Determine the minimum dimension a to the nearest mm of the beam's cross section to safely

Question

Determine the minimum dimension [latex]a[/latex] to the nearest [latex]\mathrm{mm}[/latex] of the beam's cross section to safely support the load. The wood has an allowable normal stress of [latex]\sigma_{ ext {allow }}=12 \mathrm{MPa}[/latex] and an allowable shear stress of [latex] au_{ ext {allow }}=1.5 \mathrm{MPa}[/latex].

Accepted Answer

At the supports,

[latex]V_{\max }=10 \mathrm{kN}[/latex]

Under 15-kN load,

[latex] \begin{aligned} & M_{\max }=5 \mathrm{kN} \cdot \mathrm{m} \ & I=\frac{1}{12}(a)(2 a)^{3}=\frac{2}{3} a^{4} \ & \sigma_{ ext {allow }}=\frac{M_{\max } c}{I} ; \quad 12\left(10^{6} ight)=\frac{5\left(10^{3} ight)(a)}{\frac{2}{3} a^{4}} \ & a=0.0855 \mathrm{~m}=85.5 \mathrm{~mm} \end{aligned} [/latex]

Use [latex]a=86 \mathrm{~mm}[/latex]

[latex]I=\frac{2}{3}\left(0.086^{4} ight)=36.4672\left(10^{-6} ight) \mathrm{m}^{4}[/latex]

Top half of rectangle,

[latex] \begin{aligned} Q_{\max } & =\frac{0.086}{2}(0.086)(0.086) \ & =0.318028\left(10^{-3} ight) \mathrm{m}^{3} \ au_{\max } & =\frac{V_{\max } Q_{\max }}{I t}=\frac{10\left(10^{3} ight)\left[0.318028\left(10^{-3} ight) ight]}{\left[36.4672\left(10^{-6} ight) ight](0.086)} \ & =1.01 \mathrm{MPa}< au_{ ext {allow }}=1.5 \mathrm{MPa} (\mathrm{OK}) \end{aligned} [/latex]

Question F15.3: Determine the minimum dimension a to the nearest mm of the b......

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