curve of static stability

44
4.9 Stability at Large Angles of Inclination The transverse metacenter height is a measure of the stability under initial stability’ (aka small angle stability). When the angle of inclination exceeds 5 degrees, the metacenter can be no longer regarded as a fixed point relative to the ship. Hence, the transverse metacenter height (GM) is no longer a suitable criterion for measuring the stability of the ship and it is usual to use the value of the righting arm GZ for this purpose. 1 B

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Page 1: Curve of Static Stability

4.9 Stability at Large Angles of Inclination

The transverse metacenter height is a measure of the stability under ‘initial stability’ (aka small angle stability).

When the angle of inclination exceeds 5 degrees, the metacenter can be no longer regarded as a fixed point relative to the ship. Hence, the transverse metacenter height (GM) is no longer a suitable criterion for measuring the stability of the ship and it is usual to use the value of the righting arm GZ for this purpose.

1B

Page 2: Curve of Static Stability

•The Derivation of Atwood’s Formula : W.L. when the ship is at upright position. : W.L. when the ship is inclined at an angle θ. If the ship section is not vertically sided, the two W.L., underneath which there must be the same volume, do not intersect on the center line (as in the initial stability) but at S.

0 0W L1 1W L

0

0 0

0

sin

sin

Atwood Formula

e i

e i

v h hB R

GZ B R GBv h h GB

Page 3: Curve of Static Stability

0

0 0 0

Atwood Formula

sin sin

e i

e i

v h hB R

v h hGZ B R GB GB

Page 4: Curve of Static Stability

• GZ vs.

For each angle of θ, we compute GZ, the righting arm.

The ship is unstable beyond B. (even if the upsetting moment is removed, the ship will not return to its upright position). From 0 to B, the range of angles represents the range of stabilities.

Page 5: Curve of Static Stability

Ex. Righting arm of a ship vertically sided (A special example to compute GZ at large angle inclinations)

Transverse moment of volume shifted =

3

0 0

1 4 2tan tan tan2 3 3

L L

xy y y dx y dx I

Volume arm

3

0

23

L

xI y dx

Transverse shift of C.B.

0tan tanxI B M

Page 6: Curve of Static Stability

Ex. Righting arm of a ship vertically sided (A special example to compute GZ at large angle inclinations)

Similarly, vertical moment of volume shifted =

Volume Arm

Vertical shift of C.B.

2 3 2

0 0

1 2 1 1tan tan tan tan2 3 3 2

L L

xy y y dx y dx I

21 tan /2 xI

Page 7: Curve of Static Stability

Ex. Righting arm of a ship vertically sided (A special example to compute GZ at large angle inclinations)

0 0

20

2

20 0 0 0 0

20

when 1 small angle inclination, sin / /

1cos sin sin tan sin2

1sin 1 tan2

1 sin sin tan2

1sin tan , 2

x x

x x

x

B R I I B M

I IB R

I

GZ B R B G B M B G B M

GM B M

0

when 1 small angle inclination,

This formula is called 'walled sided formula' or the 'box formula'

where .

.

x

GZ GM

IB M

Page 8: Curve of Static Stability

• Cross Curves of StabilityIt is difficult to ascertain the exact W.L. at which a ship would float in the large angle inclined condition for the same displacement as in the upright condition. The difficulty can be avoided by obtaining the cross curves of stability (see p44).

How to Computing them•Assume the position of C.G. (not known exactly) •W.L. I - V should cover the range of various displacements which a ship may have.

Page 9: Curve of Static Stability

• Cross Curves of Stability Computation Procedures 1. The transverse section area under waterline I, II, III, IV, V

2. The moment about the vertical y-axis (passing through C.G)

3. By longitudinal integration along the length, we obtain the displacement volume, the distances from the B.C. to y-axis (i.e. the righting arm GZ) under the every W.L.

4. For every we obtain

5. Plot the cross curves of stability.

5 ,10 , , and for W.L. I - VGZ

Page 10: Curve of Static Stability

Cross Curves of Stability

These curves show that the righting arm (GZ) changes with the change of displacement given the inclination angle of the ship.

Page 11: Curve of Static Stability

For the sake of understanding ‘cross curves of stability’ clearly, here is a 3-D plot of ‘cross curves of stability.’

