Rectangular beam dimensions, depth twice breadth

Given: $L = 10$ m, $w = 20$ kN/m, $P = 50$ kN at centre, $d = 2b$, permissible $\sigma_t = 120$, $\sigma_c = 100\ \text{N/mm}^2$. Self-weight is ignored.

Maximum bending moment

$$M = \frac{wL^2}{8} + \frac{PL}{4} = \frac{20\times100}{8} + \frac{50\times10}{4} = 250 + 125 = 375\ \text{kNm}$$

Governing stress

A rectangle is symmetrical, so the tension and compression fibres have equal stress. The lower permissible value, $100\ \text{N/mm}^2$ (compression), governs.

$$Z = \frac{M}{\sigma} = \frac{375\times10^6}{100} = 3.75\times10^6\ \text{mm}^3$$

Dimensions

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

$d = 2b = 355.6$ mm.

Check with $180\times360$ mm: $Z = 3.888\times10^6\ \text{mm}^3$, $\sigma = 96.5\ \text{N/mm}^2 < 100$. Safe.

Answer: $b \approx 180$ mm, $d \approx 360$ mm (calculated 177.8 mm × 355.6 mm).