SHIP BUOYANCY AND STABILITY Lecture 3 Ship initial stability 1
Literature J. Matusiak: Laivan kelluvuus ja vakavuus Biran A. B., Ship Hydrostatics and Stability, 23 J. Matusiak: Short Introduction to Ship Theory (Part 1) Rawson, K. J J. Basic Ship Theory Volume 1. 21 Buoyancy and Stability of a Ship, Finnish text book A Ship Stability book in English. Shorten version of the Finnish textbook, in English Check the library Available on KNOVEL as e-book 2
Center of Buoyancy curve Center of Flotation Wedge method Small angle assumption BEFORE THIS LECTURE Metacenter and metacentric radius Now, you should be able to: Describe qualitatively how the center of buoyancy varies, according to draft change or heeling angle Describe the ship inclination axes, defined by the center of flotation Use the wedge method results Describe the main outcomes of the small angle assumption Define the metacenter and its relation with the curve of the center of buoyancy 3
Introduction Preparation for the laboratory test Dynamic stability Following lecturers Ship equilibrium and introduction to hydrostatics Ship initial Stability Metacentric height GM Longitudinal metacentric height GM L The stability curve (GZ curve) Second generation of intact stability criteria Ship Damage Stability Stability special topics 4
Stability of the equilibrium Stability (in general): The property, quality, or characteristic of a body, which cause it, when its equilibrium is disturbed, to develop forces or moments acting to restore its original condition HOW DO WE ASSESS THE STABILITY OF A SHIP? 5
Definitions of Ship Stability Different approaches in investigating ship stability for different conditions INITIAL STABILITY (small angles): It represents the mechanical stability of the equilibrium to small perturbations. It is based on the Euler theorem thesis and assessed by analyzing the metacentric high; 6
Definitions of Ship Stability Different approaches in investigating ship stability for different conditions STABILITY AT FINITE ANGLES : It represents the tendency of a ship to return to the initial equilibrium condition after a significant perturbation, and not capsize. It is a way to assess the stability of the ship subjected to external actions. The small angles assumptions are not valid anymore. It is asses by analyzing the GZ curves (righting arm). FINITE/LARGE ANGLES Static stability taking into account constant external moments. Dynamic stability (defined by the area under GZ curve) taking into account dynamic moments 7
= B N sin Ship stability M ulk + M st = M st = GZ = h M st 8
Ship stability-in general Ship is stable, to small or large external actions, if the moment arising M st is of restoring type. Restoring means the moment works to bring the ship in the initial condition, against the external moment. The lever of this moment is the distance between the weight line and the buoyancy line. It varies according to the angle of inclination, i.e. at each equilibrium floating position related to the heel angle. Z G 2 16 4 B 2 B B 4 B 16 K 9
CENTER of BUOYANCY curve 1
GZ curve 11
SHIP INITIAL STABILITY: SMALL PERTURBATION SMALL ANGLES 12
Initial Stability of a 6DoF Body DEFINITION: A generic body with 6 degree of freedom is stable in any direction if those conditions are satisfied: F x x x, F y y y, F z z z, M x, M y, M z Example with curvilinear abscissa The system is stable F s s s s s In the stability calculation the external actions causing the perturbations do not figure. 13
VERTICAL STABILITY OF A SHIP F x x x, F y y y, F z z z, M x, M y, M z 14
Ship stability in the vertical direction T z z, z' W G T W + W G y' T B y K B' T y T K a) b) c) z, z' W G y' B' y K W a: ship in equilibrium under weight and buoyancy force b: ship perturbed with a weight increment W that leads to a draft increment, in order to generate an increment of buoyancy equal to the weight increment: T = W g A W W = g A W T Cylindrical zones assumption! F z = Increasing the draft, means z = T When the perturbation W is removed, the restoring force Fz=, acting in the opposite direction of the perturbation, will bring the ship in the initial position Stable system F z z = g A W < 15
Ship stability in the vertical direction T z z, z' W G T W + W G y' T B y K B' T y T K a) b) c) z, z' W G y' B' y K W F z z = g A W < Regarding the vertical stability, only the floating body are affected by this phenomenon, that makes them vertically stable. The submerged body (like submarine of airship) will not experience the same phenomenon; their are characterized by neutral vertical stability. Cylindrical zones assumption :T << T. If T is small enought it possible to consider the hull portion in T cylindrical, that means the waterplan area is constant in T and so with z. dz T + T T z T = 1 g A W T z d 16
