Cantilever 5 m: maximum and 25 mm bending stress

Section (as drawn): web 50 mm wide × 120 mm high standing on a flange 200 mm × 50 mm. So the flange is at the bottom, the web top is the top surface, and overall depth is 170 mm.

 8 kN/m                   15 kN
 vvvvvvvvvvvvvvvvvvvvvvv   v
 |#########################
 A            5 m          B

Maximum bending moment (at fixed end A, hogging)

$$M_{max} = \frac{8\times5^2}{2} + 15\times5 = 100 + 75 = 175\ \text{kNm}$$

Top fibres are in tension, bottom fibres in compression.

Section properties (from base, mm)

$$\bar y = \frac{200\times50\times25 + 50\times120\times110}{10000+6000} = 56.875\ \text{mm}$$ $$I = \frac{200\times50^3}{12} + 10000(25-56.875)^2 + \frac{50\times120^3}{12} + 6000(110-56.875)^2 = 36.377\times10^6\ \text{mm}^4$$

Distance to top $= 170 - 56.875 = 113.125$ mm; to bottom $= 56.875$ mm.

Maximum stresses

$$\sigma_{top} = \frac{175\times10^6\times113.125}{36.377\times10^6} = 544.2\ \text{N/mm}^2\ \text{(T)}$$ $$\sigma_{bottom} = \frac{175\times10^6\times56.875}{36.377\times10^6} = 273.6\ \text{N/mm}^2\ \text{(C)}$$

Stress 25 mm below the top surface

$y = 113.125 - 25 = 88.125$ mm above the NA (tension):

$$\sigma = \frac{175\times10^6\times88.125}{36.377\times10^6} = 423.9\ \text{N/mm}^2\ \text{(T)}$$

Answer: maximum bending stress = 544.2 N/mm² (tension, top) at the fixed end; stress 25 mm below top = 423.9 N/mm² (tension). These stresses are far above the yield stress of mild steel (about 250 MPa), so the section as given is not adequate for these loads.