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

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) (R_{B}=\frac{P\times a + 3P\times2a+P\times3a}{2a}=\frac{(1 + 6+3)Pa}{2a}=5P) (\sum F_{y}=0), (R_{A}+R_{B}-P - 3P - P=0), (R_{A}=0)

Step2: Determine the maximum bending moment

The bending moment at (C): (M_{C}=R_{A}\times2a+P\times a=P\times a) (Since (R_{A} = 0)), (M_{C}=10\times10^{3}\times1.2 = 12\times10^{3}\text{N}\cdot\text{m})

Step3: Use the bending - stress formula

The bending - stress formula is (\sigma=\frac{M}{I/y}), for a rectangular cross - section (I=\frac{bh^{3}}{12}), (y = h/2), so (\sigma=\frac{6M}{bh^{2}}) Given (h = 2b), then (\sigma=\frac{6M}{b\times(2b)^{2}}=\frac{6M}{4b^{3}})

Step4: Solve for (b)

We know (\sigma=[\sigma]=10\times10^{6}\text{Pa}), (M = 12\times10^{3}\text{N}\cdot\text{m}) Substitute into (\sigma=\frac{6M}{4b^{3}}), (10\times10^{6}=\frac{6\times12\times10^{3}}{4b^{3}}) (4b^{3}=\frac{6\times12\times10^{3}}{10\times10^{6}}) (b^{3}=\frac{6\times12\times10^{3}}{4\times10\times10^{6}}=\frac{72\times10^{3}}{40\times10^{6}} = 1.8\times10^{-3}) (b=\sqrt[3]{1.8\times10^{-3}}\approx0.122\text{m} = 122\text{mm})

Step5: Solve for (h)

Since (h = 2b), (h=2\times0.122 = 0.244\text{m}=244\text{mm})

Answer:

The width (b\approx122\text{mm}) and the height (h\approx244\text{mm})