STABILITY AND TRIM OF MARINE VESSELS. Massachusetts Institute of Technology, Subject 2.017

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Transcription:

STABILITY AND TRIM OF MARINE VESSELS

Concept of Mass Center for a Rigid Body Centroid the point about which moments due to gravity are zero: 6 g m i (x g x i )= 0 Æ x = 6m i x i / 6m i = 6m i x i / M g 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

Center of Buoyancy A similar differential approach with displaced mass: x b = 6 ' i x i 'where ' i is incremental volume, ' is total volume Center of buoyancy is the same as the center of displaced volume: it doesn t matter what is inside the outer skin, or how it is arranged. Images removed for copyright reasons. 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.

For a submerged Righting arm: h =(z b -z g )sint body, a sufficient condition for stability is that z b is above z g. g Righting moment: U'gh B T Make (z b -z g ) 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.

In most surface vessels, righting stability is provided by the waterplane area. slice thickness dx b T submerged volume d' d' = Adx intuition: wedges RECTANULAR SECTION h g T B 1 K M b/6 y B 2 l/3 l eometry: d'/dx = bh + bl/2 or h = ( d'dx bl/2 ) / b l = b tant Vertical forces: df = Ug d' no shear) dfb1 = Ug b h dx dfb2 = Ug b l dx / 2

Moment arms: y = K sin Ty B1 = h sin T / 2 ; y B2 = (h + l/3) sin T + b cos T / 6 Put all this together into a net moment (positive anti-clockwise): dm/ug = -K d' sint + bh 2 dx sint / 2 + b l dx [ (h+l/3) sint + b cost / 6] / 2 (valid until the corner comes out of the water) Linearize (sint ~ tant ~ T), and keep only first-order terms (T): dm Ug d' >K d' / 2 b dx b dx d'@t For this rectangular slice, the sum >d' / 2 b dx + b 3 dx / 12 d'@ must exceed the distance K 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! -K + KM = M : the METACENTRIC HEIHT

M Note M does not have to stay on the centerline**! R: righting arm R damage, flooding, ice down-flood heel angle Images removed for copyright reasons. Ships in stormy waves. How much M is enough? Around 2-3m in a big boat

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 [ -K(x) d'x + d' 2 (x) / 2 b(x) dx + b 3 (x) dx / 12 ] Tor dm(x) Ug [ -K(x) A(x) dx + A 2 (x) dx / 2 b(x) + b 3 (x) dx / 12@T First term: Same as Ug K ', where ' is ship s submerged volume, and K 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 M

Weight Distribution and Trim At zero speed, and with no other forces or moments, the vessel has B (submerged) or M (surface) directly above. Too bad! M For port-stbd symmetric hulls, keep on the centerline using a tabulation of component masses and their centroid locations in the hull, i.e., 6 m i y i = 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.

Rotational Dynamics Using the Centroid Equivalent to F = ma in linear case is T = J * d 2 T/ dt 2 o where T is the sum of acting torques in roll J o 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 m i : J o = 6 m i (y i -y o ) 2 OR J g = 6 m i (y i -y g ) 2 J written in terms of component masses m i and their own moments of inertia J i (by the parallel axis theorem) : J = 6 m i (y i -y g ) 2 g + 6 J i The y i s give position of the centroid of each body, and J i s are referenced to those centroids

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 M: J d 2 T / dt 2 = - D dt /dt M U g ' T OR J d 2 T / dt 2 + D dt /dt + M U g 'T = 0 d 2 T / dt 2 + a dt /dt + bt = 0 d 2 T / dt 2 + 2 ] Z n dt /dt + Zn 2 T = 0 A second-order stable system Æ Overdamped or oscillatory response from initial conditions

