Question

example 1. a wooden beam: ( p = 10kn ), ( a = 1.2m ), ( sigma=10mpa ), ( h/b = 2 ), determine the dimensions of the section.

Explanation:

Step1: Calculate the support reactions

Taking moments about \(A\):
\(\sum M_{A}=0\), \(P\times a + 3P\times2a+P\times3a - R_{B}\times2a=0\)
\(P\times a+6Pa + 3Pa-2aR_{B}=0\)
\(10Pa=2aR_{B}\), so \(R_{B} = 5P\)
By vertical force equilibrium \(\sum F_{y}=0\), \(R_{A}+R_{B}-P - 3P - P=0\), \(R_{A}+5P-5P = 0\), so \(R_{A}=0\)

Step2: Determine the maximum bending moment

The bending moment at \(C\): \(M_{C}=R_{B}\times a=5P\times a\)
Substitute \(P = 10\times10^{3}N\) and \(a = 1.2m\), \(M_{C}=5\times10\times10^{3}\times1.2=60\times10^{3}N\cdot m\)

Step3: Use the bending - stress formula

The bending - stress formula is \(\sigma=\frac{M}{I}\times y\), for a rectangular cross - section \(I=\frac{bh^{3}}{12}\) and \(y = \frac{h}{2}\)
Since \(h = 2b\), \(I=\frac{b(2b)^{3}}{12}=\frac{8b^{4}}{12}=\frac{2b^{4}}{3}\), \(y = b\)
\(\sigma=\frac{M}{\frac{2b^{4}}{3}}\times b=\frac{3M}{2b^{3}}\)
Given \(\sigma=[\sigma]=10\times10^{6}Pa\) and \(M = M_{C}=60\times10^{3}N\cdot m\)
\(10\times10^{6}=\frac{3\times60\times10^{3}}{2b^{3}}\)
\(10\times10^{6}=\frac{90\times10^{3}}{b^{3}}\)
\(b^{3}=\frac{90\times10^{3}}{10\times10^{6}}=9\times10^{-3}\)
\(b = 0.208m\approx0.21m\)

Step4: Calculate \(h\)

Since \(h = 2b\), \(h=2\times0.21 = 0.42m\)

Answer:

The width \(b\approx0.21m\) and the height \(h = 0.42m\)