Limiting moment of T-beam

Given: $b_f=1500$ mm, $b_w=240$ mm, $D_f=150$ mm, $d=590$ mm, M20, Fe415.

Step 1: Limiting depth of neutral axis (Fe415)

$$x_{u,max}=0.48d=0.48\times590=283.2\ \text{mm}>D_f=150\ \text{mm}$$

so the neutral axis lies in the web. Also $D_f/d=150/590=0.254>0.2$, therefore the flange is not thin and IS 456 Annex G-2.2 uses the equivalent flange thickness

$$y_f=0.15x_{u,max}+0.65D_f=0.15\times283.2+0.65\times150=139.98\ \text{mm}$$

Step 2: Limiting moment of resistance

$$M_{u,lim}=0.36f_{ck}b_wx_{u,max}(d-0.416x_{u,max})+0.45f_{ck}(b_f-b_w)y_f\left(d-\frac{y_f}{2}\right)$$ $$\begin{aligned} M_{u,web}&=0.36\times20\times240\times283.2\times(590-0.416\times283.2)=231.07\ \text{kN·m}\\ M_{u,flange}&=0.45\times20\times(1500-240)\times139.98\times\left(590-\frac{139.98}{2}\right)=825.45\ \text{kN·m} \end{aligned}$$ $$M_{u,lim}=231.07+825.45=1056.52\ \text{kN·m}$$

Step 3: Limiting area of steel (from $C=T$)

$$A_{st,lim}=\frac{0.36f_{ck}b_wx_{u,max}+0.45f_{ck}(b_f-b_w)y_f}{0.87f_y}=5752\ \text{mm}^2$$

Answer: $M_{u,lim}=1056.52$ kN·m and $A_{st,lim}=5752$ mm$^2$.