MoR of T-beam by limit state

Given: $b_f=1600$ mm, $D_f=110$ mm, $b_w=300$ mm, $D=460$ mm, effective cover 40 mm so $d=420$ mm; 4-20 mm bars, $A_{st}=4\times\frac{\pi}{4}(20)^2=1256.6$ mm$^2$; M20, Fe415.

Step 1: Locate neutral axis. Assume $x_u\le D_f$ (rectangular behaviour with width $b_f$). Equating compression and tension:

$$0.36f_{ck}b_fx_u=0.87f_yA_{st}\ \Rightarrow\ x_u=\frac{0.87\times415\times1256.6}{0.36\times20\times1600}=39.38\ \text{mm}$$

$x_u=39.38<D_f=110$ mm, so the neutral axis is in the flange (assumption correct).

Step 2: Check section type. $x_{u,max}=0.48d=0.48\times420=201.6$ mm $>x_u$, so the section is under-reinforced.

Step 3: Moment of resistance

$$\begin{aligned} M_u&=0.87f_yA_{st}(d-0.42x_u)\\ &=0.87\times415\times1256.6\times(420-0.42\times39.38)\\ &=183.05\ \text{kN·m} \end{aligned}$$

Answer: $M_u=183.05$ kN·m (design moment of resistance of the T-section).