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Oct 30, 2023

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NE 224 Spring 2023 Homework 04 (Due 4/17 (Monday), Canvas upload) 1. A railcar barge (rectangular) is 52 m long, 9 m wide, and it floats in fresh water at 2.4 m draft even keel, with KG = 2 . 3 m. What will be the change in draft at each end of this barge if a 25,000 kg railcar (already on board) is rolled longitudinally toward the stern, a distance of 41 m traveled by the railcar’s center of mass? Solutions: The mass displacement of the barge: ∆ m = ρ w LBT = 1 MT / m 3 × 52 m × 9 m × 2 . 4 m = 1123 . 2 MT Height of center of buoyancy above baseline: KB = T/ 2 = 1 . 2 m Longitudinal metacentric radius BM L = I L ∇ = BL 3 12 ∇ = BL 3 12 LBT = L 2 12 T = (52 m) 2 12 × 2 . 4 m = 93 . 89 m Longitudinal metacentric height GM L = KB + BM L − KG = 1 . 2 m + 93 . 89 m − 2 . 3 m = 92 . 79 m (or use the approximation for longitudinal metacentric height: GM L ≈ BM L = 93 . 89 m). Change in trim: t = w ( x 1 − x 0 ) ∆ m GM L 100 L = 25 MT × 41 m 1123 . 2 MT × 92 . 79 m 100cm / m × 52 m = 51 . 14 cm trim by stern Since the barge is a rectangle, the center of flotation is located at the middle of the barge. Therefore, the trim is symmetric about amidship: A = F = L/ 2 = 26 m. The change in draft at FP: f = F L t = t 2 = 25 . 57 cm Change in draft at AP: a = A L t = t 2 = 25 . 57 cm 1

2. A Mariner enters port at drafts of 28’0” forward and 27’0” aft. The following cargo is discharged: 1900 tons from No. 2 hold (106.2 ft from FP) 850 tons from No. 4 hold (223.1 ft from FP) 1200 tons from No. 7 hold (469.4 ft from FP) The following cargo is loaded: 100 tons in No. 1 hold (56.6 ft from FP) 300 tons in No. 2 hold (106.2 ft from FP) 800 tons in No. 5 hold (353.6 ft from FP) Use the small weights method , determine the new drafts forward and aft after the loading and discharging are completed. Solutions: First determine the displacement according to the draft FP and AP. The mean draft T M = T F + T A 2 = 28 ft + 27 ft 2 = 27 . 5 ft From the hydrostatics properties table, at draft 27.5 ft, the following properties can be found: Displacement: ∆ = 19220 ton Moment to change trim 1inch: MT1 = 1800 ton · ft/in Tons per inch immersion: TPI = 68 . 4 ton/in. Center of flotation: LCF = 279 . 1 ft fom FP Calculate the moment created by all the weights: Weight (ton) Distance (ft) to FP Moment (ton · ft) about FP -1900 106.2 -201780 -850 223.1 -189635 -1200 469.4 -563280 100 56.6 5660 300 106.2 31860 800 353.6 282880 ∑ -2750 -634295 2

Center of gravity of all the weights: − 634295 ton · ft − 2750 ton = 230 . 65 ft From FP Distance of the discharged weight from center of flotation d = LCF − 230 . 65 ft = 279 . 1 ft − 230 . 65 ft = 48 . 45 ft So the loading/charging is equivalent to discharging 2750 ton of weight at 230.65 ft from FP. Since LCF = 279 . 1 ft > 230 . 65 ft, the weight is discharged between F and FP. Use small weight method δT x = w TPI ± X L wd MT1 In the above expression, w = − 2750 ton. At FP, the weight and FP is at the same side of center of flotation, use “+” sign: δT F = w TPI + F L wd MT1 = − 2750 ton 68 . 4 ton / in + 279 . 1 ft 528 ft − 2750 ton × 48 . 45 ft 1800 ton · ft / in = − 79 . 33 in At FP, the weight and AP is at the opposite side of center of flotation, use “-” sign: δT A = w TPI − A L wd MT1 = − 2750 ton 68 . 4 ton / in − 528 ft − 279 . 1 ft 528 ft − 2750 ton × 48 . 45 ft 1800 ton · ft / in = − 5 . 31 in New draft at FP: T ′ F = T F + δT F = 28 ft − 79 . 33 / 12 ft = 21 . 39 ft New draft at AP: T ′ A = T A + δT F = 27 ft − 5 . 31 / 12 ft = 26 . 56 ft 3

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3. A ship has the following characteristics when floating on even keel in seawater at a displacement of 37,100 MT: L = 230 m B = 30 m C B = 0 . 580 C W = 0 . 750 MTcm = 663 MT · m/cm LCF = 125 m from FP How far from the bow (FP) must a mass of 600 MT be loaded in order to increase the draft at the stern by 30 cm? What will be the resulting drafts forward and aft? Use the small weights method for the calculation. Note: C W is the “Waterplane Coefficient” and is defined by C W = A w LB , where A w is the area of the waterplane. Hint: TPC can be calculated from the waterplane area: TPC = A w / 97 . 56, where the units of TPC and A w are MT/cm and m 2 , respectively. Solution: Area of the waterplane: A w = C W LB = 0 . 75 × 230 m × 30 m = 5157 m 2 Tons per inch cm immersion: TPC = A w / 97 . 56 = 5157 / 97 . 56 = 53 . 04 MT / cm Distance from FP to center of flotation: F = LCF = 125 m. Distance from AP to center of flotation: A = L − LCF = 230 m − 125 m = 105 m. Use small weight method δT x = w TPC ± X L wd MTcm 4

