+ Fblocks
- Since ship’s weight remains constant, as hull comes out of water, submerged volume decreases, hence buoyant force decreases, and force from the blocks increases.
- (P= Fblocks)

Dry-Docking

If a list develops during docking, the increasing force from the blocks can work to capsize the ship

Solutions:

- Use side blocks to force a zero list
- Stop docking evolution and correct problem, if ship develops an increasing list

M

D

M

D

G

G

WL

WL

B

B

FB

Fblock=P=D-FB

FB

P

Disturbance

trying to roll

the ship

D0

M

Gf

G0

WL

B

P=weight

removed

FB

Impact on StabilityConsider force of blocks to be the same as a weight removal from the keel:

- What is the impact on KG and GMT?
- Df= D0-wr= D0-P
- Ship’s weight/displacement is decreased
- KGfDf= KG0D0-Kgrwr, but Kgr=0;
- KGfDf= KG0D0;
- KGf= KG0D0/Df= KG0D0/(D0-P);
- Center of Gravity moves up due to keel weight removal
- GMT= KMT – KGf
- Shorter distance between Center of Gravity and Metacenter gives less distance to develop a righting moment

Comparison to Grounding:

Same stability concerns for both evolutions although grounding is obviously not planned or controlled.

Since re-floating after grounding is generally not on level sea bed with a zero list, it should only be done at highest available tide to maximize buoyant force and righting moment and avoid capsizing.

Pulling the ship directly off the shoal.

D

M

G

WL

B

FB

Fground=P=D-FB

Floating the Ship

Undocking has the same concerns as docking plus:

- The Center of Gravity may have been shifted by the work done in dock.
- All holes in the ship below the waterline need to be confirmed properly closed.

Recovery from grounding concerns:

- The Center of Gravity may have been changed by flooded or damaged compartments.
- When ship floats again, damage previously held above the water could be submerged resulting in further damage.

Example Problem

DD963 is preparing to enter drydock. It is currently moored pier side on an even keel and a draft of 18.5 feet. To ensure that the sonar dome rests properly on the blocks, the forward draft of the ship must be Tf=17.5 feet. How much ballast must be removed from a tank located 100 feet forward of amidship? Give the answer in gallons of saltwater.

Lpp=465 feet

TPI=50LT/in

MT1”=1400ft-LT/in

LCF=25 feet aft of amidships

Example Answer

100ft

25ft

w

232.5ft

l=125ft

amidship

F

FP

Lpp=465ft

AP

Tfinal fwd=Tinitialfwd±dTps±dTfwd

dTps=w/TPI

dTfwd=dTrim×Dfwd/Lpp

dTrim=wl/MT1”

Tfinal fwd= Tinitial fwd± w/TPI ± wl/MT1”×Dfwd/Lpp

= w/TPI ± wl/MT1”×Dfwd/Lpp =(17.5ft-18.5ft)×12in/ft= -12 in

=-w/(50LT/in) – w(125ft)/(1400ft-LT/in)×257.5ft/465ft= (-)12in

-12 in = -w/(50LT/in) – w/(20.23LT/in)

= -w/(14.4LT/in)

w= -12in×(-14.4LT/in)=172.8LT

Daft=207.5ft

Dfwd=257.5ft

V=w/(rg)=172.8LT/[(64lb/ft³)×2240lb/LT×7.4805gal/ft³]=45,243gal

This is just another application of moments!

Example Problem

An FFG-7 is in the process of undocking when the evolution is halted at 10ft of water on the hull.

- If D=3600LT, how much weight is being supported by the blocks?
- If the water level is raised 1in, how much additional weight is removed from the blocks?

Example Answer

At T=10ft, FB= 62×30LT = 1860LT;

P=D-FB=3600LT-1860LT = 1740LT

At T=10ft, TPI=128×0.2LT/in = 25.6LT/in;

Raising water level 1in removes an additional 25.6LT from the blocks

Background Lab 3

Lab Objectives

- Reinforce students’ understanding of the theory behind inclining experiments
- Provide students with practical experience in conducting an inclining experiment
- Determine the KG of the 27-B-1 model for future laboratories

Background Lab 3

Terminology

- Light-ship condition
- Inclined ship condition
- Plum bob

Equations

- GinclMT= wt/tan(F)×1/D
- KGincl = KMT(from Curve of Form)–GinclMT
- KG0=KGlight=(KGinclDincl–Kginclwtswinclwts)/(Dincl–winclwts)

Apparatus

General Safety

- Minimize water on the floor

Equipment

- 27-B-1 Models
- Weights
- Plum bobs

Procedures for taking measurements

- Record measurements

Data Collection/Reduction

Data to be collected & Expected results

- 27-B-1 Model Numbers
- Weight of Models
- Drafts
- Model dimensions
- Water temperature
- tan(F)
- Where do you expect KG to be?

