02a_transverse_stability

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NE224 Spring 2023 Instructor: An Wang 3 Static Equilibrium and Transverse Stability 3.1 Total Weight and Center of Gravity In order to evaluate the stability of a ship, the total weight and the position of the center of gravity are needed. Note: Weight is a force and it has a magnitude, direction and point of application. The direction is vertically down. The point of application is at G, the center of gravity. Total weight of a ship equals the summation of the weight of the ship and everything on-board. W = W 1 + W 2 + ... + W n = n X i =1 W i The longitudinal ( x ), transverse ( y ) and vertical ( z ) positions of the center of gravity are : x G = n X i =1 W i x i n X i =1 W i , y G = n X i =1 W i y i n X i =1 W i , z G = n X i =1 W i z i n X i =1 W i , respectively, where W i is the weight of the i -th component and x i , y i and z i are the longitudinal, transverse and vertical positions of the i -th component, respectively. In stability calculations, the vertical position of the center of gravity is important. If the baseline plane is chosen as the reference ( z = 0), then z G = KG, or KG = n X i =1 W i Kg i ∆ (equation (21) of chapter 2, Zubaly) (3.1) , where ∆ is the displacement (weight), Kg i is the vertical position (relative to baseline plane) of each individual weight. Classification of weights: 1) Light ship weight or light ship displacement (its mass called light ship mass): Weight of the ship itself (considered as permanent part of the ship, exclusive of loading of any kind, such as cargo, personnel, blast, fresh water, etc.. The condition without any loading is called light ship condition): a) Hull steel: all ship structural elements; 1

NE224 Spring 2023 Instructor: An Wang b) Propulsion machinery: engines, gears, shaft, propeller and auxiliary machinery and equipment associated with propulsion plant; c) Outfit: equipment for hotel services, navigation, cargo handling, etc., furnishing, joiner work, deck covering, paint,... 2) Total deadweight ( Weight carried by the ship, vary from voyage to voyage) a) Cargo: also called cargo deadweight or payload. b) Fuel oil c) Fresh water d) Salt-water ballast e) Crew and effects f) ... Load displacement = Light ship displacement + Total deadweight Example: calculate KG for a ship (Example 3-5, Zubaly) A cargo ship 150 meters long has masses and vertical centers as follows: Steel: 4402 MT at 7.7 m above keel Machinery: 889 MT at 5.7 m above keel outfit: 1859 MT at 12.8 m above keel Determine the ship’s light mass and KG. Solution: Light mass: ∆ m = 4402 MT + 889 MT + 1859 MT = 7150 MT Height of center of gravity above keel: KG = 4402 × 7 . 7 + 889 × 5 . 7 + 859 × 12 . 8 7150 = 8 . 78 m Or use table M i kg i M i × kg i Steel 4402 7.7 33895.4 Machinery 889 5.7 5067.3 Outfit 1859 12.8 23795.2 Light ship 7150 62757.9 2

NE224 Spring 2023 Instructor: An Wang 3.2 Archimedes’s Principle and the Law of Flotation Archimedes’s principle: a body immersed in a liquid is subject to an upward force equal to the weight of the liquid displaced by the body. For a ship with a volume of displacement ∇ F B = ρg ∇ (3.2) , where ρ is the density of water and g is the gravitational acceleration, F B is the buoyant force. Note: the buoyant force is the resultant force of the hydrostatic pressure distribution around the immersed portion of the floating body. The horizontal component of the resultant force is zero. Equilibrium condition for a ship floating at water surface: 1) Total force equals zero. X i F i = 0 2) Total moment about any point is zero. X i M i = 0 The two requirements at equilibrium result in the following two conditions: 1) The weight and buoyancy are equal in magnitude and opposite in direction. W + F B = 0 , where W is the weight of the ship. 2) the center of gravity and center of buoyancy are located on the same vertical line. 3.3 Stability of a floating body When a floating body is at equilibrium condition, the weight and buoyancy are equal in magnitude. The center of gravity and the center of buoyancy is located along the same vertical line. 1) If any of the two conditions are not satisfied, an external force is required to maintain the equilibrium state. 3

