Bilging and Permeability (Derrett Stability)
A box-shaped vessel 40 m long, 6 m beam, is floating at a draft of 2 m F and A. She has an amidships compartment 10 m long which is empty. If the original GM was 0.6 m, find the new GM if this compartment is bilged.
KM = KB + BM = d/2 + b²/12d = 2/2 + 6²/12*2 = 2.5 m
KM = 2.5 m
Old KG = KM – Old GM = 2.5 m – 0.6 m = 1.9 m
Increase in draft = (Volume of lost buoyancy) / (Area of intact water plan)
Increase in draft = V / (A – a)
= (l * B * d) / ((L * B) – (l * B))
= (10 * 6 * 2) / ((40 * 6) – (10 * 6))
= 120 / 180 = 0.67 m
New draft = Old draft + Increase in draft = 2 + 0.67 = 2.67 m
New KM = New KB + New BM
= New draft/2 + b²/12d = 2.67/2 + 6²/(12*2.67)
= 1.335 + 1.124 = 2.459 m
New KM = New KG + New GM
New GM = New KM – New KG = 2.459 m – 1.9 m = 0.559 m (Ans)
The mean draft of a ship increases when a compartment is bilged. The positions of the initial metacenter and the buoyancy center are also affected by changes in the mean draft. As a result, KM is changed, and while KG remains unchanged, the GM is also changed.
Bilging and Permeability (Derrett Stability)
A box-shaped vessel 64 m * 10 m * 6 m floats in salt water on an even keel at a 5 m draft. A forward compartment 6 meters long and 10 meters wide, extends from the outer bottom to a height of 3.5 m and is full of cargo of permeability of 25 percent. Find the new drafts if this compartment is now bilged.
Mass of water entering the bilge compartment = 25% * 6 * 10 * 3.5 * 1.025 = 53.813 tonne
TPCsw = (WPA * 1.025) / 100 = ((64 * 10) * 1.025)) / 100 = 6.56 tonne
Increase in draft / bodily sinkage = w / TPC = 53.813 / 6.56 = 8.20 cm = .082 m by the head
Displacement, W = L * B * d * 1.025 = 64 * 10 * 5 * 1.025 = 3280 tonne
BML = L² / 12d = 64² / (12 * 5) = 68.27 m
MCTC = (W * BML) / 100 L = (3280*68.27) / (100 * 64) = 34.99 t-m
Change of trim = (w * d) / MCTC = (53.813 * 29) / 34.99 ( HERE, d = (64-6)/2 = 29 m )
= 44.60 cm
Change of draft aft = 1/2 * COT (Box shaped vessel)
= 1/2 * 44.60 = 22.3 cm = 0.223 m
New draft fwd = old draft fwd + bodily sinkage + change due to trim
= 5 + 0.082 + 0.223 = 5.305 m
New draft aft = Old draft aft + bodily sinkage – change due to trim
= 5 + 0.082 – .223 = 4.859 m
So, New draft F – 5.305, A – 4.859 m (Ans).
