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NE 224 Spring 2023 Homework 01 (Due 11:59PM, 2/20, Monday, Canvas submission) 1. (Modified from Zubaly chapter 3, problem 7) A rectangular barge 320 ft long, 40 ft wide and 12 ft deep is floating in seawater at a draft of 5.2 ft. At what draft would it float in fresh water if the loading remains unchanged? How many tons of cargo would be removed to make the fresh water draft equal to what it was in seawater? (Assume temperature is at 68 F, Density of Seawater: ρ s = 1 . 9882 lb · s 2 /ft 4 , Density of fresh water: ρ f = 1 . 9367 lb · s 2 /ft 4 , see Appendix B, Zubaly). Note: ton and lb are force units and slug is mass units. 1 ton = 2000 lb, 1 lb = 1 slug × ft/s 2 and g = 32 . 17 ft/s 2 is the gravitational acceleration. Solution: 1) If the loading is unchanged, the displacement ∆ (weight) remains the same: ∆ = ρ f g ∇ f = ρ s g ∇ s (1) However, the volume of displacement will change. The volume of displacement in fresh water and seawater is ∇ f = LBT f (2) ∇ s = LBT s (3) , respectively. T f and T s are the draft in fresh water and seawater, respectively. Sub- stitute equations (2) and (3) into (1). ρ f gLBT f = ρ s gLBT s we get ρ f T f = ρ s T s The draft in fresh water: T f = ρ s ρ f T s = 1 . 9882 lb · s 2 / ft 4 1 . 9367 lb · s 2 / ft 4 (5 . 2 ft) = 5 . 34 ft 1

2) If the draft is kept the same, T s = T f , the volume of displacement will remain the same, according to equations (2) and (3). However, the displacement (weight) will change. The displacement (weight) in fresh water and seawater: ∆ f = ρ f g ∇ f = ρ f gLBT f (4) ∆ s = ρ s g ∇ s = ρ s gLBT s (5) Dividing the two equation above, ∆ f ∆ s = ρ f T f ρ s T s = ρ f ρ s ∆ f = ρ f ρ s ∆ s The weight of cargo needed to be removed: w = ∆ s − ∆ f = 1 − ρ f ρ s ∆ s = 1 − ρ f ρ s ρ s gLBT s w = 1 − 1 . 9367 lb · s 2 / ft 4 1 . 9882 lb · s 2 / ft 4 (1 . 9882 lb · s 2 / ft 4 )(32 . 17 ft / s 2 )(320 ft)(40 ft)(5 . 2 ft) w = 110274 lb = 110274 / 2000 ton = 55 . 14 ton 2

2. Use the table of offsets of Mariner class (see below, also in Chapter 2, figure 2-4, Zubaly) and trapezoidal method to calculate: 1) A w (area of the waterplane at 40 ′ − 0 ′′ WL between station 0 and station 10). 2) LCF (the longitudinal position of the center of flotation of the waterplane at 40 ′ − 0 ′′ WL between station 0 and station 10). LCF is relative to amidship (station 5). 3) I L and I T (the longitudinal and transverse moment of inertia of the waterplane at 40 ′ − 0 ′′ WL between station 0 and station 10 about CF). Note: The longitudinal distance between station 0 and station 10 is L = 520 ft. You can ignore station-1/2 and station-9 1/2 in the offset table and only use the offsets at stations 0, 1, 2,..., 8, 9, 10. The format of the offsets in the table is: ft-in-1/8in. Figure 1: Table of offsets. 3