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NE224 Spring 2023 Instructor: An Wang For example, a rectangular block floats on the water surface, see figure 3.1. Block in (a) and (c) are at equilibrium state, while block in (b) and (d) are at non equilibrium state unless external forces (couple) are applied to the block to maintain equilibrium. G G G G B B B B Fg2562 Fg2562 Fg2562 Fg2562 W W W W (a) (b) (c) (d) Figure 3.1: Block at static equilibrium state in ( a ) and ( c ) and at non equilibrium state in ( b ) and ( d ). 2) Even if both conditions are satisfied and the floating body is at equilibrium state, the body may not be at a stable state. For example, in the two equilibrium states (a) and (c), from intuition, we know that (a) is more “stable” than (c). Why is that? What is stability and how is stability assessed? Stability: If the position of a body floating at equilibrium is suddenly changed (by an external force or couple), the ability to return to its original equilibrium position after the external force (or couple) is removed. When the position of a body is changed, the position of the center of buoyancy ( B ) is shifted, due to the change of the immersed shape, resulting in a new net force (or couple) acting on the body. According to the definition of stability, an equilibrium condition is said to be: 4

NE224 Spring 2023 Instructor: An Wang 1) Stable : if the new force (or couple) tends to return the body to its original equilibrium position; 2) Unstable : if the new force (or couple) tends to further increase the magnitude of the disturbed position from its original equilibrium position; 3) Neutral : if body is still at static equilibrium in the disturbed position (the new net force is zero). G B Fg2562 W G B Fg2562 W Top Top ( a ) Upright ( b ) Heeled Figure 3.2: Neutral equilibrium example: a sphere of uniform density floats on a water surface (figure3-6, Zubaly). In the following discussion of ship stability, we will classify the topic based on the orien- tation of the change in position, or the magnitude of the change in position. According to the orientation of change in position: 1) Transverse stability 2) Longitudinal stability According to the magnitude of the change in position (angle of rotation): 1) Initial stability (small angle of rotation) 2) Stability at large angles. In the following, let’s first discuss the initial transverse statical stability . Initial: very small angle of rotation, body remains nearly upright. Transverse: (Potential) rotation is in the transverse direction, i.e., about the longitu- dinal axis. Such rotation is called “heel” . Statical: Angular velocity of the rotation are considered small and the inertial effects are ignored. Only the direction and magnitude of moments are of concern. 5

NE224 Spring 2023 Instructor: An Wang 3.4 Initial transverse statical stability 3.4.1 Righting moment and righting arm How to assess the initial transverse stability quantitatively? In figure 3.1 (b) and (d), due to the change of the block’s position from equilibrium, we can notice that the net moment created by the buoyancy force and gravity is nonzero. In (b), the moment is counterclockwise, which tends to return the block to upright position. This moment is called righting moment . In (d), the moment is clockwise, which tends to further increase the angle of rotation. This moment is called heeling moment (or a negative righting moment). Clearly, the difference between (b) and (d) are that the two moments have opposite sign. It is not difficult to conclude that, the sign and magnitude of the righting moment , M R , at a given heeling angle, ϕ , determines whether a ship is stable and how stable it is at this heeling angle, respectively. Since the moment equals to M R = ∆ × GZ, it can also be concluded that at a given heeling angle ϕ , the righting arm , GZ, is the measure of stability at this heeling angle. Note: The measure of stability of ship is subject to a given heeling angle. At different heeling angles, the ability of the ship to return to upright position is different. For small heeling angle ϕ , typically smaller than 10 ◦ - 15 ◦ , some assumptions can be made to simplify the calculation. In the following, let’s first discuss the transverse stability at small heeling angle, called initial transverse statical stability. 6

