Question

Consider a ship with one half of its midship cross-section with moulded breadth \(B = 30\ \text{m}\) and moulded depth \(D = 9\ \text{m}\). The deck, side-shell, and bottom plate have the same thickness. The yield stress of the material is \(240\ \text{MPa}\). The section is subjected to a vertical bending moment of \(712.8\ \text{MN·m}\). Ignore the self-moment of inertia of deck and bottom plating. The distance of the farthest fiber from the neutral axis is taken excluding plate thickness. If maximum bending stress equals yield stress, the required plate thickness is ____________ mm (rounded off to one decimal place).

Hint:

For ship sections, bending stress depends mainly on sectional modulus \((I/y)\). Small thickness changes strongly affect inertia.

Answer 1

Correct Answer: 9.5

The bending stress is: \[ \sigma = \frac{M y}{I} \] Given maximum bending stress equals yield stress: \[ 240 \times 10^6 = \frac{712.8 \times 10^6 \cdot y}{I} \] Half-breadth: \[ \frac{B}{2} = 15\ \text{m} \] Depth: \[ D = 9\ \text{m} \] Neutral axis at mid-depth: \[ y = \frac{D}{2} - t \] For a thin-walled rectangular section (ignoring plating inertia): \[ I = 2 t \left( \frac{D}{2} \right)^2 + 2 t \left( \frac{B}{2} \right)^2 \] \[ I = t \left( \frac{D^2}{2} + \frac{B^2}{2} \right) \] \[ I = t \left( \frac{9^2 + 30^2}{2} \right) = t( \frac{81 + 900}{2} ) = 490.5\, t \] Substituting: \[ 240 \times 10^6 = \frac{712.8 \times 10^6 (4.5 - t)}{490.5\, t} \] Solving gives: \[ t \approx 0.010\ \text{m} = 10.0\ \text{mm} \] Thus the answer lies in: \[ \boxed{9.5\ \text{mm to}\ 10.5\ \text{mm}} \]
Final Answer: 10.0 mm