STABILITY AND TRIM OF MARINE VESSELS
Transcript of STABILITY AND TRIM OF MARINE VESSELS
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STABILITY AND
TRIM OF MARINE
VESSELS
Massachusetts Institute of Technology, Subject 2.017
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Concept of Mass Center for a Rigid Body
G
Centroid – the point about
which moments due to
gravity are zero:
6 g mi (xg – xi) = 0 Æ
x = 6�mi xi / 6�mi = 6�mi xi / Mg
• Calculation applies to all three body axes: x,y,z
• x can be referenced to any point, e.g., bow,
waterline, geometric center, etc.
• “Enclosed” water has to be included in the mass
if we are talking about inertia Massachusetts Institute of Technology, Subject 2.017
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Center of Buoyancy
A similar differential approach with displaced mass:
xb = 6 ' i xi ��'����where ' i is incremental volume,
' is total volume
Center of buoyancy is the
same as the center of Images removed for copyright reasons.
displaced volume: it doesn’t
matter what is inside the
outer skin, or how it is
arranged.
Calculating trim of a flooded vehicle: Use in-water weights of the
components, including the water (whose weight is then zero and can be
ignored). The calculation gives the center of in-water weight.
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•
body, a sufficient
condition for
stability is that zb
is above z g.
For a submerged
g
Righting arm:
h =(zb-z g)sinT
TRighting
moment:
U'gh
G
B
Make (zb-zg) large Æ the “spring” is large and:
• Response to an initial heel angle is fast (uncomfortable?)
• Wave or loading disturbances don’t cause unacceptably
large motions
• But this is also a spring-mass system, that will oscillate
unless adequate damping is used, e.g., sails, anti-roll planes,
etc. Massachusetts Institute of Technology, Subject 2.017
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slice thickness dx
provided by the
b
intuition: wedges• In most surface vessels, righting
Tstability is
submerged volume d'�� waterplane area. d' = Adx
B2 B1
h
l
K
l/3
b/6
G
T
M
g y
Massachusetts Institute of Technology, Subject 2.017
RECTANGULAR SECTION
Geometry:
d'/dx = bh + bl/2 or
h = ( d'�dx – bl/2 ) / b
l = b tanT
dF
dF
Vertical forces:
dFG = U�g d' �no shear)
B1 = U�g b h dx
B2 = U�g b l dx / 2
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y
Moment arms:
G = KG sin T������yB1 = h sin T / 2 ; yB2 = (h + l/3) sin T + b cos T / 6
Put all this together into a net moment (positive anti-clockwise):
dM/Ug = -KG d' sinT + bh2 dx sinT / 2 + (valid until the corner
b l dx [ (h+l/3) sinT + b cosT / 6] / 2 comes out of the water)
Linearize (sinT ~ tanT ~ T), and keep only first-order terms (T):
dM ��U�g d' �>��KG ��d' / 2 b dx ��b��dx �����d'@��T
For this rectangular slice, the sum >d' / 2 b dx + b3 dx / 12 d'@ must
exceed the distance KG for stability. This sum is called KM – the
distance from the keel up to the “virtual” buoyancy center M. M is the
METACENTER, and it is as if the block is hanging from M!
-KG + KM = GM : the METACENTRIC HEIGHT
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Note M does notM have to stay on the
centerline**!
G R: righting arm
R “down-flood” damage,
flooding, ice
Images removed for copyright reasons.heel angle Ships in stormy waves.
How much GM is enough?
Around 2-3m in a big boat
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Considering the Entire Vessel…
Transverse (or roll) stability is calculated using the same moment calculation extended on the length:
Total Moment = Integral on Length of dM(x), where (for a vessel with all rectangular cross-sections)
dM(x) = Ug [ -KG(x) d'�x� +d'2(x) / 2 b(x) dx + b3(x) dx / 12 ] T������or
dM(x) �Ug [ -KG(x) A(x) dx + A2(x) dx / 2 b(x) + b3(x) dx / 12@�T
First term: Same as –U�g KG ', where ' is ship’s submerged volume, and KG is the value referencing the whole vessel.
Second and third terms: Use the fact that d' and b are functions of x. Notice that the area and the beam count, but not the draft!
