Cantilever 3 m: bending stress 25 mm below top

Section: flange 20 cm × 5 cm on top, stem 5 cm × 12 cm below; overall depth 17 cm.

Maximum bending moment (at fixed end A, hogging)

$$M = \frac{5\times3^2}{2} + 20\times3 = 22.5 + 60 = 82.5\ \text{kNm}$$

Top fibres are in tension.

Neutral axis (from base, cm)

$$\bar y = \frac{20\times5\times14.5 + 5\times12\times6}{100+60} = \frac{1450+360}{160} = 11.31\ \text{cm}$$

Moment of inertia

$$I = \frac{20\times5^3}{12} + 100(14.5-11.31)^2 + \frac{5\times12^3}{12} + 60(6-11.31)^2 = 3637.7\ \text{cm}^4 = 3.638\times10^7\ \text{mm}^4$$

Stress 25 mm below the top

Top fibre is $17 - 11.31 = 5.69$ cm $= 56.9$ mm above the NA. The point 25 mm below the top is $y = 56.9 - 25 = 31.9$ mm above the NA (tension).

$$\sigma = \frac{My}{I} = \frac{82.5\times10^6\times 31.9}{3.638\times10^7} = 72.3\ \text{N/mm}^2\ \text{(T)}$$

For reference, the extreme fibres are 129.0 N/mm² (top, tension) and 256.6 N/mm² (bottom, compression).

Answer: bending stress at 25 mm below the top = 72.3 N/mm² (tensile).