Simply supported T-beam: max bending stresses

Reading of the figure: simple span $L = 6$ m. The 100 kN and 10 kN loads both act at 3 m from the supports, i.e. at mid-span, so together they form a 110 kN central load. The T-section has a flange 200 mm $\times$ 50 mm and a web 50 mm thick extending 200 mm below the flange (total depth 250 mm). The flange is on top.

Maximum bending moment

$$M_{max} = \frac{WL}{4} = \frac{110\times6}{4} = 165.0\ \text{kN·m} = 1.6500\times10^{8}\ \text{N·mm}$$

Section properties (distances from the bottom of the web)

$$A = 200(50) + 50(200) = 20000\ \text{mm}^2$$ $$\bar y = \frac{(200\times50)(225) + (50\times200)(100)}{20\,000} = 162.50\ \text{mm} \quad(\text{from the bottom}), \qquad H - \bar y = 87.50\ \text{mm}\ (\text{from the top})$$ $$I = \left[\frac{200(50)^3}{12} + 10\,000(225-162.50)^2\right] + \left[\frac{50(200)^3}{12} + 10\,000(162.50-100)^2\right] = 1.1354\times10^{8}\ \text{mm}^4$$

Bending stresses ($\sigma = My/I$)

$$\sigma_{compression}\ (\text{top}) = \frac{1.6500\times10^{8}\times87.50}{1.1354\times10^{8}} = 127.16\ \text{N/mm}^2$$ $$\sigma_{tension}\ (\text{bottom}) = \frac{1.6500\times10^{8}\times162.50}{1.1354\times10^{8}} = 236.15\ \text{N/mm}^2$$

Answer: maximum compressive stress $= 127.2$ N/mm² (top of the flange); maximum tensile stress $= 236.1$ N/mm² (bottom of the web).