TRANSVERSAL STABILITY OF A SHIP x x x, F y y y, F z z z, M x, M y, M z 17
Euler Theorem The intersection of two very close floating planes is also very close to their centers. That means that the axis of inclination is baricentral for both the waterplanes. F is the geometrical center of the initial waterplane F belongs to the axis of inclination x: this axis is a central axis of inertia or baricetral. After a small inclination, the center the new waterplane is very close to F They are considered to be the same The x axis is baricentral also for the new waterplane 18
Euler Theorem applied to ship A ship that inclines of small angles, rotates around a constant central axis and the immersed volume remains constant!!! Z T K Y 19
Introduction to ship initial transversal stability K K In figure (a) the ship is in equilibrium. Let s incline the ship of an angle, perturbating the equilibrium position (figure b); Let s assume a small heeling angle and the IsoHull ipothesys - For the Isohull assumption, the buoyancy remains the same of the initial floating position; - For the small angle assumptions, the center of buoyancy of the perturbated positions belongs to the YZ plane - for small heeling angles, the respective centers of buoyancy belong to a circular arc with center in M M M and BM=B M =B M radius. - We define Metacenter M the intersection point between the perturbated buoyancy line with the Z axis; - The ship will react to the perturbation, with a moment around the X axis, Mx 2
Introduction to ship initial transversal stability dφ Mx will be equal to the product of the displacement and the distance between the weight and the buoyancy line; According to the figure, the moment dm x is opposite to dφ dm x GMsin dm x Where: sinφ dφ dφ dm x dφ dm x dm x dϕ = ΔGM < If the Metacenter is above the center of gravity: GM> the moment Mx arising is of stability for the system, as it tends to keep the ship into the initial equilibrium position; (upper and left figure) If the Metacenter is under the center of gravity: GM< the system is unstable, as the moment Mx will incline further the ship; (right figure) 21
Introduction to ship initial transversal stability M x is called restoring moment because after a perturbation it tends to restore the ship in her initial position M x GMsin ( r a) sin Small angle assumption sinφ dφ GM is called Metacentric height GM KM KG KM BM KB GM BM BG BM is called Metacentric radius. In nomenclature it can be also found as: r BG is the heigth of the center of gravity above the initial center of buoyancy; in nomenclature it can be also found as: a 22
Ship initial stability h = KM sin KG sin = B M sin B G sin = GM sin The first term is called shape lever The second one, dependent upon the vertical position of the centre of gravity (VCG or KG), is called weight lever. Higher shape lever, meaning higher position of metacenter, increases stability of the ship. Higher position of VCG has a detrimental effect on stability. If VCG is raised above the metacenter M, ship gets unstable. 23
Rocking chair representation Let s compare the curve of the center of buoyancy to the legs of a rocking chair. let s think to roll on it Broader leg Narrow leg D is the maximum angle for the stability i.e. the angle at which G and B have the same height; heeling angle over this value lead to instability. The narrow ship reach this condition before the broader ship The narrow ship has a Metacenter lower that the broader ship. At the same angle of inclination, the broader ship will have higher restoring moment 24
Metacentric radius M x GMsin GM BM BG r BM I xx The x axis is a symmetry axis for the waterplane; it is a central (or baricentral) axis of inertia I xx 2 Aw y 2 da The Metacentric radius depends on the shape of ship. In particular it depends on the immersed volume and on the distribution of the halfbread y along the ship length. 2 3 L y 3 ( x) dx 25
Metacentric radius zd VCB KB L vcb( x) A( x) dx r BM I xx I xx 26
Metacentric radius: summary Initial stability: we are studying the stability of the equilibrium condition for the ship for small angles; Small angle assumptions leads to the following consideration: The ship heels around the X axis that remains the same; The curve of the center of buoyancy for small heeling angles belong to the YZ plane. The curve of the center of buoyancy for small heeling angles belong to a circular arc with center in the metacenter M, and radius r=bm=ixx/. The moment arising from the perturbated system is Mx= - GMsin The system is stable if GM>, that means the Metacenter above the center of Gravity. 27