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NE224 Spring 2023 Instructor: An Wang 3.4.2 Shift of Center of Buoyancy G G g3670 g g3670 g W g3670 W g3670 W = W g3670+ W g3671 W g3671 Figure 3.3: Example of shift of center of gravity. First, let’s look at what would happen if part of a volume is moved to another position, see figure 3.3. In the figure 3.3, the center of gravity of W 1 is moved from g to g 1 . As a result, the center of gravity of the entire volume shifts from G to G 1 . The center of gravity shift in a direction parallel to the direction of the weight shift and the moment created by the shifted weight equals to the moment of the shift in the total weight. We should have: GG 1 // gg 1 (3.3) and GG 1 W = gg 1 W 1 Or GG 1 gg 1 = W 1 W GG 1 = W 1 W gg 1 (3.4) 7

NE224 Spring 2023 Instructor: An Wang 3.4.3 Metacenter When the ship is heeled to a new position, the immersed shape changes. Since the total volume of displacement remains the same, the change can be viewed as that part of the immersed volume is shifted to another location, similar to the example illustrated in figure 3.3. The goal is to find how much the center of buoyancy of the entire displaced volume shifts ( BB 1 ). The total displacement ( ∇ ) is known and unchanged, according to equation 3.4, we first need to find out the volume of the immersed/emerged wedge, V i , and the distance between the centroid of the two volumes, B i B e . Now let’s first consider a wedge element of length dx , the volume of the wedge is dV i . The centers of gravity of the emerging wedge and the immersed wedge are b e and b i , respectively, see figure 3.4. B B g3670 W L L g3670 W g3670 ϕ be bi O y 2y/3 ϕ dx bi y tan ϕ O O’ Figure 3.4: Shift of center of buoyancy for a heeled ship. When ϕ is very small, b i O = 2 3 y dV i = 1 2 · y · y tan ϕ · dx = 1 2 y 2 tan ϕdx The moment of the immersed wedge element volume about the O - O axis (see figure 3.4) is dm i = b i OdV i = 2 3 y 1 2 y 2 tan ϕdx = 1 3 y 3 tan ϕdx 8

NE224 Spring 2023 Instructor: An Wang Thus, the moment of the total immersed volume along the length of the ship about O - O axis will be the integral of the above equation from − L/ 2 to L/ 2: M i = B i OV i = Z dm i = Z b i OdV i = Z L/ 2 − L/ 2 1 3 y 3 tan ϕdx = 1 3 tan ϕ Z L/ 2 − L/ 2 y 3 dx , where B i is the centroid of the entire immersed wedge along the length of the ship. Due to symmetry, the moment of the entire emerged wedge about O - O axis is M e = B e OV e = M i = 1 3 tan ϕ Z L/ 2 − L/ 2 y 3 dx Therefore, B i B e = 2 B i O = 2 B e O = 2 M i V i = 1 V i 2 3 tan ϕ Z L/ 2 − L/ 2 y 3 dx Recall, the transverse moment of inertia of the waterplane (about the x -axis), I T = 2 3 Z L/ 2 − L/ 2 y 3 dx Therefore, B i B e = 1 V i I T tan ϕ or B i B e V i = I T tan ϕ Apply the equation for the shift of center of gravity introduced previously, the shift of center of buoyancy of the entire volume is BB 1 = V i ∇ B i B e Therefore, the shift of the center of buoyancy BB 1 can be calculated by BB 1 = V i ∇ B i B e = I T ∇ tan ϕ When ϕ is small, tan ϕ ≈ ϕ , therefore, BB 1 = I T ∇ ϕ (3.5) When ϕ is small, BB 1 can be treated as a circular arc center at a point M , which is called the transverse metacenter , see figure 3.5. The transverse metacenter M is the 9

