Question

A simply-supported steel beam made of an I-section has a span of 8 m. The beam is carrying a uniformly distributed load of 15 kN/m. The overall depth of the beam is 450 mm. The moment of inertia of the beam section is 18000 cm\(^4\). The maximum bending stress in the beam will be \(\underline{\hspace{1cm}}\) N/mm\(^2\) (in integer).

Hint:

Bending stress is determined by the moment of inertia, the distance from the neutral axis, and the applied bending moment.

Answer 1

Correct Answer: 150

The total load on the beam is: \[ W = 15 \times 8 = 120\ \text{kN} \] The bending moment at the center is: \[ M = \frac{W \times L}{4} = \frac{120 \times 8}{4} = 240\ \text{kNm} \] Maximum bending stress is given by: \[ \sigma = \frac{M}{S} \] Where \(S = \frac{I}{c}\), and \(c = \frac{d}{2} = 0.225\ \text{m}\). Thus, the maximum bending stress: \[ \sigma = \frac{240 \times 10^3}{\frac{18000 \times 10^{-8}}{0.225}} = 150\ \text{N/mm}^2 \] Thus, the maximum bending stress is \( \boxed{150}\ \text{N/mm}^2 \).