In the above equation, replace X with A . We need to find d . If the weight is loaded between the center of flotation and FP, the draft at AP has to decrease. Therefore, the weight has to be loaded between center of flotation and AP and we should use “+” in the above equation. δT A = 30 cm = w TPC + A L wd MTcm = 600 MT 53 . 04 MT / cm + 105 m 230 m 600 MT × d 663 MT · m / cm Solve, d = 45 . 23 m aft of center of flotation. So the weight is loaded at d + LCF = 45 . 23 m + 125 m = 170 . 23 m from FP. Initial draft before loading at both FP and AP: T = ∇ C B LB = ∆ m ρC B LB = 37100 MT 1 MT / m 3 × 0 . 58 × 230 m × 30 m = 9 . 27 m The final draft at AP: T ′ A = T + δT A = 9 . 27 m + 30 / 100 m = 9 . 57 m Change in draft at FP: δT F = w TPC − F L wd MTcm = 600 MT 53 . 04 MT / cm − 125 m 230 m 600 MT × 45 . 23 m 663 MT · m / cm = − 10 . 93 cm The final draft at FP: T ′ F = T + δT F = 9 . 27 m − 10 . 93 / 100 m = 9 . 16 m 5

4. Determine the displacement and location of the center of gravity ( KG and LCG) of a Mariner vessel in seawater whose drafts are 20’0” forward and 25’0” aft, and whose transverse metacentric height ( GM T ) is 3.20 ft. Solution: The mean draft: T M = T F + T A 2 = 20 ft + 25 ft 2 = 22 . 5 ft The trim: t = T A − T F = 5 ft = 60 in by stern From the hydrostatics properties table, at 22.5 ft draft even-keel condition: Displacement: ∆ 0 = 15200 ton Center of buoyancy: LCB 0 = 265 . 87 ft from FP Center of flotation: LCF 0 = 273 . 5 ft from FP Moment to change trim 1 inch: MT1 = 1562 ton · ft/in Distance of transverse metacenter to baseline: KM T = 31 . 28 ft. The distance from center of gravity to baseline: KG = KM T − GM T = 31 . 28 ft − 3 . 20 ft = 28 . 08 ft From even-keel condition to the trimmed condition, the shift of center of gravity: GG 0 = t × MT1 ∆ 0 = 60 in × 1562 ton · ft / in 15200 ton = 6 . 166 ft Since the ship is trimmed by stern, GG 0 is positive. The distance of center of gravity to FP in the trimmed condition: LCG = LCB 0 + GG 0 = 265 . 87 ft + 0 . 514 ft = 272 . 04 ft from FP 6

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5. Do problem 2 with the large weights method and compare the drafts with those determined by the small weight method used in problem 2. Solution: Step 1: Find the displacement and LCG at the trimmed condition before loading. First determine the displacement according to the draft FP and AP. The mean draft T M = T F + T A 2 = 28 ft + 27 ft 2 = 27 . 5 ft The trim: t = T A − T F = 1 ft = 12 in by head From the hydrostatics properties table, at draft 27.5 ft even-keel condition, the follow- ing properties can be found: Displacement: ∆ = 19220 ton Moment to change trim 1inch: MT1 = 1800 ton · ft/in Center of buoyancy: LCB 0 = 268 . 05 ft from FP From even-keel condition to the trimmed condition, the shift of center of gravity: GG 0 = t × MT1 ∆ 0 = 12 in × 1800 ton · ft / in 19220 ton = 1 . 124 ft Since the ship is trimmed by head, GG 0 is negative. The distance of center of gravity to FP in the trimmed condition: LCG = LCB 0 − GG 0 = 268 . 05 ft − 1 . 124 ft = 266 . 93 ft from FP Step 2: Calculate the moment created by all the weights in the following table: 7

Weight (ton) Distance (ft) to FP Moment (ton · ft) about FP 19220 266.93 5130321 -1900 106.2 -201780 -850 223.1 -189635 -1200 469.4 -563280 100 56.6 5660 300 106.2 31860 800 353.6 282880 ∑ 16470 4496026 Total displacement after the loading: ∆ ′ = 16470 ton. Center of gravity of all the weights: LCG ′ = Total moment ∆ ′ = 4496026 ton · ft 16470 ton = 272 . 98 ft From FP Step 3: Calculate the trim after loading is completed: At the new displacement, ∆ ′ = 16470 ton, from the hydrostatic properties table: The even-keel draft: T ′ 0 = 24 . 18 ft The even-keel center of buoyancy: LCB ′ 0 = 266 . 54 ft from FP The center of flotation: LCF ′ = 275 . 3 ft from FP MT1 ′ = 1632 ton · ft/in The shift of center of gravity from the even keel condition to the trimmed condition. G ′ G ′ 0 = LCG ′ − LCB ′ 0 = 272 . 98 ft − 266 . 54 ft = 6 . 44 ft G ′ G ′ 0 > 0, so the ship will trim by stern. The trim after the loading: t ′ = ∆ ′ G ′ G ′ 0 MT1 ′ = 16470 ton × 6 . 44 ft 1632 ton · ft / in = 64 . 99 in by stern The change of draft FP: f ′ = F ′ L t ′ = LCF ′ L t = 275 . 3 ft 528 ft × 64 . 99 in = 33 . 89 in and a ′ = t ′ − f ′ = 31 . 10 in 8

The final draft at forward and aft: T ′ F = T ′ 0 − f ′ = 24 . 18 ft − 33 . 89 / 12 ft = 21 . 36 ft T ′ A = T ′ 0 + a ′ = 24 . 18 ft + 31 . 10 / 12 ft = 26 . 77 ft 9

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