Question

q=3.6kn/m
a
l=3m
b
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 Maximum Bending Moment

For a simply - supported beam with uniform load \(q\), the maximum bending moment \(M_{max}\) occurs at the mid - span. The formula for the maximum bending moment of a simply - supported beam under uniform load is \(M_{max}=\frac{qL^{2}}{8}\).
Given \(q = 3.6\space kN/m=3.6\times10^{3}\space N/m\) and \(L = 3\space m\).
Substitute the values into the formula:
\(M_{max}=\frac{3.6\times 10^{3}\times3^{2}}{8}=\frac{3.6\times10^{3}\times9}{8}=\frac{32400}{8} = 4050\space N\cdot m\)

Step2: Calculate Section Modulus

For a rectangular section with width \(b\) and height \(h\), the section modulus \(W_{z}\) is given by the formula \(W_{z}=\frac{bh^{2}}{6}\).
Given \(b = 0.12\space m\) and \(h=0.18\space m\).
Substitute the values into the formula:
\(W_{z}=\frac{0.12\times(0.18)^{2}}{6}=\frac{0.12\times0.0324}{6}=\frac{0.003888}{6}=6.48\times 10^{-4}\space m^{3}\)

Step3: Calculate Maximum Bending Stress

The formula for bending stress \(\sigma_{max}=\frac{M_{max}}{W_{z}}\).
Substitute \(M_{max} = 4050\space N\cdot m\) and \(W_{z}=6.48\times 10^{-4}\space m^{3}\) into the formula:
\(\sigma_{max}=\frac{4050}{6.48\times 10^{-4}}=\frac{4050\times10^{4}}{6.48}=\frac{40500000}{6.48}=6.25\times 10^{6}\space Pa = 6.25\space MPa\)

Step4: Check Strength

Compare the maximum bending stress \(\sigma_{max}\) with the allowable stress \([\sigma]\). Given \([\sigma]=7\space MPa\).
Since \(\sigma_{max}=6.25\space MPa<[\sigma] = 7\space MPa\), the beam meets the strength requirements.

Answer:

The maximum bending stress \(\sigma_{max}=6.25\space MPa\), which is less than the allowable stress \([\sigma]=7\space MPa\), so the beam has sufficient strength.