Submarine transversal stability alusta kallistava ulkoinen momentti a) vakaa b) epävakaa G B W z G G B z' B z' G z B W voimapari tasapainottaa alusta voimapari kallistaa alusta lisää Full submerged body does not modify the immersed volume; the center of buoyancy is the center of volume of the whole body. Heeling the body the center of buoyancy remains constantly in its initial position Metacenter and center of buoyancy for a full immersed body are the same point. In order to assess the static stability, the metacenter has to be over the center of gravity, i.e. z G < z B 28
LONGITUDINAL STABILITY OF A SHIP y y y, F z z z, M x, M y, M z 29
Variation of the center of buoyancy with the trim Let s assume fixed volume and fixed heel: = φ= Let s incline the ship B=B(, φ, θ)=b(,, θ) θ ϵ [, θ*] M L z The curve of the center of buoyancy is a curve that belongs only to the vertical plane XZ, and can be well approximated with a circumference: Ship usually does not experience great trim angles WL WL v 2 F B B v 1 K x * Where is the inclination axis? 3
Initial Longitudinal Stability The same concept seen for the transversal initial stability can be extended to the longitudinal problem. Small angle assumptions is made on the trim angle : The ship heels around the transversal axis that remains the same; The center of buoyancy curve for small heeling angles belong to the XZ plane. The center of buoyancy curve for small heeling angles belong to a circular arc with center in the longitudinal metacenter M L, and radius R=BM L =Iy / (please note y i.e the central axis!!!) The moment arising from the perturbated system is My= - GM L sin The system is stable if GM L >, that means the Longitudinal Metacenter above the center of Gravity. 31
Longitudinal Stability Trim angles of ship are usually so small that the initial stability approach is sufficient. For example, if the trim angles for a 1 m long ship is 1deg forward and aft drafts difference, i.e. trim in length or trim, is ΔT = 1.75 m. For trim angles 5 trim should be ΔT = 8.75 m This amount of trim is far from normal conditions As for the initial transverse stability, the initial longitudinal stability assumes the center of buoyancy moving longitudinally on a circular arc. That means that in longitudinal inclination, the effective metacenter remains very close to the initial one. B θ M θ B M L 32
Longitudinal metacentric height GM L Longitudinal metacentric radius BM L = I L Longitudinal stability moment that is opposite to the external moment M stl = B N B G sin B M L B G sin = GM L sin. 33
Metacentric radius approximate definition: box like shape y(x ) y(x ) X BM I T B M = B3 L 12 1 LBT = 1 12 B 2 T For a conventional ship =C B LBT and I L and I T depends on y(x) Where C B is the block coefficient B M = f C B,y(x) B2 T Y The function f(c B,y(x)) for a barge i.e. a hull of box like shape assumes the value of: f(1,b/2) = 1/12. B M L I L B M L = L3 B 12 1 LBT = 1 12 L 2 T 34
Metacenter comparison B M L = f' C B,y(x) L2 T M L z B M L B M = L B 2 B M = f C B,y(x) B2 T M L The longitudinal initial metacenter radius is definitely greater than the transversal one. Longitudinal stability is greater than the transversal stability for a conventional ship. WL WL v 2 F B B 35 K v 1 x
GM L Generally in intact case, the longitudinal metacentric height is always positive, and thus the ship is longitudinally stable. BM L is generally very high: (L/B) 2 times bigger than BM Only for the very high-speed boats and semisubmersibles, longitudinal stability may represent a problem. Therefore, for conventional displacement vessels the longitudinal stability analysis is limited to the ship trim, regarding the shift of a mass in x-direction. 36
Trim definition M L z x 2 Mulk x 1 m 2 W 1 m mg mgcos G T WL T' T A A rim F B T F L A el B K L L F T' F x WL tan = Ship trim can be defined by means of: Trim lenght: (difference of the aft and forward draft) t = T' A T' F = T F + T A Trim angle: (the tangent of the angle between the trim lenght and the m g eship L lenght. m g e L B M L B G = T F L F = tan T A = t L A L B M L B G = T F L F = T A L A = t L 37
Shift of a mass in longitudinal direction M L z x 2 Mulk x 1 m 2 W 1 m mg mgcos G T WL T' T A A rim F B T F L A el B K L L F T' F x WL m g e L cos = GM L sin T' F = T F T F, T' A = T A + T A t = T'A T' F = T F + T A Note that when ship immerses is plus (+) T A/F while is negative when emerges ( ) T A/F. The tangent of the trim angle is given by the following formula: tan = m g e L B M L B G = T F L F = T A L A = t L 38
