T-beam: bending stress 15 cm above base

Assumption: the figure gives only the flange width (10 cm), web thickness (3 cm) and overall depth (42 cm). The flange thickness is taken as 4 cm, flange on top, so the web is 38 cm deep.

Bending moment

$$M = \frac{wL^2}{8} = \frac{4\times 5^2}{8} = 12.5\ \text{kNm (sagging, at mid-span)}$$

Neutral axis (from base, cm)

$$\bar y = \frac{3\times38\times19 + 10\times4\times40}{114+40} = \frac{2166+1600}{154} = 24.45\ \text{cm}$$

Moment of inertia

$$I = \frac{3\times38^3}{12} + 114(19-24.45)^2 + \frac{10\times4^3}{12} + 40(40-24.45)^2 = 26829.5\ \text{cm}^4 = 2.683\times10^8\ \text{mm}^4$$

Stress at 15 cm above the base

Distance from the neutral axis: $y = 24.45 - 15 = 9.45$ cm $= 94.5$ mm, below the neutral axis, so the fibre is in tension (sagging).

$$\sigma = \frac{My}{I} = \frac{12.5\times10^6\times 94.5}{2.683\times10^8} = 4.40\ \text{N/mm}^2$$

Answer: bending stress at 15 cm above the base = 4.40 N/mm² (tensile).