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Figure 2: Lines drawing. Solution: Use the following table for the calculation: Station y (ft) x (ft) Multiplier II × IV II × III × IV II × III 2 × IV II 3 × IV I II III IV V VI VII VIII 0 1.729 260 0.5 0.865 224.792 58445.833 2.585 1 14.302 208 1 14.302 2974.833 618765.333 2925.485 2 26.979 156 1 26.979 4208.750 656565.000 19637.473 3 35.240 104 1 35.240 3664.917 381151.333 43761.510 4 37.979 52 1 37.979 1974.917 102695.667 54781.799 5 38.000 0 1 38.000 0.000 0.000 54872.000 6 38.000 -52 1 38.000 -1976.000 102752.000 54872.000 7 38.000 -104 1 38.000 -3952.000 411008.000 54872.000 8 37.354 -156 1 37.354 -5827.250 909051.000 52121.530 9 31.260 -208 1 31.260 -6502.167 1352450.667 30548.106 10 16.979 -260 0.5 8.490 -2207.292 573895.833 2447.480 ∑ 306.469 -7416.5 5166780.667 370841.968 In the table, column II is the half-breadth in ft converted from the column “40’-0” 4

WL” in the offset table. 1) Area of water plane: A w = 2 Z L/ 2 L/ 2 ydx ≈ 2 l 1 2 y 0 + y 1 + ... + y 9 + 1 2 y 10 = 2 l X V = 2 × 52 × 306 . 469 = 31872 . 776 ft 2 2) Longitudinal position of center of flotation: LCF = M A w = 2 Z L/ 2 L/ 2 xydx 2 Z L/ 2 L/ 2 ydx ≈ 2 l 1 2 x 0 y 0 + x 1 y 1 + ... + x 9 y 9 + 1 2 x 10 y 10 2 l 1 2 y 0 + y 1 + ... + y 9 + 1 2 y 10 = l ∑ VI l ∑ V = ∑ VI ∑ V = − 7416 . 5 306 . 469 = − 24 . 200 ft 3) Longitudinal moment of inertia about midship (station 5): I = 2 Z L/ 2 L/ 2 x 2 ydx = ≈ 2 l 1 2 x 2 0 y 0 + x 2 1 y 1 + ... + x 2 9 y 9 + 1 2 x 2 10 y 10 = 2 l X VII = 2 × 52 × 5166780 . 667 = 537 , 345 , 189 . 3 ft 4 The longitudinal moment of inertia about LCF: I L = I − A w (LCF) 2 = 537345189 . 3 ft 4 − (31872 . 776 ft 2 )(24 . 200 ft) 2 = 518 , 679 , 216 . 8 ft 4 The transverse moment of inertia about centerplane: I T = 2 3 Z L/ 2 L/ 2 y 3 dx = ≈ 2 3 l 1 2 y 3 0 + y 3 1 + ... + y 3 9 + 1 2 y 3 10 = 2 3 l X VIII = 2 3 × 52 × 370841 . 968 = 12 , 855 , 854 . 88 ft 4 5

3. The waterplane areas ( A w ) of a ship at various draft ( T ) are given in the following table. Waterplane No. 0 (baseline) 1 2 3 4 5 6 7 8 9 10 A w (m 2 ) 0 400 630 810 936 1024 1104 1164 1220 1270 1315 The waterplanes listed in the table are equally spaced vertically and the vertical dis- tance between adjacent waterplanes is δT =1.10 m. Waterplane No.0 is at the baseline. 1) Find TPC (change of displacement mass in MT per cm draft change) at each draft. Then make a plot of TPC versus T , using computer program, such as Matlab or Excel or any other program of your choice. 2) Use trapezoidal rule to calculate the volume of displacement ( ∇ ) under each water- plane shown in the table. Then make a plot of ∇ versus T , using computer program, such as Matlab or Excel or any other program of your choice. 3) When the ship floats at Waterplane No.10, use trapezoidal rule AND Simpson’s 1st rule to calculate KB (the vertical position of the center of buoyancy above the baseline). Solution: 1) TPC is a function of A w : TPC = A w 97 . 56 T(m) 0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11 A w (m 2 ) 0 400 630 810 936 1024 1104 1164 1220 1270 1315 TPC 0.00 4.10 6.46 8.30 9.59 10.50 11.32 11.93 12.51 13.02 13.48 6

