25-Nav-A2 Hydrodynamics of Ships (I)_ Resistance and Propulsion · December 2017

Question 4 of 5: Trim After Flooding a Stern-End Compartment

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Question 4: Trim After Flooding a Stern-End Compartment (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Barge as in Q1 ($L=70$ m, $B=17$ m), floating at $d_0=4$ m with $KG=2.5$ m. A full-width, empty stern-end compartment of length $\ell=8$ m, permeability $\mu=0.85$, is opened to the sea. $\rho=1.025\ \text{t/m}^3$.

Find. The new forward and aft drafts.

Approach. Use the lost-buoyancy method: the flooded compartment (weighted by its permeability) can no longer contribute buoyant waterplane area, so the ship behaves as though it were shorter by $\mu\ell$ — it must sink bodily until the reduced hull displaces the same (unchanged) volume, and its centre of flotation shifts away from the flooded end, producing a trimming moment about the new F.

1. Bodily rise (parallel sinkage). The buoyancy lost in the compartment at the original draft is $v=\mu\ell B d_0=0.85\times8\times17\times4=462.4\ \text{m}^3$, over an "excluded" waterplane area $a=\mu\ell B=0.85\times8\times17=115.6\ \text{m}^2$. The remaining (intact) waterplane, $A_W-a=1190-115.6=1074.4\ \text{m}^2$, must rise enough to replace that lost volume: $$\delta=\dfrac{v}{A_W-a}=\dfrac{462.4}{1074.4}=0.430\ \text{m}\ \Rightarrow\ d_F=d_0+\delta=4.430\ \text{m (draft at the new F)}$$
2. New centre of flotation. Taking moments of area about the original midships (positive forward), the compartment's centroid is $x_c=8/2-70/2=-31\ \text{m}$ (31 m aft of midships). Removing area $a$ at $x_c$ from the original waterplane (centroid at $0$) shifts F to $$x_F=\dfrac{-a\,x_c}{A_W-a}=\dfrac{-115.6\times(-31)}{1074.4}=3.34\ \text{m forward of midships}$$ — F moves away from the flooded end, as expected.
3. Trimming moment. The lost buoyancy is a mass-equivalent of $v\rho=462.4\times1.025=474.0\ \text{t}$, acting effectively through the flooded compartment's centroid while the compensating buoyancy is generated through the new F, a lever arm of $x_F-x_c=3.34-(-31)=34.34\ \text{m}$ away: $$M_{trim}=474.0\times34.34=16{,}274\ \text{t}\cdot\text{m}$$
4. MCT1cm at the new mean draft and change of trim. Displacement is unchanged by flooding ($\Delta=\rho A_W d_0=4879.0\ \text{t}$, weight neither added nor removed — the sea simply fills a hole). Using the new draft, $BM_L=L^2/(12d_F)=4900/(12\times4.430)=92.17\ \text{m}$, so $$MCT1cm=\dfrac{\Delta\,BM_L}{100L}=\dfrac{4879.0\times92.17}{7000}=64.24\ \text{t}\cdot\text{m/cm}$$ Change of trim: $t=M_{trim}/MCT1cm=16{,}274/64.24=253.3\ \text{cm}=2.533\ \text{m}$, by the stern (the flooded end sinks deeper).
5. New forward and aft drafts. Splitting the trim about the new F (at $3.34$ m forward of midships, i.e. $38.34$ m from AP and $31.66$ m from FP): $$d_{AP}=d_F+t\times\dfrac{38.34}{70}=4.430+2.533\times0.5477=5.82\ \text{m}$$ $$\boxed{d_{FP}=d_F-t\times\dfrac{31.66}{70}=4.430-2.533\times0.4523=3.28\ \text{m}}$$ $$\boxed{d_{AP}=5.82\ \text{m}}$$ Both remain below the 7.5 m depth, so the barge does not submerge its deck at either end.