Symmetrical I-beam: moment, flange and web share

Given: flanges 200 mm × 25 mm, overall depth 500 mm, web 20 mm thick (web depth $500 - 50 = 450$ mm). $\sigma_{max} = 40\ \text{N/mm}^2$.

 +-----------+  25
     | |
     | |       450
 +-----------+  25
     200 wide, web 20

Moment of inertia

$$I = \frac{200\times500^3 - 180\times450^3}{12} = 7.1646\times10^8\ \text{mm}^4$$

(i) Maximum bending moment

Extreme fibre distance $y = 250$ mm:

$$M = \frac{\sigma I}{y} = \frac{40\times7.1646\times10^8}{250} = 114.63\times10^6\ \text{Nmm} = 114.63\ \text{kNm}$$

(ii) Moment shared by flanges and web

Stress at distance $y$ is $\sigma = 40y/250$.

Flanges (both, from $y = 225$ to $250$):

$$M_f = 2\int_{225}^{250}\frac{40y}{250}\,200\,y\,dy = 2\times\frac{40\times200}{250}\times\frac{250^3 - 225^3}{3} = 90.33\ \text{kNm}$$

Web (from $-225$ to $225$):

$$M_w = \frac{40\times20}{250}\times\frac{2\times225^3}{3} = 24.30\ \text{kNm}$$

Check: $90.33 + 24.30 = 114.63$ kNm.

Answer: (i) $M_{max} = 114.63$ kNm; (ii) flanges carry 78.8% and the web 21.2% of the moment.