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2) The volume of displacement under a waterplane at draft T (function of T): ∇ ( T ) = Z T 0 A w dz Use trapezoidal method at each T . T = T 0 = 0: ∇ ( T 0 ) = 0 T = T 1 = 1 . 1 m: ∇ ( T 1 ) = δT 1 2 A w 0 + 1 2 A w 1 = 220 m 3 T = T 2 = 2 . 2 m: ∇ ( T 2 ) = δT 1 2 A w 0 + A w 0 + 1 2 A w 2 = 786 . 5 m 3 T = T 3 = 3 . 3 m: ∇ ( T 3 ) = δT 1 2 A w 0 + A w 1 + A w 2 + 1 2 A w 3 = 1578 . 5 m 3 ... T = T 10 = 11 m: ∇ ( T 10 ) = δT 1 2 A w 0 + A w 1 + A w 2 + ... + 1 2 A w 10 = 10137 m 3 Plot ∇ versus T : 7

T(m) 0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11 A w (m 2 ) 0 400 630 810 936 1024 1104 1164 1220 1270 1315 TPC 0.00 4.10 6.46 8.30 9.59 10.50 11.32 11.93 12.51 13.02 13.48 ∇ (m 3 ) 0 220 786.5 1578.5 2538.8 3616.8 4787.2 6034.6 7345.8 8715.3 10137.1 3) When the ship floats at water plane No. 10, T = T 10 = 11 m. The vertical position of center of buoyancy KB = Z T 10 0 zA w dz ∇ ( T 10 ) = Z T 10 0 zA w dz Z T 10 0 A w dz Use trapezoidal method: Z T 10 0 zA w dz ≈ δT 1 2 T 0 A w 0 + T 1 A w 1 + T 2 A w 2 + ... + 1 2 T 10 A w 10 = 67144 m 4 ∇ ( T 10 ) = 10137 m 3 (see part 2)) Therefore, KB trapz = Z T 10 0 zA w dz ∇ ( T 10 ) = 67144 m 4 10137 m 3 = 6 . 62 m Use Simpson’s 1st rule: Z T 10 0 zA w dz ≈ δT 1 3 T 0 A w 0 + 4 3 T 1 A w 1 + 2 3 T 2 A w 2 + ... + 4 3 T 9 A w 9 + 1 3 T 10 A w 10 = 66969 m 4 8

∇ ( T 10 ) = Z T 10 0 A w dz ≈ δT 1 3 A w 0 + 4 3 A w 1 + 2 3 A w 2 + ... + 4 3 A w 9 + 1 3 A w 10 = 10181 m 3 KB simp = Z T 10 0 zA w dz ∇ ( T 10 ) = 66969 m 4 10181 m 3 = 6 . 58 m 9

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4. At designed draft T = 20 m, the immersed midship section of a ship hull can be described by the following equation y = f ( z ) = B 2 r z T , where z is the vertical position above baseline and y is the half-breadth of the hull at any height and B = 30 m is the width of the immersed midship section. z = 0 at the baseline and z = T = 20 m at the waterplane. The length on waterline at designed draft is L = 100 m. The volume of displacement is ∇ = 35 , 000 m 3 . 1) Derive the exact function of the area ( A ) below height z : A = g ( z ). And evaluate A = g ( z ) at z = T to obtain the exact value of immersed sectional area, A M = g ( T ). Then calculate the midship section coefficient ( C M ), the prismatic coefficient ( C P ) and the block coefficient ( C B ). 2) Obtain the exact values of y = f ( z ) at 7 different heights equally spaced between 0 ≤ z ≤ T , i.e., z = [0 , 1 6 T, 2 6 T, 3 6 T, ..., 5 6 T, T ] 3) Use the 7 points ( z 0 , y 0 ), ( z 1 , y 1 ),..., ( z 6 , y 6 ) from step 2) as grid points, calculate A M using trapezoidal rule AND Simpson’s 1st rule. 4) Evaluate exact value of y = f ( z ) at 15 different heights equally spaced between 0 ≤ z ≤ T , i.e., z = [0 , 1 14 T, 2 14 T, 3 14 T, ..., 13 14 T, T ] 5) Use the 15 points ( z 0 , y 0 ), ( z 1 , y 1 ),..., ( z 14 , y 14 ) from step 4) as grid points, calculate A M using trapezoidal rule AND Simpson’s 1st rule. 6) Compare the exact solution of A M from step 1) and the approximation of A M by trapezoidal rule and Simpson’s 1st rule with 7 grid points from step 3). Repeat the comparison with 15 grid points from step 5). Which approximation is closer to the exact solution? Does the error get greater or smaller as the number of grid points increases? Solution: 1) Sectional area below height z : A = g ( z ) = 2 Z z 0 ydz = 2 Z z 0 f ( z ) dz = 2 Z z 0 B 2 r z T dz = B r 1 T Z z 0 z 0 . 5 dz 10

