I-section beam: moment of resistance and UDL

Section: flanges 100 mm × 25 mm, web 50 mm thick, web height 50 mm. Overall depth $= 25 + 50 + 25 = 100$ mm. The section is symmetrical, so the neutral axis is at mid-depth, $y = 50$ mm from top and bottom.

 |<--100-->|
 +---------+  25
    | | |
    | | |     50
 +---------+  25

Moment of inertia

$$I = \frac{100\times100^3}{12} - \frac{(100-50)\times 50^3}{12} = 8.333\times10^6 - 0.521\times10^6 = 7.8125\times10^6\ \text{mm}^4$$

Moment of resistance

Both extreme fibres are 50 mm from the neutral axis. The permissible stress in the weaker fibre governs (the smaller value, 150 MPa):

$$M_r = \frac{\sigma I}{y} = \frac{150\times 7.8125\times10^6}{50} = 23.44\times10^6\ \text{Nmm} = 23.44\ \text{kNm}$$

(At 180 MPa the other face would allow 28.1 kNm, which is more, so 150 MPa governs.)

Safe UDL

$$M_{max} = \frac{wL^2}{8} = M_r \Rightarrow w = \frac{8\times 23.44}{10^2} = 1.875\ \text{kN/m}$$

Answer: moment of resistance = 23.44 kNm; safe UDL $w = 1.875$ kN/m.