Question

example 2 a wooden beam with the section of rectangular sections (b×h = 0.12m×0.18m) is shown in the figure, σ = 7mpa. try to check the strength of the beam.

Explanation:

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

For a simply - supported beam with a uniformly distributed load \(q\), the formula for the maximum bending moment is \(M_{max}=\frac{1}{8}qL^{2}\).
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 formula for the section modulus is \(W_{z}=\frac{bh^{2}}{6}\).
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 normal stress \(\sigma_{max}\)

The formula for the normal stress in bending 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: \(\sigma_{max}=\frac{4050}{6.48\times 10^{-4}} = 6.25\times10^{6}\space Pa=6.25\space MPa\)

Step4: Check the strength

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

Answer:

The strength of the beam is sufficient.