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NE224 Spring 2023 Instructor: An Wang intersection of line of buoyancy at the upright condition and line of buoyancy at a very small angle of heel. The radius of the circle, BM , is called the transverse metacentric radius . At small ϕ , BMϕ = BB 1 ≈ BB 1 = I T ∇ ϕ Therefore, the transverse metacentric radius BM = I T ∇ (3.6) Typically, the value of KB is given or can be calculated at a given draft (see the cal- culation of KB in previous chapter), BM can be calculated according to the equation 3.6. The height of center of gravity ( KG ) is also given or can be calculated (see equation 3.1). Therefore, the metacentric height, GM , (the vertical distance between the metacenter (M) to the center of gravity (G)) can be calculated by GM = KB + BM − KG (3.7) B B g3670 G M ϕ W Fg2562 W L L g3670 W g3670 ϕ Figure 3.5: Metacenter. Therefore, according to figure 3.6, the righting moment can be calculated by M R = ∆ GZ = ∆ GM sin ϕ (3.8) 10

NE224 Spring 2023 Instructor: An Wang and the righting arm GZ = GM sin ϕ (3.9) When ϕ is small, the righting moment can be approximated by M R ≈ ∆ GMϕ (3.10) and the righting arm can be approximated by GZ ≈ GMϕ (3.11) B B g3670 G M ϕ W Fg2562 = ∆ W L L g3670 W g3670 ϕ Z GZ = GM sin ϕ Figure 3.6: Righting moment. Now, we can use the quantity GM to assess the stability of a ship at small heeling angle ϕ . ( a ) GM > 0 ( b ) GM < 0 ( c ) GM = 0 B B g3670 G M W Fg2562 W L L g3670 W g3670 ϕ ϕ Mg2578>0 B B g3670 G M W Fg2562 W L L g3670 W g3670 ϕ ϕ Mg2578<0 B B g3670 G M W Fg2562 W L L g3670 W g3670 ϕ ϕ Mg2578=0 Figure 3.7: Relative location of metacenter and center of gravity. 11

NE224 Spring 2023 Instructor: An Wang 1) G is located below M ( GM > 0, stable): direction of M R is opposite to the heeling direction. When external force is removed, the heeling angle will decrease. See figure 3.7(a). 2) G is located above M ( GM < 0, unstable): direction of M R is the same as the heeling direction. When external force is removed, the heeling angle will continue to increase. See figure 3.7(b). 3) G is located at M ( GM = 0, neutral): M R = 0. When external force is removed, the heeling angle will neither decrease nor increase. See figure 3.7(c). At a given displacement, for small heeling angle ϕ , the greater the GM is, the greater the M R is and the ship is more stable. Therefore, the metacentric height GM is the main parameter to assess the initial transverse statical stability of a ship. Example: (calculate the metacentric height, Zubaly example 3-4) Show that the 2 × 4 × 10-inch board of specific gravity 0.50 shown in figure 3.8 is stable in condition (a) and unstable in condition (b) of the figure. To do so, calculate the metacentric height ( GM ) for each condition. W L (a) (b) (c) W Fg2562 W W Fg2562 Fg2562 G G G B B B Figure 3.8: Example. Equilibrium of half-immersed wood block. Solution : Condition (a): L = 10 in, B = 4 in and H = 2 in 12

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NE224 Spring 2023 Instructor: An Wang Draft T = H/ 2 = 1 in. The transverse moment of inertia I T = LB 3 12 = 10 × 4 3 12 = 53 . 33 in 4 Volume of displacement ∇ = LBT = 10 × 4 × 1 = 40 in 3 Therefore, the metacentric radius BM = I T ∇ = 53 . 33 in 4 40 in 3 = 1 . 33 in And the position of the center of buoyancy above keel KB = T 2 = 0 . 5 in The position of the center of gravity KG = H 2 = 1 in The metacentric height GM = BM + KB − KG = 1 . 33 + 0 . 5 − 1 = 0 . 83 in GM is positive ( M is located above G ), so condition (a) is stable. For condition (b) L = 10 in, H = 4 in and B = 2 in Draft T = H/ 2 = 2 in. The transverse moment of inertia I T = LB 3 12 = 10 × 2 3 12 = 6 . 67 in 4 Volume of displacement ∇ = LBT = 10 × 2 × 2 = 40 in 3 Therefore, the metacentric radius BM = I T ∇ = 6 . 67 in 4 40 in 3 = 0 . 167 in 13

