Design of a simply supported timber beam

Design loads

$w = 7.5$ kN/m, $L = 3.6$ m.

$$\begin{aligned} M &= \frac{wL^2}{8} = \frac{7.5 \times 3.6^2}{8} = 12.15\ \text{kN·m} \\ V &= \frac{wL}{2} = \frac{7.5 \times 3.6}{2} = 13.5\ \text{kN} \end{aligned}$$

Step 1: Depth for bending

$$Z_{req} = \frac{M}{\sigma_{bc}} = \frac{12.15\times10^6}{11} = 1.105\times10^6\ \text{mm}^3$$ $$d = \sqrt{\frac{6 Z_{req}}{b}} = \sqrt{\frac{6 \times 1.105\times10^6}{100}} = 257\ \text{mm}$$

Step 2: Depth for deflection

$$\delta_{allow} = \frac{3600}{360} = 10\ \text{mm}$$ $$I_{req} = \frac{5wL^4}{384 E \delta} = \frac{5 \times 7.5 \times 3600^4}{384 \times 9800 \times 10} = 1.674\times10^8\ \text{mm}^4$$ $$d = \left(\frac{12 I_{req}}{b}\right)^{1/3} = \left(\frac{12 \times 1.674\times10^8}{100}\right)^{1/3} = 272\ \text{mm}$$

Deflection controls. Provide $100 \times 275$ mm.

Step 3: Checks for 100 mm × 275 mm

$$\begin{aligned} Z &= \frac{100 \times 275^2}{6} = 1.260\times10^6\ \text{mm}^3 \\ \sigma_b &= \frac{12.15\times10^6}{1.260\times10^6} = 9.64\ \text{N/mm}^2 < 11 \quad \text{OK} \\ \tau &= \frac{3V}{2bd} = \frac{3 \times 13\,500}{2 \times 100 \times 275} = 0.74\ \text{N/mm}^2 < 1.0 \quad \text{OK} \\ I &= \frac{100 \times 275^3}{12} = 1.733\times10^8\ \text{mm}^4 \\ \delta &= \frac{5 \times 7.5 \times 3600^4}{384 \times 9800 \times 1.733\times10^8} = 9.66\ \text{mm} < 10\ \text{mm} \quad \text{OK} \end{aligned}$$

Step 4: Bearing and lateral stability

Bearing length is chosen so that bearing stress $V/(b \times l_b)$ does not exceed the permissible compression perpendicular to grain; a bearing length of 100 mm gives about 1.35 N/mm². Since $d/b = 2.75$, the ends must be held in position and the compression edge held by floor boards.

Answer: Provide a 100 mm × 275 mm timber beam (bending stress 9.64 N/mm², shear 0.74 N/mm², deflection 9.66 mm).