The curved surface is

( , )GZ f

Page 12: Curve of Static Stability

• Curve of Static Stability ‘Curve of static stability’ is a curve of righting arm GZ as a

function of angle of inclination for a fixed displacement.

Computing it based on cross curves of stability.

1. How to determine a curve of statical stability from a 3-D of ‘cross curves of stability.’ (C.C.S.), e.g., the curve of static stability is the intersection of the curved surface and the plane of a given displacement.

2. Determining a C.S.S. from 2-D ‘C.C.S.’ is to let displcement = const., which intersects those cross curves at point A, B,…, see the figure.

Page 13: Curve of Static Stability

GZ

Page 14: Curve of Static Stability

• Influences of movement of G.C on ‘curve of static stability’

1. Vertical movement (usually due to the correction of G.C position after inclining experiment.)

1 1 1 sinG Z GZ GG

Page 15: Curve of Static Stability

• Influences of movement of G.C on ‘curve of static stability’

2. Transverse movement (due to the transverse movement of some loose weight)

1 1 1

1

cosG Z GZ GG

whGG

Weight moving from the left to the right

Page 16: Curve of Static Stability

• Features of A Curve of Static Stability1. Rises steadily from the origin and for the first few degrees is

practically a straight line. Near the origin GZ = θ * slope & slope = ?, why?

2. Usually have a point of inflexion, concave upwards and concave downwards, then reaches maximum, and afterwards, declines and eventually crosses the base (horizontal axis).

1 radian

Page 17: Curve of Static Stability

The maximum righting arm & the range of stability are to a large extent a function of the freeboard.

(the definition of freeboard)

Larger freeboard Larger GZmax & the range of stability

Using the watertight superstructures Larger GZmax & the range of stability

Page 18: Curve of Static Stability

4.10 Dynamic Stability

Static stability: we only compute the righting arm (or moment) given the angle of inclination. A true measure of stability should considered dynamically.

• Dynamic Stability: Calculating the amount of work done by the righting moment given the inclination of the ship.

1B

max

max0 0 w wW GZ d W GZ d

Page 19: Curve of Static Stability

• Influence of Wind on Stability (p70-72)Upsetting moment due to beam wind

2 2cos - an empirical coeff. - projected area of the

ship above the W.L - nominal wind velocity - distance from the half

draft to wind pressure center.

- the angle of inclin

M kAhVkA

Vh

ation w.r.t. the beam

Page 20: Curve of Static Stability

When the ship is in upright position, the steady beam wind starts to blow and the ship begins to incline. At point A, the M(wind) = M(righting), do you think the ship will stop inclining at A? Why?

The inclination will usually not stop at A. Because the rolling velocity of a ship is not equal to zero at A, the ship will continue to incline. To understand this, let’s review a simple mechanical problem

Page 21: Curve of Static Stability

Fm

R F

No Friction

X = 0

XX = X1

Hence, the block will continue to move to the right. It will not stop until

The external force F = constant The work done by it

If at the work done by the spring force R,

1FW Fx

1 1,x x R kx F

1 1 210 0

12

x x

RW Rdx kxdx kx 2 2

1 11 1The total work 2 2F R RW W Fx kx mv E

20, at R F RE W W x x

Page 22: Curve of Static Stability

In a ship-rolling case:

Work done by the upright moment

Work done by the wind force

It will stop rolling (at E)

In a static stability curve

or simply,

0

0

up w

wind wind

up wind

W GM d

W M d

W W

OAEGCO DAFGCOD

ODA AEF

Area Area

Area Area

Page 23: Curve of Static Stability

• Consideration in Design (The most sever case concerning the ship stability) Suppose that the ship is inclining at angle and begins to roll back to its upright position. Meanwhile, the steady beam wind is flowing in the same direction as the ship is going to roll.

0

θ0

Wind

max

Standards for USN warships:

1) =30 , 0.055

=45 or , 0.09

30 45 , 0.03

2) 0.2 , at =30

3) occurs at 304) 0.15

GZ

GZ

GZ

MAX

A m rad

A m rad

A m rad

GZ m

GZGM m

Page 24: Curve of Static Stability

Standards of Stability: ships can withstand

1. winds up to 100 knots;

2. rolling caused by sever waves;

3. heel generated in a high speed turn;

4. lifting weights over one side (the C.G. of the weight is acting at the point of suspension);

5. the crowding of passengers to one side.

