01a_Hull_Form_Hydrostatics

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Stevens Institute Of Technology *

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NE224

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Mechanical Engineering

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Oct 30, 2023

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NE224 Spring 2023 Instructor: An Wang 1 Hull Form and Terminologies 1.1 Reference Planes Figure 1: Reference planes. 1) Baseline plane 2) Centerline plane 3) Midship section plane ( ): distances to after perpendicular and to forward perpendic- ular are equal. 1.2 Planes parallel to reference planes Figure 2: Planes parallel to reference planes. 1

NE224 Spring 2023 Instructor: An Wang 1) Waterplanes 2) Buttock planes 3) Station planes The intersections of these planes with the hull are spatial curves called respectively 1) Waterlines 2) Buttocks 3) Stations These curves are shown in different views (plans) in the lines drawing. Their respective plans in the lines drawing are called 1) Half-breadth plan 2) Profile plan (sheer plan) 3) Body plan. 1.3 Lines drawing and table of offsets The lines drawing and table of offsets are shown below. Note: 1) Lines are smooth and continuous. 2) Due to symmetry about the centerline plane, only one side of the stations, buttocks and waterlines are shown. In the half-breadth plan, the other side is typically used to draw diagonal profiles of the hull. 3) Each plan in the lines drawing is not independent of the other two. If one of the plans is available, the other two plans can be deducted. Two or three plans can provide verification. Practice: how to use information in one of the three plans to determine the other two? 2

NE224 Spring 2023 Instructor: An Wang Figure 3: Lines drawing. If in English units, the table of offsets are given in the format: ft–in–1/8 in. Figure 4: Table of offsets. These lines are also shown below in 3D space. 3

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NE224 Spring 2023 Instructor: An Wang Figure 5: Stations, buttocks and waterlines in 3D space. 4

NE224 Spring 2023 Instructor: An Wang Figure 6: Stations, buttocks and waterlines in 3D space. 5

NE224 Spring 2023 Instructor: An Wang Mention: Lofting (“laying-off”). 1.4 Terminologies and dimensions: longitudinal Figure 7: Terminology in longitudinal direction. 1.5 Terminologies and dimensions: transverse Figure 8: Terminology in transverse direction. 6

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NE224 Spring 2023 Instructor: An Wang Figure 9: Terminology in transverse direction. 7

NE224 Spring 2023 Instructor: An Wang 2 Hydrostatic properties of a ship 2.1 Properties of the waterplanes Figure 10: Waterplane. The midship section is placed at x = 0. 2.1.1 Area of the waterplanes The area of the element shown in figure 10 is dA = ydx . Therefore, the area of the waterplane, A w , is calculated by A w = 2 dA = 2 L/ 2 − L/ 2 ydx (2.1) 2.1.2 First moment of the waterplanes about amidship The first moment of the element in figure 10 about amidship ( x = 0) is dM = xdA = xydx . Therefore, the first moment of the waterplane about amidship is calculated by M = 2 dM = 2 L/ 2 − L/ 2 xydx (2.2) 2.1.3 Center of ﬂotation ( CF ) The longitudinal position of CF relative to amidship ( x = 0) is calculated by LCF = M A w = L/ 2 − L/ 2 xydx L/ 2 − L/ 2 ydx (2.3) The transverse position of CF is located at centerline plane ( y = 0), due to symmetry of the ship. 8

NE224 Spring 2023 Instructor: An Wang 2.1.4 Longitudinal moment of inertia of the waterplane The moment of inertia of a planar shape is the second moment of area about a given axis. First, let’s calculate the longitudinal moment of inertia of the waterplane about the axis at the midship section plane ( x = 0), I . In figure 10, the moment of inertia of the element about x = 0 is dI = x 2 dA = x 2 ydx . Therefore, the total longitudinal moment of inertia of the entire water plane about amidship is calculated by I = 2 x 2 dA = 2 L/ 2 − L/ 2 x 2 ydx (2.4) However, as will be shown in Chapter 3, 4 and 5, in the analysis of ship longitudinal stability, we are more interested in the longitudinal moment of inertia about CF, instead of about amidship. Parallel axis theorem: I par = I cent + Ah 2 (2.5) , where I par is the moment of inertia about the given parallel axis, I cent is the moment of inertia about the axis through centroid, A is the area and h is the distance of the parallel axis from the centroid axis. Therefore, apply parallel axis theorem, I = I L + A w (LCF) 2 . Therefore, the longitudinal moment of inertia about CF is I L = I − A w (LCF) 2 (2.6) 2.1.5 Transverse moment of inertia of the waterplane The transverse moment of inertia of the waterplane is the second moment of area about a longitudinal axis. In the analysis of ship transverse stability, we are interested in the moment of inertia about the axis at the centerline plane ( y = 0) (symmetry plane), which is also where the CF is located. 9

