Ultimate moment of T-beam (2250 mm flange)

Given (from figure): $b_f=2250$ mm, $D_f=150$ mm, $b_w=250$ mm, $D=450$ mm, 4-25 mm bars, $A_{st}=4\times\frac{\pi}{4}(25)^2=1963.5$ mm$^2$; M20, Fe415. Effective cover is not given; assume 50 mm, so $d=450-50=400$ mm.

Step 1: Position of neutral axis. Assume it lies in the flange ($x_u\le D_f$):

$$x_u=\frac{0.87f_yA_{st}}{0.36f_{ck}b_f}=\frac{0.87\times415\times1963.5}{0.36\times20\times2250}=43.76\ \text{mm}$$

$x_u=43.76$ mm $<D_f=150$ mm, so the assumption holds. The section behaves as a rectangular beam of width 2250 mm.

Step 2: Check for under-reinforcement

$$x_{u,max}=0.48d=0.48\times400=192.0\ \text{mm}>x_u$$

The section is under-reinforced (ductile).

Step 3: Ultimate moment capacity

$$\begin{aligned} M_u&=0.87f_yA_{st}(d-0.42x_u)\\ &=0.87\times415\times1963.5\times(400-0.42\times43.76)\\ &=270.54\ \text{kN·m} \end{aligned}$$

Answer: $M_u=270.54$ kN·m.