Timber joist dimensions under UDL and point load

Given: $L = 6$ m, $w = 5$ kN/m, $P = 15$ kN at centre, $\sigma_b = 12\ \text{N/mm}^2$. Self-weight is ignored.

Maximum bending moment (at mid-span)

$$M_{max} = \frac{wL^2}{8} + \frac{PL}{4} = \frac{5\times 6^2}{8} + \frac{15\times 6}{4} = 22.5 + 22.5 = 45\ \text{kNm}$$

Required section modulus

$$Z = \frac{M}{\sigma_b} = \frac{45\times10^6}{12} = 3.75\times10^6\ \text{mm}^3$$

Dimensions

A joist is deeper than it is wide; assume $d = 2b$ (common practice).

$$Z = \frac{bd^2}{6} = \frac{b(2b)^2}{6} = \frac{2b^3}{3}$$ $$b^3 = \frac{3\times 3.75\times10^6}{2} = 5.625\times10^6 \Rightarrow b = 177.8\ \text{mm}$$

$d = 2b = 355.6$ mm.

Check with $180\times360$ mm: $Z = 180\times360^2/6 = 3.888\times10^6\ \text{mm}^3$, $\sigma = 45\times10^6/3.888\times10^6 = 11.6\ \text{N/mm}^2 < 12$, safe.

Answer: provide a joist 180 mm wide × 360 mm deep (calculated 177.8 mm × 355.6 mm).