Question 15.3: The wooden T-beam shown in Fig. 15-9a is made from two 200 m......
The wooden T-beam shown in Fig. 15-9a is made from two 200 \mathrm{~mm} \times 30 \mathrm{~mm} boards. If \sigma_{\text {allow }}=12 \mathrm{MPa} and \tau_{\text {allow }}=0.8 \mathrm{MPa}, determine if the beam can safely support the loading shown. Also, specify the maximum spacing of nails needed to hold the two boards together if each nail can safely resist 1.50 \mathrm{kN} in shear.
Shear and Moment Diagrams. The reactions on the beam are shown, and the shear and moment diagrams are drawn in Fig. 15-9b. Here V_{\max }=1.5 \mathrm{kN}, M_{\max }=2 \mathrm{kN} \cdot \mathrm{m}.
Bending Stress. The neutral axis (centroid) will be located from the bottom of the beam. Working in units of meters, we have
\begin{aligned} \bar{y} & =\frac{\Sigma \bar{y} A}{\Sigma A} \\ & =\frac{(0.1 \mathrm{~m})(0.03 \mathrm{~m})(0.2 \mathrm{~m})+0.215 \mathrm{~m}(0.03 \mathrm{~m})(0.2 \mathrm{~m})}{0.03 \mathrm{~m}(0.2 \mathrm{~m})+0.03 \mathrm{~m}(0.2 \mathrm{~m})}=0.1575 \mathrm{~m} \end{aligned}
Thus,
\begin{aligned} I= & {\left[\frac{1}{12}(0.03 \mathrm{~m})(0.2 \mathrm{~m})^{3}+(0.03 \mathrm{~m})(0.2 \mathrm{~m})(0.1575 \mathrm{~m}-0.1 \mathrm{~m})^{2}\right] } \\ & +\left[\frac{1}{12}(0.2 \mathrm{~m})(0.03 \mathrm{~m})^{3}+(0.03 \mathrm{~m})(0.2 \mathrm{~m})(0.215 \mathrm{~m}-0.1575 \mathrm{~m})^{2}\right] \\ = & 60.125\left(10^{-6}\right) \mathrm{m}^{4} \end{aligned}
Since c=0.1575 \mathrm{~m}( not 0.230 \mathrm{~m}-0.1575 \mathrm{~m}=0.0725 \mathrm{~m}), we require
\begin{gathered} \sigma_{\text {allow }} \geq \frac{M_{\max } c}{I} \\ 12\left(10^{6}\right) \mathrm{Pa} \geq \frac{2\left(10^{3}\right) \mathrm{N} \cdot \mathrm{m}(0.1575 \mathrm{~m})}{60.125\left(10^{-6}\right) \mathrm{m}^{4}}=5.24\left(10^{6}\right) \mathrm{Pa} \quad \text{OK} \end{gathered}
Shear Stress. Maximum shear stress in the beam depends upon the magnitude of Q and t. It occurs at the neutral axis, since Q is a maximum there and at the neutral axis the thickness t = 0.03 m is the smallest for the cross section. For simplicity, we will use the rectangular area below the neutral axis to calculate Q, rather than a two-part composite area above this axis, Fig. 15-9c. We have
Q={\overline{{y}}}^{\prime}A^{\prime}=\biggl({\frac{0.1575\,{\mathrm{m}}}{2}}\biggr)[(0.1575\,{\mathrm{m}})(0.03\,{\mathrm{m}})]= 0.372(10^{-3}) m^3so that
\tau_{\mathrm{alow}}\geq \frac{V_{\mathrm{max}}Q}{I t}
800(10^{3})\,\mathrm{Pa}\geq{\frac{1.5(10^{3})\,\,\mathrm{N}[0.372(10^{-3})]\,\mathrm{m}^{3}}{60.125(10^{-6})\,\mathrm{m}^{4}(0.03\,\mathrm{m})}} = 309(10^{3}) Pa \text{ OK}
Nail Spacing. From the shear diagram it is seen that the shear varies over the entire span. Since the nail spacing depends on the magnitude of shear in the beam, for simplicity (and to be conservative), we will design the spacing on the basis of V = 1.5 kN for region BC, and V = 1 kN for region CD. Since the nails join the flange to the web, Fig. 15-9d, we have
Q={\overline{{y}}}^{\prime}A^{\prime}\,=\,(0.0725\,{\mathrm{m}}\,-\,0.015\,{\mathrm{m}})[(0.2\,{\mathrm{m}})(0.03\,{\mathrm{m}})]\,=\,0.345(10^{-3})\,{\mathrm{m}}^{3}
The shear flow for each region is therefore
q_{B C}=\frac{V_{B C}Q}{I}=\frac{1.5(10^{3})\,\mathrm{N}[0.345(10^{-3})\,{ m}^{3}]}{60.125(10^{-6})\,{ m}^{4}}=8.61\,\mathrm{kN/m} \\ q_{C D}=\frac{V_{C D}Q}{I}=\frac{1(10^{3})\,\mathrm{N}[0.345(10^{-3})\,{ m}^{3}]}{60.125(10^{-6})\,{ m}^{4}}=5.74\,\mathrm{kN/m}
One nail can resist 1.50 kN in shear, so the maximum spacing becomes
s_{B C}={\frac{1.50\,\,{\mathrm{kN}}}{8.61\,\,{\mathrm{kN}}/{\mathrm{m}}}}=0.174\,{\mathrm{m}} \\ s_{C D}={\frac{1.50\,\mathrm{kN}}{5.74\,\mathrm{kN/m}}}=0.261\,\mathrm{m}
For ease of measuring, use
s_{B C}=150\,\mathrm{mm}\\ s_{C D}=250\,\mathrm{mm}