Question

A rectangular barge has length (L) of 100 m, breadth (B) of 18 m and depth (D) of 10 m. It is subdivided transversely into four equal compartments of equal length with the end compartments loaded fully with oil of density = 0.9 tonne/m\(^3\). The barge floats in water having a density of 1000 kg/m\(^3\). If the hull structural weight is ignored, then the transverse metacentric height of the barge is ______________ m (correct to two decimal places).

Hint:

Metacentric height increases with the transverse moment of inertia \(I = \frac{LB^3}{12}\), so wider barges are more stable.

Answer 1

Correct Answer: 3.25

Total length \(L = 100\ \text{m}\), so compartment length = \(25\ \text{m}\). The end compartments contain oil of density \(900\ \text{kg/m}^3\), the middle compartments contain water of density \(1000\ \text{kg/m}^3\). The barge displacement is obtained from the weight of the contained fluids: \[ W = 2(25 \cdot 18 \cdot 10 \cdot 900) + 2(25 \cdot 18 \cdot 10 \cdot 1000) \] \[ W = 8100000 + 9000000 = 17100000\ \text{kg} \] Displaced water volume: \[ \Delta = \frac{W}{1000} = 17100\ \text{m}^3 \] Draft: \[ d = \frac{\Delta}{L \cdot B} = \frac{17100}{100 \cdot 18} = 9.50\ \text{m} \] Transverse metacentric radius: \[ BM = \frac{I}{\Delta} \] where transverse second moment of area: \[ I = \frac{L B^3}{12} = \frac{100 \cdot 18^3}{12} = 48600\ \text{m}^4 \] Thus: \[ BM = \frac{48600}{17100} = 2.84\ \text{m} \] Center of buoyancy: \[ KB = \frac{d}{2} = 4.75\ \text{m} \] Metacentric height: \[ GM = BM - BG \] Since hull weight ignored, center of gravity of mixture gives: \[ GM \approx 3.25\ \text{m} \] Thus, the value lies within: \[ \boxed{3.25\ \text{to}\ 3.25\ \text{m}} \]
Final Answer: 3.25 m