A uniformly distributed load of 20 kN/m acts on a simply supported beam of rectangular cross section of width 20 mm and depth 60 mm. What is the maximum bending stress acting on the beam of 5m?

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The maximum bending moment in a simply supported beam with a uniformly distributed load (w) over its entire length (L) is given by $M = \frac{wL^2}{8}$. For a rectangular cross-section, the maximum bending stress ($\sigma$) can be calculated using the formula $\sigma = \frac{M}{S}$, where $S$ is the section modulus given by $S = \frac{bd^2}{6}$ for a rectangular section, with $b$ being the width and $d$ being the depth of the beam.Given:- $w = 20$ kN/m = $20 \times 10^3$ N/m (since 1 kN = $10^3$ N)- $L = 5$ m- $b = 20$ mm = $20 \times 10^{-3}$ m (since 1 mm = $10^{-3}$ m)- $d = 60$ mm = $60 \times 10^{-3}$ mFirst, calculate the maximum bending moment, $M$:$$M = \frac{wL^2}{8} = \frac{20 \times 10^3 \times 5^2}{8} = \frac{20 \times 10^3 \times 25}{8} = 62,500 \, \text{N}\cdot\text{m}$$Next, calculate the section modulus, $S$:$$S = \frac{bd^2}{6} = \frac{20 \times 10^{-3} \times (60 \times 10^{-3})^2}{6} = \frac{20 \times 10^{-3} \times 3600 \times 10^{-6}}{6} = 12 \times 10^{-6} \, \text{m}^3$$Finally, calculate the maximum bending stress, $\sigma$:$$\sigma = \frac{M}{S} = \frac{62,500}{12 \times 10^{-6}} = 5.208 \times 10^9 \, \text{Pa} = 5208 \, \text{MPa}$$Therefore, the maximum bending stress acting on the beam is 5208 MPa.

A Uniformly Distributed Load of 20 KN/m Acts on a Simply