Beam Analysis: Maximum Bending Stress Calculation

The problem asks for the maximum bending stress ($\sigma_{max}$) in a simply-supported steel I-beam under a uniformly distributed load (UDL). We need to calculate this value based on the given span, load, moment of inertia, and beam depth.

Load and Span Conversion

First, convert all units to a consistent system, preferably millimeters (mm) and Newtons (N) for stress calculation in N/mm$^2$.

• Span, $L = 8 \text{ m} = 8000 \text{ mm}$
• Uniformly Distributed Load, $w = 15 \text{ kN/m} = 15 \times \frac{1000 \text{ N}}{1000 \text{ mm}} = 15 \text{ N/mm}$
• Moment of Inertia, $I = 18000 \text{ cm}^4 = 18000 \times (10 \text{ mm})^4 = 1.8 \times 10^8 \text{ mm}^4$
• Overall Depth, $D = 450 \text{ mm}$

Maximum Bending Moment Calculation

For a simply-supported beam with a UDL, the maximum bending moment ($M_{max}$) occurs at the center and is calculated using the formula:

$ M_{max} = \frac{wL^2}{8} $

Substituting the values:

$ M_{max} = \frac{(15 \text{ N/mm}) \times (8000 \text{ mm})^2}{8} $

$ M_{max} = \frac{15 \times 64,000,000}{8} \text{ N-mm} $

$ M_{max} = 15 \times 8,000,000 \text{ N-mm} = 120,000,000 \text{ N-mm} $

$ M_{max} = 1.2 \times 10^8 \text{ N-mm} $

Maximum Stress Determination

The maximum bending stress ($\sigma_{max}$) is given by the flexure formula:

$ \sigma_{max} = \frac{M_{max} \cdot y_{max}}{I} $

Where $y_{max}$ is the distance from the neutral axis to the extreme fiber. For a symmetrical I-section, this is half the overall depth:

$ y_{max} = \frac{D}{2} = \frac{450 \text{ mm}}{2} = 225 \text{ mm} $

Now, calculate the maximum bending stress:

$ \sigma_{max} = \frac{(1.2 \times 10^8 \text{ N-mm}) \times (225 \text{ mm})}{1.8 \times 10^8 \text{ mm}^4} $

$ \sigma_{max} = \frac{1.2 \times 225}{1.8} \text{ N/mm}^2 $

$ \sigma_{max} = \frac{270}{1.8} \text{ N/mm}^2 $

$ \sigma_{max} = 150 \text{ N/mm}^2 $

The maximum bending stress is 150 N/mm$^2$. Since the question asks for the answer as an integer, the value is 150.