Homogeneous Underdamped Second-Order Systems x + ax + bx = 0; write as x + 2]Z n x + Z 2 x = 0 n Let x = X e st Î (s 2 + 2]Z n s + Zn 2 ) e st = 0 OR s 2 + 2]Z n s + Zn 2 = 0 Î s = [-2]Z n +/- sqrt(4] 2 Zn 2 4Zn 2 )] / 2 = Z n [-] +/- sqrt] 2-1)] from quadratic equation s 1 and s 2 are complex conjugates if ] < 1, in this case: s 1 = -Z n] + iz d, s 2 = -Z n] iz d where Z d = Z n sqrt(1-] 2 ) Recalling e r+it = e r ( cost + i sint, we have x = e -]Znt [ (X 1 r + ix 1i )(cosz d t + isinz d t) + (X 2 r + ix 2i )(cosz d t isinz d t) ] AND

x = -]Z n x+ Z d e -]Znt [ (X 1 r + ix 1i )(-sinz d t + icosz d t) + (X 2 r + ix 2i )(-sinz d t icosz d t) ] Consider initial conditions x (0) = 0, x(0) = 1: x(t=0) = 1 means X r 1 + X r 2 = 1 (real part) and X i 1 + X i 2 = 0 (imaginary part) x (t=0) = 0 means X r 1 -X r 2 = 0 (imaginary part) and -]Z n + Z d (X i 2 X i 1 ) = 0 (real part) Combine these and we find that X r 1 = X r 2 = ½ X i 1i = -X 2 = -]Z n / 2 Z d Plug into the solution for x and do some trig: x = e -]Znt sin(z d t + k) / sqrt(1-] 2 ), where k = atan(z d /]Z n )

]= 0.0 has fastest rise time but no decay ] = 0.2 gives about 50% overshoot ] = 0.5 gives about 15% overshoot ] = 1.0 gives the fastest response without overshoot ] > 1.0 is slower

STABILITY REFERENCE POINTS M etacenter ravity Buoyancy Keel CL

LINEAR MEASUREMENTS IN STABILITY M M BM KM K B K CL

THE CENTER OF BUOYANCY WATERLINE B

CENTER OF BUOYANCY WL BBBB B

CENTER OF BUOYANCY Reserve Buoyancy Freeboard BB B B Draft -Massachusetts The freeboard Institute of Technology, and reserve Subject 2.017 buoyancy will also change

MOVEMENTS IN THE CENTER OF RAVITY MOVES TOWARDS A WEIHT ADDITION 1 K o K 1

MOVEMENTS IN THE CENTER OF RAVITY MOVES AWAY FROM A WEIHT REMOVAL 1 K o K 1

MOVEMENTS IN THE CENTER OF RAVITY 2 MOVES IN THE DIRECTION OF A WEIHT SHIFT

DISPLACEMENT = SHIP S WIEHT B - if it floats, B always equals 20

METACENTER BBBBBBB

METACENTER M B SHIFTS

+M 0 o -7/10 o M B CL

+M M 20 M B B 20 CL

+M M 20 M 45 M B B 20 B 45 CL

neutral M M 20 M 45 B M 70 M CL B 20 B45 B 70

-M M 45 M 20 CL B M 90 M 70 B 20 B 45 B 70 B 90 M

MOVEMENTS OF THE METACENTER THE METACENTER WILL CHANE POSITIONS IN THE VERTICAL PLANE WHEN THE SHIP'S DISPLACEMENT CHANES THE METACENTER MOVES IAW THESE TWO RULES: 1. WHEN B MOVES UP M MOVES DOWN. 2. WHEN B MOVES DOWN M MOVES UP.

M 1 M 1 1 M 1 B 1 M B

Righting Arm MM M Z BB 1 B 11 K K K CL CL CL CL CL

Righting Arm RIHTIN ARMS (FT) MAXIMUM RIHTIN ARM DANER ANLE ANLE OF MAXIMUM RIHTIN ARM MAXIMUM RANE OF STABILITY 0 10 20 30 40 50 60 70 80 90 ANLE OF HEEL (DEREES) WL WL 20 WL Z B 40 Z B 60 Z B Z = 1.4 FT Z = 2.0 FT Z = 1 FT

6 5 Righting Arm for LSD MAX Z Light Ship Full Load Damage/Full Load 4 Righting Arm (ft) 3 2 1 LIST REDUCED MAX ANLE OF HEEL 0 0 10 20 30 40 50 60-1 Angle of Heel (deg)

THINS TO CONSIDER Effects of: Weight addition/subtraction and movement Ballasting and loading/unloading operations Wind, Icing Damage stability result in an adverse movement of or B sea-keeping characteristics will change compensating for flooding (ballast/completely flood a compartment) maneuvering for seas/wind

References NSTM 079 v. I Buoyancy & Stability NWP 3-20.31 Ship Survivability Ship s Damage Control Book Principles of Naval Architecture v. I