2 / , - distance between C.G to the half draft

- radius of turning circle, - mass, - velocity of ship.turningM MV h r h

r M V

Page 25: Curve of Static Stability

4.11 Flooding & Damaged Stability

So far we consider the stability of an intact ship. In the event of collision or grounding, water may enter the ship. If flooding is not restricted, the ship will eventually sink. To prevent this, the hull is divided into a number of watertight compartments by watertight bulkheads. (see the figure)

Transverse (or longitudinal) watertight bulkheads can • Minimize the loss of buoyancy

• Minimize the damage to the cargo

• Minimize the loss of stability

1B

Page 26: Curve of Static Stability
Page 27: Curve of Static Stability

Too many watertight bulkheads will increase cost & weight of the ship. It is attempted to use the fewest watertight bulkheads to obtain the largest possible safety (or to satisfy the requirement of rule).

Forward peak bulkhead (0.05 L from the bow)After peak bulkheadEngine room: double bottom Tanker: (US Coast Guard) Double Hull (anti pollution)

This section studies the effects of flooding on the

1) hydrostatic properties 2) and stability

Page 28: Curve of Static Stability

• Trim when a compartment is open to Sea

0 0

1 1

W.L. before the damageW.L. after the damage

W LW L

If W1L1 is higher at any point than the main deck at which the bulkheads stop (the bulkhead deck) it is usually considered that the ship will be lost (sink) because the pressure of water in the damaged compartments can force off the hatches and unrestricted flooding will occur all fore and aft.

Page 29: Curve of Static Stability

• (1) Lost buoyancy method

0

0 0

(

1) Computing the ( ) - volume of lost buoyancy up to - intact . . area; - area of the damage

2) Find the (excluding damage aera ) area , C.F., m y m

vy A av W LA W L a

a A I

parallel sinkage

midway W.L.

) 0 at draft 0.5

( - draft before the damage). cf d y

d

Page 30: Curve of Static Stability

2

2

13) Update the sinkage .

4) Find the trim , Trim (in)420

distance between the C.F. of mean . & centroid of lost buoyancy of .

(0.5 )5) Draft aft T

y

y

m

mfc wd

d

m

vy AI v xL

x W Lv

v LdA L

MTIMTI

rim

(0.5 ) Draft forward Trim

is the distance between C.F. & midshipAassuming that C.F. is aft of midship & the damage takes place of forward.

m

v LdA L

Page 31: Curve of Static Stability

: update the midway . . 0.5 / , then find new , and

repeat the procedures given in the previouspage.

The iteration may be stopped if the results areconvergent, e.g. the mid

m m

W Ld v A ASecond iteration

way . . of the previous iteration & present iteration are different by an amount smaller than a prescribed tolerance error.

W L

Page 32: Curve of Static Stability

Ex. p121-123

A vessel of constant rectangular cross-section L = 60 m, B = 10 m, T = 3 m. ZG = 2.5 m l0 = 8 m.

L1

L0

l0 = 8 m

w1

w0

0

3

2

20

2

1) 3 8 10

240

60 10 600

8 10 80

520

w

w

v T l B

m

A L B m

a l B m

A A a m

2) Parallel sinkage

0240 0.46520

vy mA

Page 33: Curve of Static Stability

3) Draft at midway between W0L0 – W1L1 : 3.232yT m

21 0520 , 0.46 , same as ,

no iteration is necessary in this case. A m y m y

2

3 4

4) Computing , , .

