Question

A vessel of 100 m length has a constant triangular cross-section with a depth of 12 m and breadth of 15 m as shown in following figure. The vessel has a vertical center of gravity (KG) = 6.675 m. The minimum draft (d), at which the vessel will become stable is .................... m (round off to one decimal place) 

 

Hint:

For stability problems, break them down into finding the components of KM: KB and BM. Remember that for simple geometric shapes, KB is at the centroid of the submerged volume. BM depends on the waterplane shape (\(I_T\)) and submerged volume (\(\nabla\)).

Answer 1

Step 1: Understanding the Concept:
For a vessel to be stable, its initial metacentric height (GM) must be non-negative (\(GM \geq 0\)). The metacentric height is the vertical distance between the center of gravity (G) and the transverse metacenter (M). We need to find the minimum draft \(d\) that satisfies this condition.
Step 2: Key Formula or Approach:
The stability condition is \(GM = KM - KG \geq 0\), which means \(KM \geq KG\).
\(KM\) is the height of the metacenter above the keel, given by \(KM = KB + BM\).
- \(KB\): Height of the center of buoyancy (centroid of the submerged volume) above the keel. For a triangle with vertex at the keel, \(KB = \frac{2}{3}d\). - \(BM\): Transverse metacentric radius, given by \(BM = \frac{I_T}{\nabla}\), where \(I_T\) is the second moment of area of the waterplane and \(\nabla\) is the displaced volume.
Step 3: Detailed Explanation or Calculation:
Given values:
Total depth D = 12 m, Total breadth B = 15 m, Length L = 100 m, KG = 6.675 m.
1. Express waterplane breadth (b) in terms of draft (d): By similar triangles: \[ \frac{b}{d} = \frac{B}{D} = \frac{15}{12} = \frac{5}{4} \implies b = \frac{5}{4}d \] 2. Calculate the second moment of waterplane area (\(I_T\)): The waterplane is a rectangle of length L and breadth b. \[ I_T = \frac{1}{12} L b^3 = \frac{1}{12} (100) \left(\frac{5}{4}d\right)^3 = \frac{100}{12} \frac{125}{64} d^3 = \frac{25}{3} \frac{125}{64} d^3 = \frac{3125}{192} d^3 \] 3. Calculate the displaced volume (\(\nabla\)): The submerged section is a triangular prism. \[ \nabla = \text{Area}_{\text{submerged}} \times L = \left(\frac{1}{2} b d\right) \times L = \frac{1}{2} \left(\frac{5}{4}d\right) d \times 100 = \frac{125}{2} d^2 \] 4. Calculate BM: \[ BM = \frac{I_T}{\nabla} = \frac{\frac{3125}{192}d^3}{\frac{125}{2}d^2} = \frac{3125}{192} \times \frac{2}{125} d = \frac{25}{96}d \] 5. Apply the stability condition: \[ KB + BM \geq KG \] \[ \frac{2}{3}d + \frac{25}{96}d \geq 6.675 \] Combine the terms with d: \[ \left(\frac{2 \times 32}{3 \times 32} + \frac{25}{96}\right)d \geq 6.675 \] \[ \left(\frac{64}{96} + \frac{25}{96}\right)d \geq 6.675 \] \[ \frac{89}{96}d \geq 6.675 \] Solve for d: \[ d \geq 6.675 \times \frac{96}{89} \approx 6.675 \times 1.07865 \approx 7.1998 \text{ m} \] Step 4: Final Answer:
Rounding to one decimal place, the minimum draft for stability is 7.2 m.
Step 5: Why This is Correct:
The calculation correctly formulates the stability requirement \(KM \geq KG\) in terms of the draft \(d\). Solving the inequality gives the minimum draft required for the metacenter to be at or above the center of gravity, which is the condition for neutral/stable equilibrium. The result of 7.2 m matches the provided answer range.