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NE224 Spring 2023 Instructor: An Wang ( a ) ( b ) Figure 11: ( a ) the moment of inertia of a rectangle about its horizontal centerline. ( b ) The element used to calculate the transverse moment of inertia of the waterplane about the symmetry axis ( x -axis). For a rectangle with width ( b ) and height ( h ), see figure 11( a ), the moment of in- ertia about its horizontal central axis equals to bh 3 / 12. Therefore, if we choose an el- ement illustrated in figure 11( b ), the moment of inertia of the element about x -axis is dI T = dx (2 y ) 3 / 12 = (2 / 3) y 3 dx . Therefore, the total transverse moment of inertia of the waterplane about the centerline plane is I T = dI T = 2 3 L/ 2 − L/ 2 y 3 dx (2.7) 2.1.6 Waterplane area versus draft curve Figure 12: Waterplane area versus draft curve. A w is a function of z . If we plot A w versus z , the area between the curve and the z -axis is the volume of displacement ∇ , which can be calculated by 10

NE224 Spring 2023 Instructor: An Wang ∇ = T 0 A w dz Example: (Properties of the waterplane) At the design draft, the waterline of a ship can be described by the following equation: y = ± B 2 cos πx L , where L equals the length between perpendiculars (LBP), x is the longitudinal distance from amidship ( x = 0) and AP and FP are located at x = − L/ 2 and x = L/ 2, respectively. B is the width of the waterplane at amidship. Find: 1) Area of the waterplane, A w 2) First moment of the waterplane about amidship, M . 3) Longitudinal center of ﬂotation (LCF) of the waterplane. 4) Transverse and longitudinal moment of inertia of the waterplane, I T and I L , respec- tively. Solution: 1) A w = 2 L/ 2 − L/ 2 ydx = 2 L/ 2 − L/ 2 B 2 cos πx L dx = BL π sin πx L L/ 2 − L/ 2 = BL π (1 − ( − 1)) = 2 BL π 2) M = 2 L/ 2 − L/ 2 xydx = 2 L/ 2 − L/ 2 x B 2 cos πx L dx Integration by parts: 11

NE224 Spring 2023 Instructor: An Wang M = B L/ 2 − L/ 2 x L π d sin πx L = BL π x sin πx L L/ 2 − L/ 2 − BL π L/ 2 − L/ 2 sin πx L dx = BL π x sin πx L L/ 2 − L/ 2 + BL π L π cos πx L L/ 2 − L/ 2 = BL π L 2 (1) − − L 2 ( − 1) + BL π L π (0 − 0) = 0 3) LCF = M A w = 0 4) Moment of inertia about amidship ( x = 0) is I = 2 L/ 2 − L/ 2 x 2 ydx = 2 L/ 2 − L/ 2 x 2 B 2 cos πx L dx = B L/ 2 − L/ 2 x 2 cos πx L dx 12

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NE224 Spring 2023 Instructor: An Wang Integration by parts: I = B L/ 2 − L/ 2 x 2 L π d sin πx L = BL π x 2 sin πx L L/ 2 − L/ 2 − BL π L/ 2 − L/ 2 sin πx L d ( x 2 ) = BL π x 2 sin πx L L/ 2 − L/ 2 − 2 BL π L/ 2 − L/ 2 x sin πx L dx = BL π x 2 sin πx L L/ 2 − L/ 2 + 2 BL 2 π 2 L/ 2 − L/ 2 xd cos πx L = BL π x 2 sin πx L L/ 2 − L/ 2 + 2 BL 2 π 2 x cos πx L L/ 2 − L/ 2 − 2 BL 2 π 2 L/ 2 − L/ 2 cos πx L dx = BL π x 2 sin πx L L/ 2 − L/ 2 + 2 BL 2 π 2 x cos πx L L/ 2 − L/ 2 − 2 BL 3 π 3 sin πx L L/ 2 − L/ 2 = BL π x 2 sin πx L + 2 BL 2 π 2 x cos πx L − 2 BL 3 π 3 sin πx L L/ 2 − L/ 2 = BL π L 2 4 (1) + 0 − 2 BL 3 π 3 (1) − BL π L 2 4 ( − 1) + 0 − 2 BL 3 π 3 ( − 1) = BL 3 2 π − 4 BL 3 π 3 = 1 2 π − 4 π 3 BL 3 ≈ 0 . 03015 BL 3 Longitudinal moment of inertia about CF is I L = I − A w (LCF) 2 = I 13

NE224 Spring 2023 Instructor: An Wang Transverse moment of inertia about CF is I T = 2 3 L/ 2 − L/ 2 y 3 dx = 2 3 L/ 2 − L/ 2 B 2 cos πx L 3 dx = B 3 12 L/ 2 − L/ 2 cos πx L 1 − sin 2 πx L dx = B 3 12 L/ 2 − L/ 2 cos πx L dx − B 3 12 L/ 2 − L/ 2 cos πx L sin 2 πx L dx = B 3 12 L/ 2 − L/ 2 cos πx L dx − B 3 L 12 π L/ 2 − L/ 2 sin 2 πx L d sin πx L = B 3 L 12 π sin πx L L/ 2 − L/ 2 − B 3 L 36 π sin 3 πx L L/ 2 − L/ 2 = B 3 L 6 π − B 3 L 18 π = B 3 L 9 π 2.2 Properties of the Stations Figure 13: Station area integration. 2.2.1 Immersed Station Area The area of the element is dA s = ydz . Therefore, the immersed station area ( A s ) for a given station is calculated by A s = 2 T 0 ydz (2.8) A s : immersed station area, or sectional area; y : half-breadth of the station; 14