520 ,

New (damaged) . 's. C.F. 4 & 301 (52) 10 longitudinal moment of inertia

12

w MFC

w

MFC

A a I

A A a m

W L m x m

I m

Because

MFC MFCL B G

MFCG B

I IGM Z Z

IZ Z

Page 34: Curve of Static Stability

Moment for Trim per meter:

33

3

231

12

2

1mMTI(m)(m)

1 (52) 10 kNm 1.025 9.81 12 60 m19,637 kN-m

30 240 60343 0.46MTI 60 52 10

3 0.46 2.09 5.55m

1.86

MFC wL w

Lw w

Fw

Lw

A

IGM

L L

vxT T yL

vxT T y mL MCT

Page 35: Curve of Static Stability

(0) (0), , wv x A a

(())(0) (0)w

vyA a

(0 )

12

(0)2

yd d (0)mA

(0)ma

(1)(0) (0)m m

vyA a

Page 36: Curve of Static Stability

(0) (1)y y if

Find trim.

MTI ( at )

(1)y(1)

2yd

420mfcI

MTIL

2(1)

2(1)

L

F

L

A

xvT d yL MTI

xvT d yL MTI

Page 37: Curve of Static Stability

• (2) Added Weight Method (considering the loss of buoyancy as added weight)

also a Trial – error (iterative) method

1) Find added weight v under W0L0. Total weight = W + v

2.) According to hydrostatic curve , determine W1L1 (or T) & trim (moment caused by the added weight & MTI).

3.) Since we have a larger T, and v will be larger, go back to

step 1) re-compute v.

The iterative computation continues until the difference between two added weights v obtained from the two consecutive computation is smaller than a prescribed error tolerance.

Page 38: Curve of Static Stability

• Stability in damaged condition

1

1

Increase in draft increases ,

where is the of the lost buoyancy, is the of the buoyancy recovered on

above the original W.L, the porosity coeff.of the damaged compartment.The

B

B

ZvbbZ

bb

C.G C.G

1

loss of the moment of inertia w.r.t to thelongitudinal axis , thus

(intact) (intact)

d

d

d d

II IKM

I IvbbGM GM GM

Page 39: Curve of Static Stability

• Asymmetric flooding

1. If the inclination angle is large, then the captain should let the corresponding tank flooding. Then the flooding is symmetric.

2. If the inclination angle is small,

w

Lost buoyancy : Heeling moment:

sin ,

sin ,

is the inclination angle.

w

w

w

vv y

GM v yvyGM

Page 40: Curve of Static Stability

•Floodable length and its computation

Floodable Length: The F.L. at any point within the length of the ship is the maximum portion of the length, having its center at the point which can be symmetrically flooded at the prescribed permeability, without immersing the margin line.

Page 41: Curve of Static Stability

Bulkhead deck: The deck tops the watertight bulkhead

Margin line: is a line 75 mm (or 3”) below the bulkhead at the side of a ship

Without loss of the ship: When the W.L. is tangent to the margin line.

Floodable length (in short) The length of (part of) the ship could be flooded without loss of the ship.

• Determine Floodable length is essential to determine1. How many watertight compartments (bulkheads) needed 2. Factor of subdivision (How many water compartments

flooded without lost ship)

Page 42: Curve of Static Stability

0 0 0 0

1 1

1 1 1 1

1 0

Steps for computing the F.L. given , , or 1) Obtain a limit W.L.: , 2) using the Bonjean Curves to obtain and under . 3) , is the loss of buoyancy due to the flooding.

B W LW L

B W Lv v

1 0 11 0 1 1 0

1 0

0

1 0 0 0

4) ,

is the distance of the center of the lost buoyancy from .5) , is the area of the cross section, is the F.L.Setting a half of on either

B BB B v x x x

x Bv A l A l

l

0

sideof the center of lost Buoyancy. Near the ends of a ship, is changing rapidly then using the iterative process to determine .

A

l

Page 43: Curve of Static Stability

6) Repeating (1) to (5) for a series of . .s tangent to the margin line at different positions in the length of the ship. Then a series of values of the F.L. can be obtained for different positions a

W L

longthe ship.7) Considering the different permeability coefficients at different positions along the ship.

Page 44: Curve of Static Stability

8) Factor of Subdivision F

Factor of subdivision is the ratio of a permissible length to the F.L.

For example, if F is 0.5, the ship will still float at a W.L. under the margin line when any two adjacent compartments of the ship are flooded. If F is 1.0, the ship will still float at a W.L. under the margin line when any one compartment of the ship is flooded.

Rules and regulations about the determination of F are set by

many different bureaus all over the world (p126-127)