NE224 Spring 2023 Instructor: An Wang And the position of the center of buoyancy above keel KB = T 2 = 1 in The position of the center of gravity KG = H 2 = 2 in The metacentric height GM = BM + KB − KG = 0 . 167 + 1 − 2 = − 0 . 83 in GM is negative ( M is located below G ), so condition (b) is unstable. Please work on condition (c) yourself. 3.5 Effect of moving weight 3.5.1 Moving weights in the vertical direction Moving weights in the vertical direction does not change the total displacement or the geometry of the immersed volume. Therefore, the location of the center of buoyancy ( B ) remains the same. The value of BM = I T / ∇ also remains the same because the waterlane and displacement are not changed. Shown in figure 3.9, a weight w is moving from z 0 to z 1 . As a result, the center of gravity moves from G to G 1 . B and M remain unchanged. B G g3670 G M W L z g3669 z g3670 w w Figure 3.9: Moving weight w vertically from z 0 to z 1 . The center of gravity moves from G to G 1 . 14

NE224 Spring 2023 Instructor: An Wang However, the center of gravity changes due to the vertically shifted weight. According to equation (3.4), the shift of the center of gravity is GG 1 = w ( z 1 − z 0 ) ∆ (3.12) The change in the vertical location of the center of gravity causes change in metacentric height. The new metacentric height G 1 M = GM − GG 1 or G 1 M = GM − w ( z 1 − z 0 ) ∆ (3.13) 1) if z 1 > z 0 (weight is moved up), G 1 M < GM , the stability of the ship decreases. 1) if z 1 < z 0 (weight is moved down), G 1 M > GM , the stability of the ship increases. Example: (Zubaly) 3.5.2 Moving weights in the transverse direction Moving weights along transverse direction does not change total displacement. But it generates a heeling angle for the ship so the geometry of the immersed volume is no longer the same. As a result, both the center of buoyancy ( B ) and center of gravity ( G ) changes. Shown in figure 3.10, a weight w is moving from y 0 to y 1 . As a result, the center of gravity moves from G to G 1 and center of buoyancy moves from B to B 1 . M remain unchanged. 15

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NE224 Spring 2023 Instructor: An Wang B B g3670 G M W L L g3670 W g3670 ϕ y g3670 y g3669 w w G g3670 Figure 3.10: Moving weight w transversely from y 0 to y 1 . The center of gravity moves from G to G 1 . The center of buoyancy moves from B to B 1 . The shift in center of gravity can be calculated by GG 1 = w ( y 1 − y 0 ) ∆ Note: In figure 3.10, y 0 is a negative value. Shown in figure 3.10, GG 1 and GM should satisfy the following: GG 1 = GM tan ϕ . Therefore, we have: GM tan ϕ = w ( y 1 − y 0 ) ∆ Thus, we can calculate the angle of heel after the transverse shift of weight: tan ϕ = w ( y 1 − y 0 ) ∆ GM (3.14) 3.5.3 Moving weights in any direction within a station Step 1: move the weight vertically. New metacentric height G 1 M = GM − w ( z 1 − z 0 ) ∆ 16

NE224 Spring 2023 Instructor: An Wang Step 2: move the weight transversely: Tangent of the heeling angle: tan ϕ = w ( y 1 − y 0 ) ∆ G 1 M Sign convention: y is positive from the centerline toward starboard. z is positive above keel. ϕ is positive when the ship heels toward the starboard. 3.6 Loading and discharging small weight to any location on a ship When a small weight w is loaded/discharged to/from a ship at height z (measured from baseline) and y from the centerline plane, how does the stability of the ship change and how much the angle of heel does the loading/discharging create? 3.6.1 Load/discharge weight vertically above/below G First, let’s consider the case that the weight is loaded/discharged vertically above/below the original center of gravity of the ship G , i.e., the transverse distance between the loaded/discharged weight and G is zero. The loading/discharging process can be considered as a two-step process. The first step is to load/discharge the weight to/from the original center of gravity, G , see figure 3.11. The second step is to move the weight vertically to its final height. Step 1: Load/discharge weight at G 17