Question 15.2: The laminated wooden beam shown in Fig. 15-8a supports a uni......
The laminated wooden beam shown in Fig. 15-8a supports a uniform distributed loading of 12 \mathrm{kN} / \mathrm{m}. If the beam is to have a height-to-width ratio of 1.5 , determine its smallest width. Take \sigma_{\text {allow }}=9 \mathrm{MPa}, and \tau_{\text {allow }}=0.6 \mathrm{MPa}. Neglect the weight of the beam.
Shear and Moment Diagrams. The support reactions at A and B have been calculated, and the shear and moment diagrams are shown in Fig. 15-8b. Here V_{\max }=20 \mathrm{kN}, M_{\max }=10.67 \mathrm{kN} \cdot \mathrm{m}.
Bending Stress. Applying the flexure formula,
S_{\text {req'd }}=\frac{M_{\text {max }}}{\sigma_{\text {allow }}}=\frac{10.67\left(10^{3}\right) \mathrm{N} \cdot \mathrm{m}}{9\left(10^{6}\right) \mathrm{N} / \mathrm{m}^{2}}=0.00119 \mathrm{~m}^{3}
Assuming that the width is a, then the height is 1.5 a, Fig. 15-8a. Thus,
\begin{aligned} S_{\text {req'd }} & =\frac{I}{c}=0.00119 \mathrm{~m}^{3}=\frac{\frac{1}{12}(a)(1.5 a)^{3}}{(0.75 a)} \\ a^{3} & =0.003160 \mathrm{~m}^{3} \\ a & =0.147 \mathrm{~m} \end{aligned}
Shear Stress. Applying the shear formula for rectangular sections (which is a special case of \tau_{\mathrm{max}}=V Q/I t, as shown in Example 12.2), we have
\tau_{\mathrm{max}}=1.5{\frac{V_{\mathrm{max}}}{A}}= (1.5)\frac{20(10^{3})\,\mathrm{N}}{(0.147\,\mathrm{m})(1.5)(0.147\,\mathrm{m})} \\ =0.929\,\mathrm{MPa}\gt 0.6\,\mathrm{MPa}Since the design based on bending fails the shear criterion, the beam must be redesigned on the basis of shear.
\begin{aligned} \tau_{\text {allow }} & =1.5 \frac{V_{\max }}{A} \\ 600 \mathrm{kN} / \mathrm{m}^{2} & =1.5 \frac{20\left(10^{3}\right) \mathrm{N}}{(a)(1.5 a)} \\ a & =0.183 \mathrm{~m}=183 \mathrm{~mm} \end{aligned}
This larger section will also adequately resist the bending stress.