example 2 a wooden beam with the section of rectangular sections (b×h = 0.12m×0.18m) is shown in the figure…

Explanation:

Step1: Calculate the maximum bending moment (M_{max})

For a simply - supported beam with a uniformly distributed load (q), the maximum bending moment (M_{max}=\frac{1}{8}qL^{2}). Given (q = 3.6\times10^{3}\space N/m) and (L = 3\space m), then (M_{max}=\frac{1}{8}\times3.6\times 10^{3}\times3^{2}=\frac{1}{8}\times3.6\times10^{3}\times9 = 4050\space N\cdot m)

Step2: Calculate the section modulus (W_{z})

For a rectangular cross - section, the section modulus (W_{z}=\frac{bh^{2}}{6}). Given (b = 0.12\space m) and (h = 0.18\space m), then (W_{z}=\frac{0.12\times0.18^{2}}{6}=\frac{0.12\times0.0324}{6}=6.48\times 10^{-4}\space m^{3})

Step3: Calculate the maximum bending stress (\sigma_{max})

The formula for bending stress is (\sigma_{max}=\frac{M_{max}}{W_{z}}). Substitute (M_{max}=4050\space N\cdot m) and (W_{z}=6.48\times 10^{-4}\space m^{3}) into the formula: (\sigma_{max}=\frac{4050}{6.48\times 10^{-4}}=\frac{40500000}{6.48}=6.25\times 10^{6}\space Pa = 6.25\space MPa)

Step4: Check the strength

Compare (\sigma_{max}) with ([\sigma]). Given ([\sigma]=7\space MPa), and (\sigma_{max}=6.25\space MPa)

Answer:

Since (\sigma_{max}=6.25\space MPa<[\sigma]=7\space MPa), the strength of the beam is sufficient.