Sources of error

- Measurement error
- Round off

Review of Chapters 1-3for Six Week Exam

- Chapter 1: Engineering Fundamentals
- Chapter 2: Hull Form and Geometry
- Chapter 3: Hydrostatics
- Review Equation & Conversion Sheet

Chapter 1: Engineering Fundamentals

- Drawings, sketches, graphs
- Dependent/independent variables
- Region under and slope of a curve
- Unit analysis
- Significant figures
- Linear interpolation
- Forces, moments, couples, static equilibrium, hydrostatic pressure, mathematical moments
- Six degrees of freedom
- Bernoulli’s Equation

Chapter 1: Engineering Fundamentals

- Force × distance
- Equal and opposite forces applied with an offset distance to produce a rotation
- åF=0; åM=0
- P= rgz
- Mx=òydA
- Translational: heave, surge, sway
- Rotational: roll, pitch, yaw
- List, trim, heel
- p/r+V²/2+gz=constant

Chapter 2: Hull Form and Geometry

- Categorizing ships
- Ways to represent the hull form
- Table of Offsets
- Hull form characteristics
- Centroids
- Center of Flotation, Center of Buoyancy
- Simpson’s Rule
- Curves of Form

Chapter 2: Hull Form and Geometry

Plans

Body: Section Lines

Sheer: Buttock Lines

Half-Breadth: Waterlines

Depth(D), draft(T), beam(B), freeboard

Centroid (location): LCF=(2/AWP)*òxdA

Center of waterplane area

Center of submerged volume

òydx=Dx/3*[1y0+4y1+2y2+4y3+…+2yn-2+4yn-1+1yn]

D, LCB, KB, TPI, AWP, LCF, MT1”, KML, KMT

Draft->proper curve, proper axis, proper multiple/units

Y

y(x)

Simpson IntegralsHalf-

Breadths

(feet)

dx=Station Spacing

X

0

Stations

See your “Equations and Conversions” Sheet

Waterplane Area

AWP=2òydx; where integral is half breadths by station

Sectional Area

Asect=2òydz; where integral is half breadths by waterline

Z

(Body Plan)

dz=Waterline Spacing

Water

lines

y(z)

Y

0

Half-Breadths (feet)

Simpson Integrals

Asect

A(x)

Sectional

Areas

(feet²)

dx=Station Spacing

See your “Equations and Conversions” Sheet

Submerged Volume

- ÑS=òAsectdx; where integral is sectional areas

by station

Longitudinal Center of Floatation

LCF=(2/AWP)*òxydx; where integral is product of distancefrom FP & half breadths by station

X

0

Stations

(Half-Breadth Plan)

Y

y(x)

Half-

Breadths

(feet)

dx=Station Spacing

x

X

0

Stations

Chapter 3: Hydrostatics

- Archimedes Principle/Static Equilibrium
- Impact to G of weight addition, removal, movement
- Metacenter
- Angle of list
- Inclining Experiment
- Trim calculations
- Drydocking

Chapter 3: Hydrostatics

The Buoyant Force on an object is equal to the weight of the volume of the water displaced by the object: FB=rgV

For box shaped barge, FB= rgV = P×Awp= rgzAwp

åF=0; åM=0

Center of Gravity (G)

Df= D0+Swa-Swr

KGfDf= KG0D0+SKgawa-SKgrwr

TCGfDf= TCG0D0+STcgawa-STcgrwr

gf

Gf

WL

Gi

G0

g0

G moves parallel to

weight shift

BL

K

CL

MT

F2(-)

F1(+)

GMT

WL

WL

BMT

G

KMT

B2

B1

G0

B

KG

FB

KB

B0

BL

BL

K

K

CL

TCG/TCB (-)

CL

TCG/TCB (+)

Chapter 3: HydrostaticsM

tan(F) = wt/(DG0MT)

To find KG:

- Plot wt vs. tan(F); divide slope by Dincl to get GinclMT
- KGincl = KMT(from Curve of Form)–GinclMT
- KG0=KGlight=(KGinclDincl–Kginclwtswinclwts)/(Dincl–winclwts)

Chapter 3: Hydrostatics

Trim Equations:

- dTPS=w/TPI
- dTrim=wl/MT1”
- dTfwd/aft/dfwd/aft =dTrim/Lpp
- Tfinal fwd/aft=Tinitial fwd/aft±dTPS±dTfwd/aft

Weight

Added

dTPS

F

l

w

dTfwd

q

dTrim

dTaft

Tfinal fwd

daft

dfwd

Ap

Lpp

Fp

Tfinal aft

General Problem Solving Technique

Write down applicable reference equation which contains the desired “answer variable”.

Solve the reference equation for the “answer variable”.

Write down additional reference equations and solve for unknown variables in the “answer variable” equation, if needed.

Draw a quick sketch to show what information is given and needed and identify variables, if applicable.

Rewrite “answer variable” equation, substituting numeric values with units for variables.

Simplify this expanded equation, including units, to arrive at the final answer.

Check the answer:

Do units match answer?

Is the answer on the right order of magnitude?

Summary

- Equation Sheet
- Assigned homework problems
- Additional homework problems
- Example problems worked in class
- Example Problems worked in text

Download Presentation

Connecting to Server..