Longitudinal (or pitch) stability is similarly calculated, but it is usually secondary, since the waterplane area is very long Æ very high GM
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Weight Distribution and Trim
• At zero speed, and with no
other forces or moments,
the vessel has B
(submerged) or M (surface)
directly above G.
G
M
Too bad!
For port-stbd symmetric hulls, keep G on the centerline using a
tabulation of component masses and their centroid locations in the
hull, i.e., 6 mi yi = 0
Longitudinal trim should be zero relative to center of waterplane
area, in the loaded condition.
Pitch trim may be affected by forward motion, but difference is
usually only a few degrees.
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Rotational Dynamics Using the Centroid
Equivalent to F = ma in linear case is
T = J * d2T�/ dt2 o
where T is the sum of acting torques in roll
Jo is the rotary moment of inertia in roll,
referenced to some location O
T is roll angle (radians)
J written in terms of incremental masses mi :
J = 6 mi (yi-yo )2 OR J = 6 mi (yi-yg )
2 o g
J written in terms of component masses mi and their own moments of
inertia Ji (by the parallel axis theorem) :
J = 6 mi (yi-yg )2 + 6 Jig
The yi’s give position of the centroid of each body, and Ji’s are
referenced to those centroids
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What are the acting torques T ?
• Buoyancy righting moment – metacentric height
• Dynamic loads on the vessel – e.g., waves, wind, movement of
components, sloshing
• Damping due to keel, roll dampers, etc.
• Torques due to roll control actuators
An instructive case of damping D, metacentric height GM:
J d2T / dt2 = - D dT /dt – GM U g ' T OR
J d2T / dt2 + D dT /dt + GM U g '�T = 0
d2T / dt2 + a dT /dt + bT = 0
d2T / dt2 + 2 ] Zn dT /dt + Z 2T = 0n
A second-order stable system Æ Overdamped or oscillatory
response from initial conditions
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Homogeneous Underdamped Second-Order Systems
x’’ + ax’ + bx = 0; write as x’’ + 2]Znx’ + Z 2x = 0n
Let x = X est Î
(s2 + 2]Zns + Z 2) est = 0 OR s2 + 2]Zns + Z 2 = 0 În n
s = [-2]Zn +/- sqrt(4]2Z 2 – 4Z 2)] / 2n n
= Zn[-] +/- sqrt�]2-1)] from quadratic equation
s1 and s2 are complex conjugates if ] < 1, in this case:
s1 = -Z ] + iZd, s2 = -Z ] – iZd where Zd = Zn sqrt(1-]2)n n
Recalling er+iT = er ( cosT + i sinT��, we have
x = e-]Znt [ (X1r + iX1
i)(cosZdt + isinZdt) +
(X2r + iX2
i)(cosZdt – isinZdt) ] AND
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x’ = -]Znx + Zde-]Znt [ (X1
r + iX1i)(-sinZdt + icosZdt) +
(X2r + iX2
i)(-sinZdt – icosZdt) ]
Consider initial conditions x’(0) = 0, x(0) = 1:
x(t=0) = 1 means X1 r + X2
r = 1 (real part) and
X1 i + X2
i = 0 (imaginary part)
x’(t=0) = 0 means X1 r - X2
r = 0 (imaginary part) and
-]Zn + Zd(X2 i – X1
i) = 0 (real part)
Combine these and we find that
X1r = X2
r = ½iX1
i = -X2 = -]�Zn / 2 Zd
Plug into the solution for x and do some trig:
x = e -]Znt sin(Zdt + k) / sqrt(1-]2), where k = atan(Zd/]Z n)
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]= 0.0 has
but no decay
] = 0.2 gives
about 50%
overshoot
] about 15%
overshoot
] the fastest
response
without
overshoot
] > 1.0 is
slower
fastest rise time
= 0.5 gives
= 1.0 gives
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STABILITY REFERENCE POINTS
M
G
B
K
etacenter
ravity
Buoyancy
Keel
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LINEAR MEASUREMENTS IN STABILITY
KM
M
G
B
K
GM
KG
BM
Massachusetts Institute of Technology, Subject 2.017 CL
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THE CENTER OF BUOYANCY
WATERLINE
B
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Massachusetts Institute of Technology, Subject 2.017
BBBB
CENTER OF BUOYANCY
WLWLWLWLWL
B
G