Moment to trim one meter The moment required to trim the ship by 1 meter is called i.e. moment to trim one meter). The trim angle ', due to M TM will depend on the length of the ship L, recalling the tangent formula tan ' = 1 [m]/l[m]. The unitary moment is given by: M TM = m g e L 1 m = B M L B G L = GM L L Being B M L very huge compared to GB, then is possible to approximate the longitudinal metacentric height with the metacentric radius: GM L BM L BM L = I L / M TM g I L N m L m The moment to trim one meter depends only on ship shape and dimensions; anyway this moment varies with the ship draft. Using this moment it is possible to evaluate with approximate formula, the draft variation due to weight shifting longitudinally GM L = M TM L 39
Trim caused by loading and un-loading a ship. T' F = T F + T T F T' A = T A + T + T A When loading and un-loading a ship we are interested in the distance x m from the center of floatation x F in order to evaluate the arm for the longitudinal inclining moment: M ulk = -e L m g cos where e L = x m - x F x F in this case represent the indifferent or neutral point, meaning that if we load a weigth in x F the ship won t change her trim. 4
DRAFT VARIATION? T=. Exercise T=5.91m Initial draft, even-keel A wl =44 m 2 waterplane area g=1.25 tons/m 3 sea water Q=1 tons added weight L A =83.761 m (center of floatation longitudinal coordinate from AP) L G =87.426 m (center of gravity longitudinal coordinate from AP) L B =87.426 m (center of buoyancy longitudinal coordinate from AP) Z G =1. m (center of gravity vertical coordinate from BL) Z B =3.352 m (center of buoyancy vertical coordinate from BL) WEIGHT LONGITUDINAL COORDINATE, TO REMAIN EVEN-KEEL? L Q =.. 41
Introduction to stability at large angles Transversal stability 42
Ship stability at large (or finite) angles M ulk It is usually that a ship experiences heeling angles over than 1. In that case the initial stability approach does not work properly, as the centers of buoyancy do not belong to a circular arc anymore. Ship under wind and wave external loads will have high heeling angle, for which the initial stability approach is no more valid. The centers of buoyancy move on a curve that is not circular and do not belong to the YZ plan. The shape of this curve depends on the hull shape. From the equilibrium of the momentum of the forces W-, let s evaluate the righting moment arising at high or finite angle. M ulk + M st = f WL WL f B K W G M f M h N f Z B f D 43
Transverse stability at large heel angles M st = GZ = h = GN sin = B N sin B G sin. Prometacenter or false metacenter lim N M The B-curve is not any more a circular arc. Neither metacenter remains at the same position. For a normal ship forms (vertical sides), large inclination is followed by a shift of metacenter ship s plane off the symmetry plane and upwards. Only ships with circular sides will continue to have the B-curve as a precise circular arc 44
We defined the metacentric radius B M for the initial equilibrium condition with no trim and no heel as the radius of the circumference, with center in M, on which laid the center of buoyancy for small heeling angle. For finite heeling angles, the center of buoyancy usually do not belong to the YZ plan and do not laid on the circumference of the metacentric method. Is possible to evaluate for each heeled condition the relative metacenter (on the actual buoyancy line, not the z axix) and metacentric radius. Let s define metacentric evolute the locus of the metacenters at different heeling angles. Evolute in geometry is defined as the locus of the centers of curvature of another curve. Metacenter evolute For the metacenter evolute properties, the tangent to this curve in a generic point M is the buoyancy direction. On this direction the quantity B M define the current center of buoyancy B. The intersection of the buoyancy direction with the vertical axis, for large angles identifies the false prometacenter N Only for small angle prometacenter N and metacenter M coincides with the initial WL WL WL' g 1 M' M M F B B M = I T B B' g 2 metacenter M 45
46 SUMMARY Ship initial stability Small angle assumptions Metacentric height Transversal stability VS Longitudinal stability Now, you should be able to: Define the concept of ship stability and how to use it Define the metecentric height and why the metacenter has to be above the center of gravity Describe the metacenter evolute Motivate why generally transversal stability is more important than the longitudinal one. Calculate how the trim changes by moving/loading weights longitudinally 46