Bending stresses in a T-section beam

(a) Neutral axis and I

Measure $\bar y$ from the bottom of the web.

$$\bar y = \frac{931500}{8100} = 115\ \text{mm from bottom}, \quad 65\ \text{mm from top}$$

Using the parallel axis theorem:

$$I = \left[\frac{120\times30^3}{12} + 3600(165-115)^2\right] + \left[\frac{30\times150^3}{12} + 4500(115-75)^2\right]$$ $$I = (0.27 + 9.0)\times10^6 + (8.4375 + 7.2)\times10^6 = 24.91\times10^6\ \text{mm}^4$$

(b) Stresses at mid-span

$$M_{max} = \frac{wL^2}{8} = \frac{12\times4^2}{8} = 24\ \text{kN m} = 24\times10^6\ \text{N mm}$$

The load sags the beam, so the top (flange) is in compression and the bottom is in tension.

$$\sigma_c = \frac{M\,y_{top}}{I} = \frac{24\times10^6\times65}{24.91\times10^6} = 62.6\ \text{MPa}$$ $$\sigma_t = \frac{M\,y_{bot}}{I} = \frac{24\times10^6\times115}{24.91\times10^6} = 110.8\ \text{MPa}$$

(c) Safe UDL

Limit moment from tension: $M_t = \dfrac{\sigma_{t,all} I}{y_{bot}} = \dfrac{40\times24.91\times10^6}{115} = 8.66\times10^6$ N mm.

Limit moment from compression: $M_c = \dfrac{150\times24.91\times10^6}{65} = 57.5\times10^6$ N mm.

Tension governs (the bottom fibre is farther from the neutral axis and cast iron is weak in tension): $M = 8.66$ kN m.

$$w = \frac{8M}{L^2} = \frac{8\times8.66}{16} = 4.33\ \text{kN/m}$$

Answer: NA 115 mm above the bottom; $I = 24.91\times10^6\ \text{mm}^4$; $\sigma_t = 110.8$ MPa and $\sigma_c = 62.6$ MPa (the 12 kN/m load is unsafe in tension); safe load $w = 4.33$ kN/m. Placing the flange at the bottom (tension side) would be more suitable for cast iron.