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Massachusetts Institute of Technology, Subject 2.017
CENTER OF BUOYANCY
BBBBBBB B
B
- The freeboard and reserve buoyancy will also change
Reserve Buoyancy
Draft
Freeboard
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Massachusetts Institute of Technology, Subject 2.017
G
KGo
G1
KG1
G MOVES TOWARDS A WEIGHT ADDITION
MOVEMENTS IN THE CENTER OF GRAVITY
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Massachusetts Institute of Technology, Subject 2.017
GG
MOVEMENTS IN THE CENTER OF GRAVITY
G MOVES AWAY FROM A WEIGHT REMOVAL
GGG G
G1
KG1
KGo
G
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Massachusetts Institute of Technology, Subject 2.017
G G2
G MOVES IN THE DIRECTION OF A WEIGHT SHIFT
MOVEMENTS IN THE CENTER OF GRAVITY
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DISPLACEMENT = SHIP’S WIEGHT
20
G
B
- if it floats, B always equals G
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Massachusetts Institute of Technology, Subject 2.017
METACENTER
BBBBBBBBBBBBB
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Massachusetts Institute of Technology, Subject 2.017
METACENTER
B SHIFTS
M
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Massachusetts Institute of Technology, Subject 2.017
+GM 0o-7/10o
CL
B
M
G
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Massachusetts Institute of Technology, Subject 2.017
+GM
CL
B B20
M
M20
G
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Massachusetts Institute of Technology, Subject 2.017
CL
M
M20
M45
B
B20 B45
G
+GM
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Massachusetts Institute of Technology, Subject 2.017
G
CL
B
B20
B45
M
M20
M45
B70
M70
neutral GM
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Massachusetts Institute of Technology, Subject 2.017
CL
M20M45
M70
M90
B
B20
B45 B70
B90
M G
-GM
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MOVEMENTS OF THE
METACENTER
THE METACENTER WILL CHANGE POSITIONS IN THE
VERTICAL PLANE WHEN THE SHIP'S DISPLACEMENT
CHANGES
THE METACENTER MOVES IAW THESE TWO
RULES:
1. WHEN B MOVES UP M MOVES DOWN.
2. WHEN B MOVES DOWN M MOVES UP.
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Massachusetts Institute of Technology, Subject 2.017
M
G
B
M
G
B
G
M
B
M1
B1
G
M
B
M1
B1
G
M
B
M1
B1
G
M
B
M1
B1
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Massachusetts Institute of Technology, Subject 2.017 CL
K
B
G
MM
CL
G
B
K
B1
CL
M
G
B
K
B1
CL
M
G
B
K
B1
CL
K
B
G
M
B1
Z
Righting Arm
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Righting ArmANGLE OF
RIG
HT
ING
AR
MS
(F
T) MAXIMUM RIGHTING
ARM MAXIMUM RIGHTING
ARM
DANGER ANGLE
MAXIMUM RANGE OF STABILITY
0 10 20 30 40 50 60 70 80 90
ANGLE OF HEEL (DEGREES)
WL
WL
20°
G
B
Z
WL
40°
G
B
Z
60°
G
B
Z
GZ = 1.4 FT GZ = 2.0 FT GZ = 1 FT Massachusetts Institute of Technology, Subject 2.017
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Massachusetts Institute of Technology, Subject 2.017
-1
0
1
2
3
4
5
6
0
MAX GZ
REDUCED MAX
Righting Arm for LSD
10 20 30 40 50 60
Angle of Heel (deg)
Rig
hti
ng
Arm
(ft
)
Light Ship
Full Load
Damage/Full Load
LIST ANGLE OF HEEL
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THINGS TO CONSIDER
• Effects of:
– Weight addition/subtraction and movement
– Ballasting and loading/unloading operations
– Wind, Icing
– Damage stability
• result in an adverse movement of G or B
• sea-keeping characteristics will change
• compensating for flooding (ballast/completely flood a
compartment)
• maneuvering for seas/wind
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References
• NSTM 079 v. I Buoyancy & Stability
• NWP 3-20.31 Ship Survivability
• Ship’s Damage Control Book
• Principles of Naval Architecture v. I
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