A = g ( z ) = 2 B 3 r z 3 T At the designed draft, z = T , plug into the equation above to get the exact solution of immersed sectional area: A M = g ( T ) = 2 B 3 r T 3 T A M exact = g ( T ) = 2 B 3 T = 2 3 (30 m)(20 m) = 400 . 0000 m 2 Midship section corfficient: C M = A M BT = 400 . 0000 m 2 (30 m)(20 m) = 0 . 6667 Prismatic coefficient: C P = ∇ A M L = 35000 m 3 (400 m 2 )(100 m) = 0 . 875 Block coefficient: C B = C M × C P = 0 . 6667 × 0 . 875 = 0 . 583 2) Evaluate y = f ( z ) at z = [0 , 1 6 T, 2 6 T, 3 6 T, ..., 5 6 T, T ] we get: y = B 2 r z T = [0 . 0000 , 6 . 1237 , 8 . 6603 , 10 . 6066 , 12 . 2474 , 13 . 6931 , 15 . 0000] 3) Use trapezoidal method to calculate the immersed sectional area A M : A M trapz7 = 2 Z T 0 ydz = 2 δT 1 2 y 0 + y 1 + y 2 + ... + 1 2 y 6 = 2 T 6 1 2 × 0 + 6 . 1237 + 8 . 6603 + ... + 1 2 × 15 = 392 . 2073 m 2 11

Use Simpson’s 1st rule to calculate the immersed sectional area A M : A M simp7 = 2 Z T 0 ydz = 2 δT 1 3 y 0 + 4 3 y 1 + 2 3 y 2 + ... + 1 3 y 6 = 2 T 6 1 3 × 0 + 4 3 × 6 . 1237 + 2 3 × 8 . 6603 + ... + 1 3 × 15 = 396 . 6866 m 2 4) Evaluate y = f ( z ) at z = [0 , 1 14 T, 2 14 T, 3 14 T, ..., 13 14 T, T ] we get: y = B 2 r z T = [0 . 0000 , 4 . 0089 , 5 . 6695 , 6 . 9437 , 8 . 0178 , 8 . 9642 , 9 . 8198 , 10 . 6066 , 11 . 3389 , 12 . 0268 , 12 . 6773 , 13 . 2961 , 13 . 8873 , 14 . 4544 , 15 . 0000] 5) Use trapezoidal method to calculate the immersed sectional area A M : A M trapz15 = 2 Z T 0 ydz = 2 δT 1 2 y 0 + y 1 + y 2 + ... + 1 2 y 14 = 2 T 14 1 2 × 0 + 4 . 0089 + 5 . 6695 + ... + 1 2 × 15 = 397 . 7464 m 2 Use Simpson’s 1st rule to calculate the immersed sectional area A M : A M simp15 = 2 Z T 0 ydz = 2 δT 1 3 y 0 + 4 3 y 1 + 2 3 y 2 + ... + 1 3 y 14 = 2 T 14 1 3 × 0 + 4 3 × 4 . 0089 + 2 3 × 5 . 6695 + ... + 1 3 × 15 = 399 . 0701 m 2 6) Compare: A M exact = 400 . 0000 m 2 A M trapz7 = 392 . 2073 m 2 A M simp7 = 396 . 6866 m 2 A M trapz15 = 397 . 7464 m 2 12

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A M simp15 = 399 . 0701 m 2 With the same number of points, Simpson’s 1st rule is closer to the exact solution than the trapezoidal method. For either trapezoidal method or Simpson’s 1st rule, solution with more gird points is closer to the exact solution. 13