Design of a timber beam

Reactions and maximum bending moment

          12 kN
  6 kN/m   |
 _vvvvvvvvvvvvvv_
 A                B     L = 5 m

Total load $= 6\times5 + 12 = 42$ kN, so $R_A = R_B = 21$ kN. Maximum shear force $V_{max} = 21$ kN at the supports. Maximum bending moment at mid-span:

$$M_{max} = \frac{wL^2}{8} + \frac{PL}{4} = \frac{6\times25}{8} + \frac{12\times5}{4} = 18.75 + 15 = 33.75\ \text{kN m}$$

Design for bending

$$Z_{req} = \frac{M_{max}}{\sigma_{all}} = \frac{33.75\times10^6}{10} = 3.375\times10^6\ \text{mm}^3$$

With $h = 2b$:

$$Z = \frac{bh^2}{6} = \frac{b(2b)^2}{6} = \frac{2b^3}{3}$$ $$\frac{2b^3}{3} = 3.375\times10^6 \;\Rightarrow\; b^3 = 5.0625\times10^6 \;\Rightarrow\; b = 171.7\ \text{mm}$$

Adopt $b = 175$ mm and $h = 350$ mm.

Check bending stress with the adopted section:

$$Z = \frac{175\times350^2}{6} = 3.573\times10^6\ \text{mm}^3, \qquad \sigma = \frac{33.75\times10^6}{3.573\times10^6} = 9.45\ \text{MPa} < 10\ \text{MPa}\ \checkmark$$

Check for shear

$$\tau_{max} = \frac32\,\frac{V}{bh} = \frac32\times\frac{21\times10^3}{175\times350} = 0.514\ \text{MPa} < 1.0\ \text{MPa}\ \checkmark$$

The shear stress is only about half the allowable value, so bending governs the design (the beam is long, $L/h = 14$).

Answer: provide a 175 mm $\times$ 350 mm section ($\sigma_b = 9.45$ MPa, $\tau_{max} = 0.51$ MPa). Deflection should also be checked.