Question

An 80 m long barge with a rectangular cross-section of 12 m beam and 4 m draft floats at even keel. The transverse metacenter (KM) above the keel is ………. m.
 

Hint:

To calculate the transverse metacenter (KM), ensure that the values for \( KB \) and \( BM \) are derived accurately using the geometry of the hull and displacement volume.

Answer 1

Step 1: Recall the formula for the transverse metacenter height (KM). 
The transverse metacenter height (KM) above the keel is given by: \[ KM = KB + BM, \] where: - \( KB \) is the distance from the keel to the center of buoyancy, - \( BM \) is the metacentric radius. 
Step 2: Calculate \( KB \). 
For a rectangular cross-section, the center of buoyancy (KB) is located at half the draft: \[ KB = \frac{\text{Draft}}{2} = \frac{4}{2} = 2 \, \text{m}. \] 
Step 3: Calculate \( BM \). 
The metacentric radius (\( BM \)) is calculated using the formula: \[ BM = \frac{I}{V}, \] where: - \( I \) is the second moment of area of the waterplane about the centerline, given by \( \frac{B^3 \cdot L}{12} \), - \( V \) is the volume of displacement, given by \( B \cdot L \cdot \text{Draft} \). Substitute the values: \[ I = \frac{B^3 \cdot L}{12} = \frac{12^3 \cdot 80}{12} = 11,520 \, \text{m}^4, \] \[ V = B \cdot L \cdot \text{Draft} = 12 \cdot 80 \cdot 4 = 3,840 \, \text{m}^3. \] Thus: \[ BM = \frac{I}{V} = \frac{11,520}{3,840} = 3 \, \text{m}. \] Step 4: Calculate \( KM \). 
Substitute the values of \( KB \) and \( BM \) into the formula for \( KM \): \[ KM = KB + BM = 2 + 3 = 5 \, \text{m}. \] Conclusion: The transverse metacenter (KM) above the keel is \( 5 \, \text{m} \).