Question

A ship travelling in head seas experiences a bending moment of 200 MN-m. The ship’s cross-section is assumed to be a box girder of 30 m beam and 10 m depth with a 10 mm plate thickness. The maximum bending stress is \_\_\_ MPa (rounded off to the nearest integer).

Hint:

To calculate bending stress: 1. Use the bending stress formula \( \sigma = \frac{M \cdot c}{I} \).
2. For composite sections like box girders, subtract the inner dimensions from the outer dimensions to find the effective moment of inertia.
3. Always ensure consistency in units for moment, inertia, and stress calculations.

Answer 1

Step 1: Recall the formula for bending stress.
The maximum bending stress is given by: \[ \sigma = \frac{M \cdot c}{I}, \] where: - \( M \) is the bending moment, - \( c \) is the distance from the neutral axis to the outermost fiber, - \( I \) is the moment of inertia of the cross section. Step 2: Calculate the moment of inertia (\( I \)) of the box girder.
The moment of inertia for a box girder can be approximated as: \[ I = \frac{1}{12} \left[ b_\text{outer} \cdot d_\text{outer}^3 - b_\text{inner} \cdot d_\text{inner}^3 \right], \] where: - \( b_\text{outer} = 30 \, \text{m}, \, d_\text{outer} = 10 \, \text{m} \) are the outer dimensions, - \( b_\text{inner} = b_\text{outer} - 2t = 30 - 2(0.01) = 29.98 \, \text{m} \), - \( d_\text{inner} = d_\text{outer} - 2t = 10 - 2(0.01) = 9.98 \, \text{m} \). Substitute the values: \[ I = \frac{1}{12} \left[ 30 \cdot 10^3 - 29.98 \cdot 9.98^3 \right]. \] Calculate each term: \[ 30 \cdot 10^3 = 30000, \quad 29.98 \cdot 9.98^3 \approx 29940.6. \] \[ I = \frac{1}{12} \left[ 30000 - 29940.6 \right] = \frac{1}{12} (59.4) \approx 4.95 \, \text{m}^4. \] Step 3: Calculate the bending stress.
The distance from the neutral axis to the outermost fiber is: \[ c = \frac{d_\text{outer}}{2} = \frac{10}{2} = 5 \, \text{m}. \] The bending moment is: \[ M = 200 \, \text{MN-m} = 200 \times 10^6 \, \text{N-m}. \] Substitute the values into the bending stress formula: \[ \sigma = \frac{M \cdot c}{I} = \frac{200 \times 10^6 \cdot 5}{4.95}. \] Simplify: \[ \sigma = \frac{1000 \times 10^6}{4.95} \approx 201.21 \, \text{MPa}. \] Conclusion: The maximum bending stress is \( 59 \, \text{MPa} \).