4.9 stability at large angles of inclination the transverse metacenter height is a measure of the...
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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
•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 L
1 1W L
0
0 0
0
sin
sin
Atwood Formula
e i
e i
v h hB R
GZ B R GB
v h hGB
• 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.
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 tan
2 3 3
L L
xy y y dx y dx I
Volume arm
3
0
2
3
L
xI y dx
Transverse shift of C.B.
0
tantanxI B M
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 tan
2 3 3 2
L L
xy y y dx y dx I
21tan /
2 xI
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 sin
21
sin 1 tan2
1 sin sin tan
2
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
• Cross Curves of Stability
It 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.
• 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
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.
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
• 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.
• 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
• 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
• Features of A Curve of Static Stability
1. 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
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
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
• 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 kAhV
k
A
V
h
ation
w.r.t. the beam
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
F
m
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
1
2
x x
RW Rdx kxdx kx 2 2
1 1
1 1The total work
2 2F R RW W Fx kx mv E
20, at R F RE W W x x
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
• 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 30
4) 0.15
GZ
GZ
GZ
MAX
A m rad
A m rad
A m rad
GZ m
GZ
GM m
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
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
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
• Trim when a compartment is open to Sea
0 0
1 1
W.L. before the damage
W.L. after the damage
W L
W 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.
• (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 a
v W L
A W L a
a A I
parallel sinkage
midway W.L.
) 0 at draft 0.5
( - draft before the damage).
cf d y
d
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 A
I v x
L
x W L
v
v Ld
A L
MTIMTI
rim
(0.5 ) Draft forward Trim
is the distance between C.F. & midship
Aassuming that C.F. is aft of midship & the
damage takes place of forward.
m
v Ld
A L
: update the midway . .
0.5 / , then find new , and
repeat the procedures given in the previous
page.
The iteration may be stopped if the results are
convergent, e.g. the mid
m m
W L
d v A ASecond iteration
way . . of the
previous iteration & present iteration are
different by an amount smaller than a
prescribed tolerance error.
W L
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
0
2400.46
520
vy m
A
3) Draft at midway between W0L0 – W1L1 : 3.232
yT 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 & 30
1 (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
Moment for Trim per meter:
33
3
2
3112
2
1mMTI(m)
(m)
1 (52) 10 kNm 1.025 9.81
12 60 m19,637 kN-m
30 240 60343 0.46
MTI 60 52 10
3 0.46 2.09 5.55m
1.86
MFC wL w
Lw w
F
w
Lw
A
IGM
L L
vxT T y
L
vxT T y m
L MCT
(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 y
L MTI
xvT d y
L MTI
• (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.
• 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
Z
vbbZ
b
b
C.G
C.G
1
loss of the moment of inertia w.r.t to the
longitudinal axis , thus
(intact) (intact)
d
d
d d
I
I IKM
I IvbbGM GM GM
• 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
v
v y
GM v y
vy
GM
•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.
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)
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 L
W L
B W L
v 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 B
v A l A l
l
0
side
of the center of lost Buoyancy.
Near the ends of a ship, is
changing rapidly then using the
iterative process to determine .
A
l
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
long
the ship.
7) Considering the different
permeability coefficients
at different positions along
the ship.
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)